Physics Book - User contributions [en]

Potential Difference at One Location

2015-12-05T21:41:14Z

Agxgeorge: /* References */

Written by Alex George

==The Main Idea==

Potential difference is usually defined as the change in the potential between two distinct points. However, it is also useful to determine potential at a single location of interest.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field, not a vector field.

===A Mathematical Model===

'''The Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#'''How is this topic connected to something that you are interested in?''' This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#'''How is it connected to your major?''' My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#'''Is there an interesting industrial application?''' Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

The SI unit for electric potential is the volt, which is named in honor of the Italian physicist [[Count Alessandro Volta]] (1745-1827), who invented the voltaic pile, the first chemical battery.

Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.

Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.

More information on Alessandro Volta can be found at Wikipedia [https://en.wikipedia.org/wiki/Alessandro_Volta] .

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

[https://en.wikipedia.org/wiki/Alessandro_Volta Alessandro Volta]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

Wikipedia.org

[[Category:Electric Potential]]

Potential Difference at One Location

2015-12-05T21:40:40Z

Agxgeorge: /* A Mathematical Model */

Potential Difference at One Location

2015-12-05T21:38:03Z

Agxgeorge: /* The Main Idea */

Written by Alex George

==The Main Idea==

Potential difference is usually defined as the change in the potential between two distinct points. However, it is also useful to determine potential at a single location of interest.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field, not a vector field.

===A Mathematical Model===

'''The Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#'''How is this topic connected to something that you are interested in?''' This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#'''How is it connected to your major?''' My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#'''Is there an interesting industrial application?''' Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

The SI unit for electric potential is the volt, which is named in honor of the Italian physicist [[Count Alessandro Volta]] (1745-1827), who invented the voltaic pile, the first chemical battery.

Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.

Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.

More information on Alessandro Volta can be found at Wikipedia [https://en.wikipedia.org/wiki/Alessandro_Volta] .

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

[https://en.wikipedia.org/wiki/Alessandro_Volta Alessandro Volta]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Electric Potential]]

Potential Difference at One Location

2015-12-05T21:32:19Z

Agxgeorge:

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#'''How is this topic connected to something that you are interested in?''' This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#'''How is it connected to your major?''' My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#'''Is there an interesting industrial application?''' Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

The SI unit for electric potential is the volt, which is named in honor of the Italian physicist [[Count Alessandro Volta]] (1745-1827), who invented the voltaic pile, the first chemical battery.

Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.

Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.

More information on Alessandro Volta can be found at Wikipedia [https://en.wikipedia.org/wiki/Alessandro_Volta] .

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

[https://en.wikipedia.org/wiki/Alessandro_Volta Alessandro Volta]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Electric Potential]]

Potential Difference at One Location

2015-12-05T21:30:56Z

Agxgeorge: /* Connectedness */

Potential Difference at One Location

2015-12-05T21:30:09Z

Agxgeorge:

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

The SI unit for electric potential is the volt, which is named in honor of the Italian physicist [[Count Alessandro Volta]] (1745-1827), who invented the voltaic pile, the first chemical battery.

Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.

Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.

More information on Alessandro Volta can be found at Wikipedia [https://en.wikipedia.org/wiki/Alessandro_Volta] .

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

[https://en.wikipedia.org/wiki/Alessandro_Volta Alessandro Volta]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T21:29:25Z

Agxgeorge: /* History */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

The SI unit for electric potential is the volt, which is named in honor of the Italian physicist [[Count Alessandro Volta]] (1745-1827), who invented the voltaic pile, the first chemical battery.

Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.

Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.

More information on Alessandro Volta can be found at Wikipedia [https://en.wikipedia.org/wiki/Alessandro_Volta] .

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T21:28:30Z

Agxgeorge: /* History */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

The SI unit for electric potential is the volt, which is named in honor of the Italian physicist [[Alessandro Volta]] (1745-1827), who invented the voltaic pile, the first chemical battery.

Volta had developed the voltaic pile using an effective pair of metals (zinc and silver) to create a steady electric current in the 1880s.

Today, the "conventional" volt is used and was defined in 1988 by the 18th general conference on Weights and Measures and adopted in 1990. The Josephson effect was employed to produce an exact frequency to voltage conversion and a cesium frequency standard is used to define the modern volt.

More information on Alessandro Volta can be found at Wikipedia [https://en.wikipedia.org/wiki/Alessandro_Volta] .

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T21:01:24Z

Agxgeorge: /* External links */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[https://youtu.be/70SsJNE3VFE Youtube Video on Electric Potential of a Point Charge]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T21:00:26Z

Agxgeorge: /* A Visual Model */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [https://youtu.be/70SsJNE3VFE].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:58:53Z

Agxgeorge:

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

Here is a video explaining the above concepts with more examples [http://www.scientificamerican.com/article/bring-science-home-reaction-time/].

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:56:45Z

Agxgeorge: /* Connectedness */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological systems. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. A particular implementation in medicine would be a defibrillator.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:50:23Z

Agxgeorge: /* See also */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological system. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. In any project involving electrical work, electric potential and its implications on the system are undoubtedly considered.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

[[Electric Potential]]

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Potential Difference of point charge in a non-Uniform Field]]

[[Sign of Potential Difference]]

[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:50:05Z

Agxgeorge: /* See also */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological system. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. In any project involving electrical work, electric potential and its implications on the system are undoubtedly considered.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

[[Electric Potential]]
[[Potential Difference in a Uniform Field]]
[[Potential Difference Path Independence]]
[[Potential Difference of point charge in a non-Uniform Field]]
[[Sign of Potential Difference]]
[[Potential Difference in an Insulator]]

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:42:52Z

Agxgeorge: /* Connectedness */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge. I am interested in bio instrumentation, so this concept has direct relevance in terms of circuit analysis and design.

#How is it connected to your major? My major is biochemistry and electric potential has several applications in biological system. For instance, in mechanisms such as the firing of a neuron's action potential, the electric potential across the neuronal membrane must reach certain thresholds in order to propagate the neural signal.

#Is there an interesting industrial application? Electric potential has various applications in industry, particularly in the aspects of circuit design and analysis. In any project involving electrical work, electric potential and its implications on the system are undoubtedly considered.

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:37:29Z

Agxgeorge: /* Connectedness */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in? This topic is related to the concept of [[Potential Energy]], or the capacity to do work in a system. [[Electric Potential]] is a derivation of potential energy and is a convenient way to express potential in terms of electric potential energy per unit charge.

#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:34:31Z

Agxgeorge: /* Connectedness */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

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

This topic is related to the concept of

#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:30:02Z

Agxgeorge: /* Further reading */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Chabay, Ruth W. ''Matter and Interactions, Ch. 16.7.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics, Ch. 19.3''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:26:32Z

Agxgeorge: /* Difficult Examples */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math> where <math> z </math> is the distance from the center of the ring, <math> R </math> is radius of the ring, and <math> Q </math> is the charge on the ring.

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {Qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {Q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:24:51Z

Agxgeorge: /* Difficult Examples */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math>

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

<math> V_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {q}{\sqrt{R^{2}+z^{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:23:37Z

Agxgeorge: /* Difficult Examples */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math>

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int_{\infty}^{z} \frac{1}{4\pi\varepsilon _{0} } \frac {qz}{(R^{2}+z^{2})^{\frac {3}{2}}} dz</math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:22:17Z

Agxgeorge: /* Difficult Examples */

Written by Alex George

==The Main Idea==

Usually, we are interested in the potential difference between two different points. However, it is also useful to determine potential at a single location.

In this case, we must define the potential at a location (ex. Location A) as the potential difference between a point infinitely far from all other charged particles and the location we defined.

As potential has a value at all locations at space, it is a scalar field in it of itself.

===A Mathematical Model===

'''Potential at One Location'''

<math>V_{a} = V_{a} - V_{\infty }</math> where '''<math>V_{a}</math>''' is the potential location of interest and '''<math>V_{\infty }</math>''' is the potential at a location infinitely far away.

This equation is only valid when <math>V_{\infty }</math> is equal to zero.

'''Potential near a Point Charge'''

Parameters: A point charge with charge <math>q</math>. We are a distance <math>x</math> from this point charge and we want to find the potential difference between our point charge and the potential at infinity.

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } (\frac{q}{x}-\frac{q}{\infty}) = \frac{1}{4\pi\varepsilon _{0}} \frac{q}{x} </math>

The sign of the potential depends on the sign of the charge on the particle:

if <math> q < 0, V_{x} < 0 </math>

if <math> q > 0, V_{x} > 0 </math>

'''Potential Energy at a Single Location'''

Once the value of the potential at one location is known, the potential energy of a system can be calculated.

If a charged object was placed at the location where the potential is known, then the potential energy of the system would be given the the equation:

<math> U_{A} = qV_{A} </math>

Substituting in the derived equation from the previous section for <math>V_{A}</math>, we obtain the following:

<math> U_{A} = q_{1} (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{2}}{x}) = (\frac{1}{4\pi\varepsilon _{0}} \frac{q_{1}q_{2}}{x}) </math> where <math> x </math> is the separation between the two particles, and <math> q_{1} </math> and <math> q_{2} </math> are the respective charges of the particles.

===A Visual Model===

A visualization of Distance vs. Potential from a point charge is given by Figure 1:

[[File:Potential_vs_Distance.png]]

'''Figure 1.''' Potential vs. Distance for a Point charge with Charge <math> e </math>

The electric field of the point charge can be obtained by differentiation. The negative gradient of the potential with respect to distance would yield the electric field:

<math> E_{x} = -\frac{\partial V}{\partial x} </math>

==Examples==

Here are a couple of problems that illustrate the concepts presented above.

===Simple Examples===

'''Example 1 - Potential at a Location Near a Sphere'''

What is the electric potential 5 cm away from a 1 cm diameter metal sphere that has a -3.00 nC static charge?

'''Solution:'''

As the distance from the center of the sphere is greater than the radius, we can treat the metal sphere as a point charge.

From the above derivation of the potential for a single point charge, we have:

<math>V_{x} = V_{x} - V_{\infty } = -\int_{\infty }^{x} \frac{1}{4\pi\varepsilon _{0} } \frac{q}{x^2} dx = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}-\frac{q}{\infty} \bigg) </math>

<math>V_{x} = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \bigg(\frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m}-\frac{-3.00\times 10^{-9}\ C}{\infty}\bigg) = 9\times 10^{9} \frac{N\cdot m^{2}}{C^{2} } \frac{-3.00\times 10^{-9}\ C}{5.00\times 10^{-2}\ m} = -539 V </math>

'''Example 2 - Charge Required to Produce a Specific Potential'''

What is the excess charge on a Van de Graff generator that has a 25.0 cm diameter metal sphere that produces a voltage of 50 kV near the surface?

'''Solution'''

The potential of the surface will be the same as that of a point charge from the center of the sphere, 12.5 cm away. The excess charge can be derived from the equation for potential a distance away from a point charge:

<math> V_{x} = \frac{1}{4\pi\varepsilon _{0} } \bigg(\frac{q}{x}\bigg) </math>

<math> q = {4\pi\varepsilon _{0}} x V_{x} </math>

<math> q = \frac{1}{9\times 10^{9}} \frac {N\cdot m^{2}}{C^{2}} (12.5 \times 10^{-2} \ m) (50 \times 10^{3} \ V) = 6.94 \times 10^{-7} \ C </math>

===Difficult Examples===

'''Example 1 - Potential Along the Axis of a Ring'''

What is the potential a distance <math> z </math> from the center of the ring?

'''Solution:'''

The electric field of a ring is given by the following equation:

<math> E_{z,ring} = \frac{1}{4\pi\varepsilon _{0} } \frac {qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math>

This equation can be integrated with respect to <math> z </math> and the appropriate bounds to obtain the potential at a location <math> z </math> from the center of the ring:

<math> V_{z,ring} = -\int{\infty}{z} \frac{1}{4\pi\varepsilon _{0} } \frac {qz}{(R^{2}+z^{2})^{\frac {3}{2}}} </math>

==Connectedness==
#How is this topic connected to something that you are interested in?
#How is it connected to your major?
#Is there an interesting industrial application?

==History==

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

== See also ==

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

===Further reading===

Books, Articles or other print media on this topic

===External links===
[http://www.scientificamerican.com/article/bring-science-home-reaction-time/]

==References==

Chabay, Ruth W. ''Matter and Interactions.'' 4th ed. New York: Wiley, 2015. Print.

Urone, Paul Peter., et al. ''College Physics''. Houston, Texas: OpenStax College, Rice University, 2013. Print.

[[Category:Which Category did you place this in?]]

Potential Difference at One Location

2015-12-05T20:20:00Z

Agxgeorge: /* Difficult Examples */

Potential Difference at One Location

2015-12-05T20:19:24Z

Agxgeorge: /* Difficult Examples */

Potential Difference at One Location

2015-12-05T20:18:32Z

Agxgeorge: /* Difficult Examples */

Potential Difference at One Location

2015-12-05T20:18:05Z

Agxgeorge:

Potential Difference at One Location

2015-12-05T20:15:33Z

Agxgeorge: /* A Visual Model */

Potential Difference at One Location

2015-12-05T20:14:25Z

Agxgeorge: /* A Mathematical Model */

Potential Difference at One Location

2015-12-05T20:14:09Z

Agxgeorge: /* A Mathematical Model */

Potential Difference at One Location

2015-12-05T20:13:46Z

Agxgeorge: /* Difficult Examples */

Potential Difference at One Location

2015-12-05T20:11:22Z

Agxgeorge: /* Simple Examples */

Potential Difference at One Location

2015-12-05T20:10:19Z

Agxgeorge: /* Simple Examples */

Potential Difference at One Location

2015-12-05T20:09:04Z

Agxgeorge: /* References */

Potential Difference at One Location

2015-12-05T20:08:00Z

Agxgeorge: /* References */

Potential Difference at One Location

2015-12-05T20:05:38Z

Agxgeorge: /* Difficult */

Potential Difference at One Location

2015-12-05T20:05:22Z

Agxgeorge: /* Simple */

Potential Difference at One Location

2015-12-05T20:04:53Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T20:04:39Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T20:04:23Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T20:02:01Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:59:52Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:57:28Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:57:01Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:55:39Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:55:29Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:55:10Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:54:41Z

Agxgeorge: /* Simple */

Potential Difference at One Location

2015-12-05T19:49:53Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:49:26Z

Agxgeorge: /* Examples */

Potential Difference at One Location

2015-12-05T19:48:51Z

Agxgeorge: /* Examples */