Physics Book - User contributions [en]

Potential Difference in an Insulator

2016-11-27T05:21:34Z

Yzhang922: /* Round Trip Potential Difference */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Round Trip Potential Difference==

Although the pattern of the electric field created by the dipoles inside the insulator is complex, the pattern of the electric field outside the insulator is much easier to tell. According to the diagram, when you travel in a circular path, the direction of the electric field of the dipoles and the direction of the change in distance are never opposite to each other. Thus, their dot product will always be a positive quantity. The round-trip potential difference, which is the round-trip path integral of the electric field, will always add up to zero.

[[File:screenshotwiki.png]]

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

The cases we discussed above only concern the spaces occupied by the insulator. If the insulator only fits part of the capacitor, the spaces it filled will be reduced by the factor K but the rest of the place will hardly be influenced.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

File:Screenshotwiki.png

2016-11-27T05:21:08Z

Yzhang922:

Potential Difference in an Insulator

2016-11-27T05:20:45Z

Yzhang922: /* Round Trip Potential Difference */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Round Trip Potential Difference==

Although the pattern of the electric field created by the dipoles inside the insulator is complex, the pattern of the electric field outside the insulator is much easier to tell. According to the diagram, when you travel in a circular path, the direction of the electric field of the dipoles and the direction of the change in distance are never opposite to each other. Thus, their dot product will always be a positive quantity. The round-trip potential difference, which is the round-trip path integral of the electric field, will always add up to zero.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

The cases we discussed above only concern the spaces occupied by the insulator. If the insulator only fits part of the capacitor, the spaces it filled will be reduced by the factor K but the rest of the place will hardly be influenced.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:57:08Z

Yzhang922:

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Round Trip Potential Difference==

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

The cases we discussed above only concern the spaces occupied by the insulator. If the insulator only fits part of the capacitor, the spaces it filled will be reduced by the factor K but the rest of the place will hardly be influenced.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:54:52Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

The cases we discussed above only concern the spaces occupied by the insulator. If the insulator only fits part of the capacitor, the spaces it filled will be reduced by the factor K but the rest of the place will hardly be influenced.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:51:08Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

[[File:wikinewi.jpg]]

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:49:56Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

File:Wikinewi.png

2016-11-27T04:49:38Z

Yzhang922:

Potential Difference in an Insulator

2016-11-27T04:48:44Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

[[File:wikinewi.png]]

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:46:46Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

[[File:wiki_2455.png]]

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:45:44Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

[[File:wiki.png]]

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

Potential Difference in an Insulator

2016-11-27T04:44:10Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

[[File:Screen Shot 2016-11-26 at 23.40.30]]

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

File:Screen Shot 2016-11-26 at 23.40.30.png

2016-11-27T04:41:31Z

Yzhang922:

Potential Difference in an Insulator

2016-11-27T04:38:42Z

Yzhang922: /* Dielectric Constant */

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
Placing an insulator between the plates of a capacitor would decrease the electric field inside the insulator and decrease potential difference across the insulator. When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. The dielectric constant is related to the atomic polarizability. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

\

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

RC Circuit

2016-11-27T04:05:57Z

Yzhang922:

BY MARK RUSSELL SPRING 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems

<math>{ΔV}_{round trip} = {0} </math>

<math>{ΔV} = {I}{R} </math>

<math>{Q} = {C}{V} </math>

In a loop of a circuit, the change of potential difference has to be zero. The energy equation for the RC Circuit in the figure at the top of the page is:

<math>{ΔV}_{round trip} = emf-RI-Q/C = 0</math>

At the final state of the circuit after the current has dropped to zero and the capacitor is fully charged, <math>{RI} = {0} </math>. The new equation for the final state is:

<math>{V}_{round trip} = emf-Q/C = 0</math>

<math>{Q} = emf*C</math>

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

You know that <math>{I} = {\frac{dQ}{dt}}</math> due to the rate at which charge builds up the positive capactior plate.

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

If you exponentiate both sides, the following equation is achieved.

<math>{\frac{IR}{emf}} = {e^{\frac{-t}{RC}}}</math>

<math>{I} = {\frac{emf}{R}e^{\frac{-t}{RC}}={\frac{dQ}{dt}}}</math>

<math>{dQ} = \int_0^t{\frac{emf}{R}e^{\frac{-t}{RC}}\,\mathrm{d}t}</math>

<math>{Q} = {C(emf)(1-e^{\frac{-t}{RC}})}</math>.

Since <math>{V} = {Q/C}</math>, thus

<math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>.

This is the formula for the change in voltage of a series RC circuit with respect to time.

The RC time constant formula is:

<math>{τ} = {RC}</math>

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

Potential Difference in an Insulator

2016-11-27T04:05:17Z

Yzhang922:

'''Claimed by Yunshu Zhang Fall 2016'''

[http://www.physicsbook.gatech.edu/Potential_Difference_in_a_Uniform_Field Potential difference] is defined as a scalar quantity that measures the difference of energy per unit charge. This page will not go over how to calculate electric potential in a conductor (because other pages cover this topic), but rather, how to find the electric potential in an insulator given the potential difference in a vacuum.

==Potential Difference==
Although this section will not go in depth into how to calculate potential difference, the following analysis requires the knowledge that potential difference equals the dot product of the electric field vector and distance vector between two points. Understanding that potential difference is dependent on the distance between two points is an important prerequisite to comprehending how to find the potential difference inside an insulator.

==Net Electric Field Inside an Insulator==
[[File:Electric_field_Wiki.png]]

Between two points inside a metal object in equilibrium, the potential difference is zero. However, this is untrue for an insulator that is polarized by an applied electric field (such as that inside a capacitor). In order to quantitatively predict the potential difference between two points in an insulator, we first must understand how this applied electric field polarizes molecules in the surrounding objects. Typically, this electric field creates induced dipoles inside the insulating materials. These induced dipoles contribute their own electric field to the net field. Because dipoles create such a unique pattern of electric field, it is slightly more complex to find the electric field and potential difference inside the insulator.

The net electric field inside the insulating material is the sum of the applied electric field (often due to a capacitor) and the electric field produced by the induced dipoles. The field from the induced dipoles always points in the direction opposite to the applied electric field (as shown in Diagram 1). Consequently, the net electric field is in the direction of the applied field/capacitor, but it is weaker in magnitude.

==Dielectric Constant==
When solving for the potential difference in an insulator, we define the constant ''K'' as the dielectric constant. This quantity represents the amount by which the net electric field is "weakened" due to the induced dipoles. It is a value typically known through previous scientific experimentation (specifically, measuring the effect of an insulator on the potential difference between two charged objects).

====A Mathematical Model for the Electric Field Inside an Insulator====
<math>\vec {E}_{insulator} = \frac{\vec{E}_{applied}}{K}</math>

The dielectric constant ''K'' is always bigger than one if an insulator is present because the induced dipoles in the polarized insulator always weaken the net electric field. When an insulating substance is easy t polarize, ''K'' will be large because the induced dipoles will create a weaker net electric field. If there is no insulating material between the charged objects (the space is a vacuum), ''K'' equals one.

==Relating Electric Field to Potential Difference==

Because the relationship between the electric field and potential difference is proportional, potential difference will also decrease by a value ''K''. That is, placing an insulator in between two charged objects like the plates of a capacitor also decreases potential difference across the insulator. It is important to note that if the insulator does not fill the gap between objects, the electric field and potential difference inside the insulator are still reduced by a factor ''K''. However, the areas that are not filled by the insulator are not affected since the electric field inside the insulator is negligibly small.

====A Mathematical Model for the Potential Difference Inside an Insulator====
<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

==Examples==

Be sure to show all steps in your solution and include diagrams whenever possible

===Simple===
A capacitor is known to create a uniform electric field of magnitude 200 N/C. The capacitor plates are 5mm apart. A plastic slab (whose dielectric constant is 5) is carefully inserted into the gap between the capacitor. Calculate the potential difference across the insulator.
[[File:Wiki_2_CB.png]]

'''Step 1:''' Find <math>\Delta{V}_{vacuum}</math>

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V}_{vacuum} = 200*\frac{5}{1000}</math>

<math>\Delta{V}_{vacuum} = 1 V</math>

'''Step 2:''' Use the potential difference in a vacuum and the dielectric constant to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{1}{5}</math>

<math>\Delta{V}_{insulator} = 0.2V</math>

The potential difference inside the plastic slab is 0.2 V.

===Middling===
[[File:Wiki_2_CB3.png]]
A capacitor's left plate has charge +Q. It's right plate has charge -Q. As shown in the diagram, the plates are separated by a distance ''s''. Both plates have a length ''L'' and a width ''W''. An insulator with dielectric constant ''K'' is inserted into the gap. Find the potential difference across the capacitor.

'''Step 1''': Find the electric field of the capacitor.

<math>{E}_{capacitor} = \frac{Q/A}{\epsilon_0 }</math>

<math>{E}_{capacitor} = \frac{Q/LW}{\epsilon_0 }</math>

'''Step 2''': Find the potential difference across the capacitor if there was no insulator.

<math>\Delta{V}_{vacuum} = E●\vec{dl}</math>

<math>\Delta{V} = \frac{Q/LW}{\epsilon_0}*s</math>

'''Step 3''': Use the dielectric constant to find the potential difference across the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>\Delta{V}_{insulator} = \frac{Q/LW}{\epsilon_0}*\frac{s}{K} = \frac{Qs}{LWK\epsilon_0}</math>

===Difficult===
[[File:Wiki_2_CB2.png]]
A capacitor originally has a potential difference of 500V. The capacitor plates are 2mm apart. A 1mm thick plastic slab (whose dielectric constant is 5) is inserted into the gap between the capacitor, but it does not fill the gap. It is placed 0.3mm from the positively charged left plate and 0.7mm from the negatively charged left plate. Calculate the following potential differences: <math>{V}_{1}-{V}_{2}</math>, <math>{V}_{2}-{V}_{3}</math>, <math>{V}_{3}-{V}_{4}</math>, and <math>{V}_{1}-{V}_{4}</math>.

'''Step 1''': Find <math>{V}_{1}-{V}_{2}</math>

Originally, the potential difference in the capacitor is 500V. We must solve for the electric field in the capacitor (without the insulator) first.
<math>E = \frac{\Delta{V}}{\vec{dl}} = \frac{500}{0.002} = 250000 N/C</math>

Because the insulator is not present in the area between points 1 and 2, we can find the potential difference from point 1 to point 2 by multiplying the electric field by the distance between points.

<math>\Delta{V} = E●\vec{dl}</math>

<math>{V}_{1}-{V}_{2} = 250000*.0003 = 75V</math>

'''Step 2''': Find <math>{V}_{2}-{V}_{3}</math>
In order to find this quantity, we must first find the potential difference that would occur in a vacuum. To do this, we take the electric field we found in step 1 and multiply by the distance.

<math>\Delta{V} = E●\vec{dl}</math>

<math>\Delta{V} = 250000*.001 = 250V</math>

Next, we must use the dielectric constant of plastic to find the potential difference in the insulator.

<math>\Delta{V}_{insulator} = \frac{\Delta{V}_{vacuum}}{K}</math>

<math>{V}_{2}-{V}_{3} = \frac{250}{5} = 50V</math>

'''Step 3''': Find <math>{V}_{3}-{V}_{4}</math>

Because the insulator is not present in the area between points 3 and 4, we can find the potential difference from point 3 to point 4 by multiplying the electric field by the distance between points.

<math>{V}_{3}-{V}_{4} = E●\vec{dl} = 250000*0.007 = 175V</math>

'''Step 4''': Find <math>{V}_{1}-{V}_{4}</math>

In order to find the potential difference between points 1 and 4, we simply need to add up the potential differences found in steps 1-3.

<math>{V}_{1}-{V}_{4} = 75V + 50V + 175V = 300V</math>

==Connectedness==
Understanding insulation is particularly important in fully comprehending circuits and electricity. Although not used as commonly in experimental entry-level physics (such as Physics 2211 or 2212) as elements of circuits like capacitors and resistors, insulators have a more practical application. As evidenced by several of the examples on this page, the potential difference across an insulator is always significantly less than through a vacuum. Because they do not let electric charge flow easily from one atom to another, insulators are used to coat wires in order to protect human users from the dangerously high voltage produced by some electric current.

==History==

Potential difference is a phenomenon that was first discovered by Alessandro Volta, an Italian physicist who lived from 1745 to 1827. Through his work with Luigi Galvani, Volta detected the flow of electric current through various conducting materials. This led to the discovery of electromotive force (what we more commonly refer to as emf) and eventually allowed Volta to create the first battery. Slightly later, Georg Ohm, a German physicist, began studying Volta's research. Through experimentation, Ohm was the first to uncover the relationship between the potential difference applied through a conductor and the resultant electric current. These findings and scientists were the most crucial to discovering the relationships between current, resistance, electric field, and voltage/potential difference.

== See also ==
In order to more fully understand potential difference in an insulator, a basic understanding of potential difference through a vacuum is necessary. These other Physics Book Wiki pages should further one's understanding of this topic.

[[Potential Difference in a Uniform Field]]

[[Potential Difference Path Independence]]

[[Sign of Potential Difference]]

===External links===
https://www.youtube.com/watch?v=TFGpgfe3Q2g

http://physics.bu.edu/~duffy/semester2/c08_dielectric.html

==References==

http://www.physics.sjsu.edu/becker/physics51/capacitors.htm
https://www.nde-ed.org/EducationResources/HighSchool/Electricity/conductorsinsulators.htm
http://www.physics.utah.edu/~woolf/2220_buehler/electricpotential.pdf
Matter and Interactions, 4th Edition by: Ruth Chabay and Bruce Sherwood

[[Category:Fields]]

-cbrogan7

RC Circuit

2016-11-26T00:51:26Z

Yzhang922:

Claimed by Yunshu Zhang Fall 2016

BY MARK RUSSELL SPRING 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems

<math>{ΔV}_{round trip} = {0} </math>

<math>{ΔV} = {I}{R} </math>

<math>{Q} = {C}{V} </math>

In a loop of a circuit, the change of potential difference has to be zero. The energy equation for the RC Circuit in the figure at the top of the page is:

<math>{ΔV}_{round trip} = emf-RI-Q/C = 0</math>

At the final state of the circuit after the current has dropped to zero and the capacitor is fully charged, <math>{RI} = {0} </math>. The new equation for the final state is:

<math>{V}_{round trip} = emf-Q/C = 0</math>

<math>{Q} = emf*C</math>

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

You know that <math>{I} = {\frac{dQ}{dt}}</math> due to the rate at which charge builds up the positive capactior plate.

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

If you exponentiate both sides, the following equation is achieved.

<math>{\frac{IR}{emf}} = {e^{\frac{-t}{RC}}}</math>

<math>{I} = {\frac{emf}{R}e^{\frac{-t}{RC}}={\frac{dQ}{dt}}}</math>

<math>{dQ} = \int_0^t{\frac{emf}{R}e^{\frac{-t}{RC}}\,\mathrm{d}t}</math>

<math>{Q} = {C(emf)(1-e^{\frac{-t}{RC}})}</math>.

Since <math>{V} = {Q/C}</math>, thus

<math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>.

This is the formula for the change in voltage of a series RC circuit with respect to time.

The RC time constant formula is:

<math>{τ} = {RC}</math>

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]

RC Circuit

2016-11-26T00:51:13Z

Yzhang922:

Claimed by Yunshu Zhang Fall 2016
BY MARK RUSSELL SPRING 2016
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==

An RC circuit is a circuit that contains a battery with a known emf, a resistor (R), and a capacitor (C). An RC circuit can be in either series or parallel. The figure in the top right of the page shows an RC circuit. The capacitor stores electric charge (Q)

RC Circuits use a DC (direct current) voltage source and the capacitor is uncharged at its initial state. In the figure below, you see an RC circuit with a switch. When the switch is closed, the capacitor will begin to charge as the current can now flow throughout the circuit. To discharge the capacitor, you simply disconnect the switch.

[[File:Rc_switch.JPG|400px|center|]]

Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.

[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.

[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.

===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems

<math>{ΔV}_{round trip} = {0} </math>

<math>{ΔV} = {I}{R} </math>

<math>{Q} = {C}{V} </math>

In a loop of a circuit, the change of potential difference has to be zero. The energy equation for the RC Circuit in the figure at the top of the page is:

<math>{ΔV}_{round trip} = emf-RI-Q/C = 0</math>

At the final state of the circuit after the current has dropped to zero and the capacitor is fully charged, <math>{RI} = {0} </math>. The new equation for the final state is:

<math>{V}_{round trip} = emf-Q/C = 0</math>

<math>{Q} = emf*C</math>

To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

You know that <math>{I} = {\frac{dQ}{dt}}</math> due to the rate at which charge builds up the positive capactior plate.

<math>{I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}}</math>

If you exponentiate both sides, the following equation is achieved.

<math>{\frac{IR}{emf}} = {e^{\frac{-t}{RC}}}</math>

<math>{I} = {\frac{emf}{R}e^{\frac{-t}{RC}}={\frac{dQ}{dt}}}</math>

<math>{dQ} = \int_0^t{\frac{emf}{R}e^{\frac{-t}{RC}}\,\mathrm{d}t}</math>

<math>{Q} = {C(emf)(1-e^{\frac{-t}{RC}})}</math>.

Since <math>{V} = {Q/C}</math>, thus

<math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>.

This is the formula for the change in voltage of a series RC circuit with respect to time.

The RC time constant formula is:

<math>{τ} = {RC}</math>

===A Computational Model===

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)

[[File:rcvoltage.gif|300px|center|]]

For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

[[File:rccurrent.gif|300px|center|]]

==Examples==

===Simple===
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

<math> {RC}= {12*1} = {12 seconds}</math>

===Middling===
Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

===Difficult===
Suppose one wished to capture the picture of a bullet (moving at 0.04 m/s ) that was passing through an orange. The duration of the flash is related to the RC time constant, τ . What size capacitor would one need in the RC circuit to succeed, if the resistance of the flash tube was 10.0 Ω? Assume the oragne is a sphere with a diameter of 0.08 m.

Solution:

You know the velocity of the bullet and the distance. You can find the time using Physics I principles such as <math>{time} = {\frac{distance}{velocity}} = {\frac{.08}{.04}} = {.0032 seconds} </math>

the time becomes equal to τ, so:

<math>{C} = {\frac{τ}{v}} = {\frac{0.0032}{10Ω}} = {32μF} </math>

==Connectedness==
1. How is this topic connected to something that you are interested in?
:: I like listening to music and RC currents are used in subwoofers in my car and other audio equipment.
2. How is it connected to your major?
:: I'm an ISyE major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled.
3. Is there an interesting industrial application?
:: RC circuits are used to hear certain sounds so that in an assembly line something can be sorted or passed through to another area.

==History==

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists.

== See also ==

===Further reading===
[[Charge in a RC Circuit]]

[[Current in a RC circuit]]

[[Loop Rule]]

[[Node Rule]]

[[Current]]

===External links===

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

==References==

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

[[Category:Simple Circuits]]