Physics Book - User contributions [en]

Point Charge

2018-11-25T21:26:04Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

The electric field is an inherent quality of a charged particle. It is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward. The magnitude of the electric field decreases with increasing distance from the point charge. The source of the electric field in this article is the point charge.

[[File:Point_charge.GIF|600px]]

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

<math>Q_1, Q_2</math>= The charge of two particles of interest

r = The distance between the two particles

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{\mid\vec r\mid ^2} \hat r</math>

Which is derived from [[Gauss's Law]], modeling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

====Connection Between Electric Field and Force====

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This force is represented by Coulomb's Law which was described above.

By solving for the electric field in <math> F = Eq </math>, with F modeled by Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1q_2}{r^2}\frac{1}{q_2}\hat r = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to a code which can help visualize the Electric Field at various observation locations due to a proton. Notice how the arrows decrease in size by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets farther from the proton. The magnitude of the electric field decreases as the distance to the observation location increases.

https://trinket.io/glowscript/725d552305

Two adjacent point charges of opposite sign exhibit an electric field pattern that is characteristic of a dipole. This interaction is displayed in the code below. Notice how the electric field points towards the negatively charged point charge (blue) and away from the positively charged point charge (red).

https://trinket.io/glowscript/f29782cb9d

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

The electric field of a single point charge is the most basic application of the fundamental property. It is the combined electric field that results from many point charges packed in the same region that causes atomic level changes such as polarization. I am interested in the topic of polarization because it is what directs the movement of current through conductors.

2.How is it connected to your major?

As a BME major, I learn to work with many types of electronics that involve circuits, capacitors, and resistors, to build medical devices. The movement of current through these electronics is dependent on the presence of an electric force (caused by the electric field) on a negatively charged point charge. Specifically, my BMED 2250 group is currently designing a device that includes a capacitive touch sensor. This sensor works due to the charge transfer between two charged plates of a capacitor. The charge transfer is driven by the attraction of oppositely charged point charges and their corresponding electric fields.

3.Is there an interesting industrial application?

Electric fields are actually used in tissue engineering to determine the properties of cells and tissues. They are also used in electrokinetics to assemble artificial tissues from cells.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T21:00:38Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

The electric field is an inherent quality of a charged particle. It is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward. The magnitude of the electric field decreases with increasing distance from the point charge. The source of the electric field in this article is the point charge.

[[File:Point_charge.GIF|600px]]

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

<math>Q_1, Q_2</math>= The charge of two particles of interest

r = The distance between the two particles

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{\mid\vec r\mid ^2} \hat r</math>

Which is derived from [[Gauss's Law]], modeling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

====Connection Between Electric Field and Force====

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This force is represented by Coulomb's Law which was described above.

By solving for the electric field in <math> F = Eq </math>, with F modeled by Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1q_2}{r^2}\frac{1}{q_2}\hat r = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to a code which can help visualize the Electric Field at various observation locations due to a proton. Notice how the arrows decrease in size by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets farther from the proton. The magnitude of the electric field decreases as the distance to the observation location increases.

https://trinket.io/glowscript/725d552305

Two adjacent point charges of opposite sign exhibit an electric field pattern that is characteristic of a dipole. This interaction is displayed in the code below. Notice how the electric field points towards the negatively charged point charge (blue) and away from the positively charged point charge (red).

https://trinket.io/glowscript/f29782cb9d

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T20:49:58Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

The electric field is an inherent quality of a charged particle. It is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward. The magnitude of the electric field decreases with increasing distance from the point charge. The source of the electric field in this article is the point charge.

[[File:Point_charge.GIF|600px]]

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

<math>Q_1, Q_2</math>= The charge of two particles of interest

r = The distance between the two particles

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{\mid\vec r\mid ^2} \hat r</math>

Which is derived from [[Gauss's Law]], modeling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

====Connection Between Electric Field and Force====

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law which was described above.

By solving for the electric field in <math> F = Eq </math>, with F modeled by Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{\mid q_1q_2 \mid}{r^2}\frac{1}{q_2}\hat r = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T20:47:05Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

The electric field is an inherent quality of a charged particle. It is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward. The magnitude of the electric field decreases with increasing distance from the point charge. The source of the electric field in this article is the point charge.

[[File:Point_charge.GIF|600px]]

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

<math>Q_1, Q_2</math>= The charge of two particles of interest

r = The distance between the two particles

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{\mid\vec r\mid ^2} \hat r</math>

Which is derived from [[Gauss's Law]], modeling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

====Connection Between Electric Field and Force====

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law which was described above.

<math> F =

By solving for the electric field in <math> F = Eq </math>, with F modeled by Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{Q_1 Q_2}{r^2} \hat r \frac{q_2}{\hat r} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T20:42:29Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

The electric field is an inherent quality of a charged particle. It is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward. The magnitude of the electric field decreases with increasing distance from the point charge. The source of the electric field in this article is the point charge.

[[File:Point_charge.GIF|600px]]

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

<math>Q_1, Q_2</math>= The charge of two particles of interest

r = The distance between the two particles

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{\mid\vec r\mid ^2} \hat r</math>

Which is derived from [[Gauss's Law]], modeling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T20:39:10Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

The electric field is an inherent quality of a charged particle. It is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward. The magnitude of the electric field decreases with increasing distance from the point charge. The source of the electric field in this article is the point charge.

[[File:Point_charge.GIF|600px]]

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

Q_1, Q_2 = The charge of two particles of interest
r = The distance between the two particles

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modeling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T05:52:21Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|600px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=\frac{1}{4 \pi \epsilon_0 } \frac{\mid Q_1Q_2 \mid}{r^2}</math>

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T05:51:26Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|600px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<math>\mid\vec F\mid=<\frac{1}{4 \pi \epsilon_0 } \frac{\mid\Q_1\Q_2\mid}{r^2}</math>

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T05:49:45Z

Sindhu K: /* Coulomb Force Law for Point Charges */

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|600px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the electrical force of attraction or repulsion between two given point charges. The following formula describes the magnitude of this force:

<\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-25T00:37:09Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|600px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

This law defines the force of attraction or repulsion between two given point charges.

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:55:04Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|600px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]] 
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:50:43Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|800px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|600px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:50:22Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|1000px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF|1000px]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:50:02Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another. This pattern of attraction and repulsion is shown below.

[[File:Phys 1.png|1000px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:49:32Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another.

[[File:Phys 1.png|1000px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:49:21Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another.

[[File:Phys 1.png|500px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:49:07Z

Sindhu K: /* The Main Idea */

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another.

[[File:Phys 1.png|250px]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

File:Phys 1.png

2018-11-24T23:46:27Z

Sindhu K:

Point Charge

2018-11-24T23:45:48Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C). Like point charges, such as two positive or two negative charges, repel one another; opposite charges are attracted to one another.

[[File:Phys 1.png]]

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model===

====Coulomb Force Law for Point Charges====

====Electric Field due to Point Charge====

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:28:29Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point in space. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C).

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:28:14Z

Sindhu K: /* See also */

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C).

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]
[[Electric Dipole]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:24:01Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C).

Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.

[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T23:23:37Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
A point charge refers to any particle that has a significantly small radius relative to the distance between it and any observation point of interest. This allows the generalization that all of the particle's charge and mass are localized at a specific point. The two known elementary charges are the proton and electron, both of which are modeled as point charges. A single proton holds a charge of +e (1.6e-19C), while an electron holds a charge of -e (-1.6e-19C).
Every charge creates its own electric field. As shown below, the electric field of a positive point charge points radially outward and that of a negative point charge points radially inward.
[[File:Point_charge.GIF]]

The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-24T22:38:04Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
Every charge creates its own electric field. The two known elementary charges are the proton and electron, both of which are modeled as point charges. The electric field for point charges points radially outward in the case of the proton, and radially inward in the case of the electron. The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

Below is a visual representation of electric field lines due to a positive and negative point charge

[[File:Point_charge.GIF]]

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Chabay. (2000-2018). ''Matter & Interactions'' (4th ed.). John Wiley & Sons.

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-22T07:42:22Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018)'''

==The Main Idea==
Every charge creates its own electric field. The two known elementary charges are the proton and electron, both of which are modeled as point charges. The electric field for point charges points radially outward in the case of the proton, and radially inward in the case of the electron. The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

Below is a visual representation of electric field lines due to a positive and negative point charge

[[File:Point_charge.GIF]]

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Matter and Interactions Vol. II

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]

Point Charge

2018-11-22T07:42:10Z

Sindhu K:

This page is all about the [[Electric Field]] due to a Point Charge.'''CLAIMED BY Sindhu Kannappan 11-22-2018 (Fall 2018'''

==The Main Idea==
Every charge creates its own electric field. The two known elementary charges are the proton and electron, both of which are modeled as point charges. The electric field for point charges points radially outward in the case of the proton, and radially inward in the case of the electron. The field is calculated using Coulomb's Law. Indeed, the electric field is defined as a vector field that associates each point in space with the electric force that an infinitesimal test charge would experience. The source of the electric field in this article is the point charge.

===A Mathematical Model of Electric Field due to Point Charge===

The Electric Field of a Point Charge can be found by using the following formula:

<math>\vec E=\frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} \hat r</math>

Which is derived from [[Gauss's Law]], modelling the point charge as a sphere.

<math>{\epsilon_0} </math> is a constant representing vacuum permittivity, the permittivity of free space, or the electric constant, and is approximately <math>8.854*10^{-12}\frac{C^2}{N m^2}</math>

<math>\frac{1}{4 \pi \epsilon_0 } </math> is known as Coulomb's Constant and is approximately <math>8.987*10^{9}\frac{N m^2}{C^2} </math>

'''''r''''' is the magnitude of the distance between the observation location and the source location
, '''''q''''' is the charge of the particle
and <math>\hat r </math> is the unit vector in the direction of the distance from the source location to the observation point.

The force on a given test charge is governed by <math> F = Eq </math> where '''''E''''' is the electric field and '''''q''''' is the charge of a test charge in Coulombs. This can be represented by Coulomb's Law:

<math> F = \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{r^2} \hat r </math>

By solving for the electric field in <math> F = Eq </math> modeled as Coulomb's Law, one obtains the equation for the electric field of the point charge:

<math> E = \frac{F}{q_2} = \frac{1}{4 \pi \epsilon_0 } \frac{q_1}{r^2} \hat r </math>

Below is a visual representation of electric field lines due to a positive and negative point charge

[[File:Point_charge.GIF]]

===A Computational Model===

Below is a link to some code which can help visualize the Electric Field due to a proton.

https://trinket.io/glowscript/725d552305

Notice how the arrows grow by a factor of <math> \frac{1}{r^{2}} </math> as the observation location gets closer to the proton.

==Examples==

===Simple===
There is an electron at the origin. Calculate the electric field at (4,-3,1)m.

'''Step 1:''' Find <math>\hat r</math>

Find <math>\vec r_{obs} - \vec r_{electron} ((4,-3,1) - (0,0,0) = <4,-3,1> </math>m

Calculate the magnitude of <math>\vec r</math>. (<math>\sqrt{4^2+(-3)^2+1^2}=\sqrt{26}</math>

From <math>\vec r</math>, find the unit vector <math>\hat{r}.</math> <math> <\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> </math>

'''Step 2:''' Find the magnitude of the Electric Field

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26} </math>

'''Step 3:''' Multiply the magnitude of the Electric Field by <math>\hat{r}</math> to find the Electric Field

<math>E = \frac{1}{4 \pi \epsilon_0 } \frac{-1.6 * 10^{-19}}{26}*<\frac{4}{\sqrt{26}},\frac{-3}{\sqrt{26}},\frac{1}{\sqrt{26}}> = <-4.34*10^{-11},3.26*10^{-11},-1.09*10^{-11}> N/C </math>

===Middling===
A particle of unknown charge is located at (-0.21, 0.02, 0.11) m. Its electric field at point (-0.02, 0.31, 0.28) m is <math><0.124, 0.188, 0.109> </math> N/C. Find the magnitude and sign of the particle's charge.

Given both an observation location and a source location, one can find both r and <math>\hat{r}</math> Given the value of the electric field, one can also find the magnitude of the electric field. Then, using the equation for the magnitude of electric field of a point charge,<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math> one can find the magnitude and sign of the charge.

'''Step 1:'''
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (-0.02,0.31,0.28) m - (-0.21,0.02,0.11) m = <0.19,0.29,0.17> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><0.19,0.29,0.17></math>

<math>\sqrt{0.19^2+0.29^2+0.17^2}=\sqrt{0.1491}= 0.39</math>

'''Step 2:''' Find the magnitude of the Electric Field

<math>E= <0.124, 0.188, 0.109> N/C</math>

<math>E_{mag} = (\sqrt{0.124^2+0.188^2+0.109^2}=\sqrt{0.0626}=0.25</math>

'''Step 3:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{0.39^2}*0.25 </math>

<math> q= + 4.3*10^{-12} C </math>

===Difficult===
The electric force on a -2mC particle at a location (3.98 , 3.98 , 3.98) m due to a particle at the origin is <math>< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}></math> N. What is the charge on the particle at the origin?

Given the force and charge on the particle, one can calculate the surrounding electric field. With this variable found, this problem becomes much like the last one.

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r_{mag}^2} </math>

to find the rmag value. To find <math>\hat r</math> we can find the direction of the electric field as that is obviously going to be in the same direction as <math>\hat r</math>. Then, once we find <math>\hat r</math>, all that is left to do is multiply <math>\hat r</math> by rmag and that will give us the <math> r</math> vector. We can then find the location of the particle as we know <math>r=r_{observation}-r_{particle}</math>

'''Step 1''':
Find the magnitude of the Electric field.

<math> F = Eq </math>

<math> <5.5*10^{3}, -7.6*10^{3}, 0> = E * -2mC </math>

<math> E = \frac{< -5.5*10^{3} , -5.5*10^{3}, -5.5*10^{3}>}{-2mC} = <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

'''Step 2:''':
Find <math>\vec r_{obs} - \vec r_{particle} </math>.

<math>\vec r = (3.98 , 3.98 , 3.98) m - (0 , 0 , 0) m = <3.98 , 3.98 , 3.98> m </math>

To find <math>\vec r_{mag} </math>, find the magnitude of <math><3.98 , 3.98 , 3.98></math>

<math>\sqrt{3.98^2+3.98^2+3.98^2}=\sqrt{47.52}= 6.9</math>

'''Step 3:''' Find the magnitude of the Electric Field

<math>E= <2.75*10^{6} , 2.75*10^{6} , 2.75*10^{6}> </math> N/C

<math>E_{mag} = (\sqrt{(2.75*10^{6})^2+2.75*10^{6})^2+2.75*10^{6})^2}=\sqrt{2.27*10^{13}}=4.76*10^{6}</math>

'''Step 4:''' Find '''''q''''' by rearranging the equation for <math>E_{mag}</math>

<math> E_{mag}= \frac{1}{4 \pi \epsilon_0 } \frac{q}{r^2} </math>

By rearranging this equation we get

<math> q= {4 pi * epsilon_0 } *{r^2}*E_{mag} </math>

<math> q= {1/(9*10^9)} *{6.9^{2}}*4.76*10^{6} </math>

<math> q= + 0.253 C </math>

==Connectedness==
1.How is this topic connected to something that you are interested in?

This topic is the most very basic aspect of physics relating to electricity and magnetism. As someone who is very interested in physics, this topic is important to understanding how more complex ideas that come in later on chapters work.

2.How is it connected to your major?

As a mechanical engineer, static charges don't relate to what I do all that much, but the vector algebra certainly does.

3.Is there an interesting industrial application?

The electric field of a point particle and the [[Superposition Principle]] are integral in understanding the interactions of many charges in various situations, such as a charged sphere or rod. Understanding the way charges behave is important in working with electricity.

==History==
[[File:CoulombCharles300px.jpg]]
''Charles de Coulomb''

Charles de Coulomb was born in June 14, 1736 in central France. He spent much of his early life in the military and was placed in regions throughout the world. He only began to do scientific experiments out of curiously on his military expeditions. However, when controversy arrived with him and the French bureaucracy coupled with the French Revolution, Coulomb had to leave France and thus really began his scientific career.

Between 1785 and 1791, de Coulomb wrote several key papers centered around multiple relations of electricity and magnetism. This helped him develop the principle known as Coulomb's Law, which confirmed that the force between two electrical charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This is the same relationship that is seen in the electric field equation of a point charge.

== See also ==

[[Electric Field]] 
[[Electric Force]] 
[[Superposition Principle]]

===Further reading===

Principles of Electrodynamics by Melvin Schwartz
ISBN: 9780486134673

Electricity and Magnetism: Edition 3 , Edward M. Purcell David J. Morin

===External links===

Some more information : http://hyperphysics.phy-astr.gsu.edu/hbase/electric/epoint.html

http://www.physics.umd.edu/courses/Phys260/agashe/S10/notes/lecture18.pdf

==References==

Matter and Interactions Vol. II

PY106 Notes. (n.d.). Retrieved November 27, 2016, from http://physics.bu.edu/~duffy/py106.html

Retrieved November 28, 2016, from http://www.biography.com/people/charles-de-coulomb-9259075#controversy-and-absolution

[[Category:Fields]]