# Do charges always lie on the inner surface of capacitors?

• phantomvommand
In summary: I am just studying undergraduate level physics (eg University Physics level). It seems to me that you are discussing the 2 plate situation, which I see why charges would all lie on 1 side of the plate. Does Gauss's law still apply if the charges are on both sides of the plate?In summary, Gauss's law still applies if the charges are on both sides of the plate.
phantomvommand
Parallel plates A, B are 5mm apart, with charges +1C and -1C respectively. Parallel plates C, D are 2mm apart, with charges +1C and -1C respectively. Capacitor CD is slid between capacitor AB. Find the potential difference between AB.

The key idea to solving this problem is to suppose that +1C lies on the inner surface of A, thus, -1C is induced on the outer surface of C. Since C has charge +1C, its inner surface charge is +2C. Following this logic, D has inner surface charge of -2C, outer surface charge of +1C, B has inner surface charge of -1C.

My question is: Why must all the +1C charge lie on the inner surface of A? Why can't some of it lie on the outer surface of plate A?

After sliding the second capacitor in and waiting a short time you have an electrostatic situation. The key thing to remember is that in electrostatics the charges arrange themselves so that there is no E field in the conductor.

DaveE and vanhees71
Dale said:
After sliding the second capacitor in and waiting a short time you have an electrostatic situation. The key thing to remember is that in electrostatics the charges arrange themselves so that there is no E field in the conductor.
Thanks for the reply. Could you please elaborate on this further? If I have charges that lie on both the outer and inner surface of plate A, there is still 0 electric field in plate A. Why must all the charges lie only one 1 surface then?

phantomvommand said:
If I have charges that lie on both the outer and inner surface of plate A, there is still 0 electric field in plate A.
This is not correct. Don’t forget the charges on the other plates.

phantomvommand
Dale said:
This is not correct. Don’t forget the charges on the other plates.

If I have the following charges on the outer and inner surfaces of each plate (A, C, D, B):

2 - Q
Q - 1

1-Q
Q

-Q
Q-1

1-Q
Q-2

Are you saying that the 1-Q charge (and other charges outside) creates some electric field in plate A? How does that reconcile with Gauss's Law, which suggests that because there are no charges inside plate A, the electric field is 0?

You can also think about it in terms of the interactions between the charges: The positive charges on one plate attract the negative charges at the other. Thus the charges tend to get as close to each other as they can, which also minimizes the potential energy as it must be in the static situation.

phantomvommand
vanhees71 said:
You can also think about it in terms of the interactions between the charges: The positive charges on one plate attract the negative charges at the other. Thus the charges tend to get as close to each other as they can, which also minimizes the potential energy as it must be in the static situation.
This was how I resolved it intuitively. However, I have yet to grasp Dale's idea that there is E-field in the conductors. Do you mind explaining why there would be an E-field in the conductor if the charges lie on both exterior surfaces, instead of just 1 surface?

Just start by thinking about the electrostatic field of a homogeneously charged plane (i.e., a plane carrying constant surface charge). I don't know at which level you are, but one simple way is to think about the symmetry of this configuration and make an ansatz for the electrostatic field due to this symmetry and then use Gauss's Law to get the numbers out.

Then just put another infinitely large plate parallel to the first with the opposite surface-charge density and write down its field and then add it to the field of the first plate.

If you have some more vector calculus at your disposal you can also look at the problem by solving Poisson's equation for the electrostatic potential using the appropriate boundary conditions at the plates.

vanhees71 said:
Just start by thinking about the electrostatic field of a homogeneously charged plane (i.e., a plane carrying constant surface charge). I don't know at which level you are, but one simple way is to think about the symmetry of this configuration and make an ansatz for the electrostatic field due to this symmetry and then use Gauss's Law to get the numbers out.

Then just put another infinitely large plate parallel to the first with the opposite surface-charge density and write down its field and then add it to the field of the first plate.

If you have some more vector calculus at your disposal you can also look at the problem by solving Poisson's equation for the electrostatic potential using the appropriate boundary conditions at the plates.
I am just studying undergraduate level physics (eg University Physics level). It seems to me that you are discussing the 2 plate situation, which I see why charges would all lie on 1 side of the plate. Does your argument extend to the case of multiple plates, in which the topmost and bottom most plate must all have their charges lying on 1 surface of the plate?

Dale said:
The key thing to remember is that in electrostatics the charges arrange themselves so that there is no E field in the conductor.
phantomvommand said:
However, I have yet to grasp Dale's idea that there is E-field in the conductors.

Did you miss the word "no" in what Dale said?

Well, if you have a third plate somewhere between the two plates, at this plate there must be surface charges on both surfaces.

anorlunda said:
Did you miss the word "no" in what Dale said?
Sorry, I meant why there would be E-field if the charges are not fully on 1 side.

vanhees71 said:
Well, if you have a third plate somewhere between the two plates, at this plate there must be surface charges on both surfaces.
Why is it not possible that the third plate causes a redistribution of the charge in the 2 other plates, such that there would now be charges on both surfaces of those 2 plates?

phantomvommand said:
Why is it not possible that the third plate causes a redistribution of the charge in the 2 other plates, such that there would now be charges on both surfaces of those 2 plates?
I think the energy argument is most straightforward. This is (intuitively) not favourable energetically. Is there another way to see this?

As I said, you can argue with symmetry arguments how the electric field should look like (for very large plates such that you can neglect edge effects) and then use Gauss's Law together with the contraint that there must be no electric field inside of conductors.

phantomvommand
vanhees71 said:
As I said, you can argue with symmetry arguments how the electric field should look like (for very large plates such that you can neglect edge effects) and then use Gauss's Law together with the contraint that there must be no electric field inside of conductors.
yes, I did not realize the symmetry idea. Thank you!

phantomvommand said:
How does that reconcile with Gauss's Law, which suggests that because there are no charges inside plate A, the electric field is 0?
Be careful here. It is not Gauss' law that tells us the electric field inside the plates are 0. We know that the field is 0 inside the plates due to Ohm's law, ##\vec J = \sigma \vec E## and the electrostatic condition, ##\vec J = 0##.

Once we have that information that inside the conductor the field is zero then we can use that together with Gauss' law to calculate the field between the plates. Usually we use a simplifying assumption about the symmetry, namely that the plates are infinitely large. This allows us to know by symmetry that the E-field is purely normal to the plates. We can then draw cylindrical Gaussian surfaces with the ends parallel to the plates and the sides along the normal direction. Then we know that there is no flux through the sides and all of the flux is at the ends.

Putting one end inside conductor A and moving the other end inside conductor C we can determine that the charge on the right side of A is equal and opposite the charge on the left side of C. Similarly for C and D and then for D and B. That gives us three equations.

We get four equations from the conservation of charge on each plate. So we have 7 equations in 8 unknowns. We need one more equation. That is energy minimization. The energy density is proportional to the square of the E-field. So the energy is equal to the energy density times the volume. The volume outside the outer plates is infinitely larger than the volume between the inner plates, so it is the only one that matters. So the final equation is obtained by minimizing the field outside the outer plates. This is proportional to the square of the charge on the left side of A plus the square of the charge on the right side of B. In this case that is 0.

phantomvommand
Dale said:
Be careful here. It is not Gauss' law that tells us the electric field inside the plates are 0. We know that the field is 0 inside the plates due to Ohm's law, ##\vec J = \sigma \vec E## and the electrostatic condition, ##\vec J = 0##.

Once we have that information that inside the conductor the field is zero then we can use that together with Gauss' law to calculate the field between the plates. Usually we use a simplifying assumption about the symmetry, namely that the plates are infinitely large. This allows us to know by symmetry that the E-field is purely normal to the plates. We can then draw cylindrical Gaussian surfaces with the ends parallel to the plates and the sides along the normal direction. Then we know that there is no flux through the sides and all of the flux is at the ends.

Putting one end inside conductor A and moving the other end inside conductor C we can determine that the charge on the right side of A is equal and opposite the charge on the left side of C. Similarly for C and D and then for D and B. That gives us three equations.

We get four equations from the conservation of charge on each plate. So we have 7 equations in 8 unknowns. We need one more equation. That is energy minimization. The energy density is proportional to the square of the E-field. So the energy is equal to the energy density times the volume. The volume outside the outer plates is infinitely larger than the volume between the inner plates, so it is the only one that matters. So the final equation is obtained by minimizing the field outside the outer plates. This is proportional to the square of the charge on the left side of A plus the square of the charge on the right side of B. In this case that is 0.
Thank you so much for explaining this is such great detail :)

vanhees71
Or:

1. Assign unknown charge density ##\sigma_i## , i=1 to 8, to the 8 sides.
2. ##\sigma## on sides of facing surfaces must have equal and opposite value (why?). This reduces the number of equations (in ##\sigma##) from 8 to 5.
2. For each plate the sum of ##\sigma## on both faces (times plate area) must equal the total charge on each plate. That gives you 4 equations in 5 unknowns.
3. The fifth equation is most easily found by mentally placing a small test charge in one of the plates and vectorially adding the contributions of all 8 sides to the test charge so as to make it see zero force (zero E field).

There is one and only one configuration for any combinatiuon of charge that will satisfy these equations.
Knowing the ##\sigma##s you know D eveywhere, thus E, then integrate from A to B for your potential.

phantomvommand
rude man said:
Or:

1. Assign unknown charge density ##\sigma_i## , i=1 to 8, to the 8 sides.
2. ##\sigma## on sides of facing surfaces must have equal and opposite value (why?). This reduces the number of equations (in ##\sigma##) from 8 to 5.
2. For each plate the sum of ##\sigma## on both faces (times plate area) must equal the total charge on each plate. That gives you 4 equations in 5 unknowns.
3. The fifth equation is most easily found by mentally placing a small test charge in one of the plates and vectorially adding the contributions of all 8 sides to the test charge so as to make it see zero force (zero E field).

There is one and only one configuration for any combinatiuon of charge that will satisfy these equations.
Knowing the ##\sigma##s you know D eveywhere, thus E, then integrate from A to B for your potential.
Thanks for this, I suppose the only configuration is the one where the charges all lie on the inner surface of the topmost and bottommost plate? This is a general result, right?

phantomvommand said:
Thanks for this, I suppose the only configuration is the one where the charges all lie on the inner surface of the topmost and bottommost plate? This is a general result, right?
No. Remember what I said about any inside face pairs?
In general there will be charges on all sides.
In your case (assuming unit area),
Q1 = ##\sigma1+\sigma2##
Q2 = ##-\sigma2+\sigma3## (remember about inside faces)
Q3 = ##-\sigma3+\sigma4##
Q4= ##-\sigma4+\sigma5##

For the last equation (putting the test source inside plate 1),
##\sigma1-\sigma5 = 0## (the inside charges always cancel).
From this,
##\sigma1 = \sigma5 (outside faces) = 1/2 (Q1+Q2+Q3+Q4) ##
which of course can be non-zero.
Etc.
To get the E fields, ##\sigma = D = \epsilon E ##
Then integrate to get ##V_1-V_2##.

phantomvommand
rude man said:
No. Remember what I said about any inside face pairs?
In general there will be charges on all sides.
In your case (assuming unit area),
Q1 = ##\sigma1+\sigma2##
Q2 = ##-\sigma2+\sigma3## (remember about inside faces)
Q3 = ##-\sigma3+\sigma4##
Q4= ##-\sigma4+\sigma5##

For the last equation (putting the test source inside plate 1),
##\sigma1-\sigma5 = 0## (the inside charges always cancel).
From this,
##\sigma1 = \sigma5 (outside faces) = 1/2 (Q1+Q2+Q3+Q4) ##
which of course can be non-zero.
Etc.
To get the E fields, ##\sigma = D = \epsilon E ##
Then integrate to get ##V_1-V_2##.
thanks for this!

BTW one more thing - if you think about it you will realize that the charge on the outer two surfaces will always be the same - in magnitude and polarity. So actually that reduces the number of equations to 4.

## 1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, one plate becomes positively charged and the other becomes negatively charged, creating an electric field between them. This electric field stores the electric charge, allowing the capacitor to hold energy.

## 2. Why do charges accumulate on the inner surface of a capacitor?

Charges accumulate on the inner surface of a capacitor because of the electric field created between the two plates. The electric field causes the charges to be attracted to the inner surface of the plates, where they can be stored. This creates a potential difference between the plates, allowing the capacitor to store energy.

## 3. Do charges always lie on the inner surface of a capacitor?

Yes, charges always lie on the inner surface of a capacitor. This is because the electric field created by the charges on the plates is strongest between the two plates. The charges are attracted to the inner surface of the plates, where they can be stored and contribute to the overall capacitance of the capacitor.

## 4. Can charges accumulate on the outer surface of a capacitor?

No, charges cannot accumulate on the outer surface of a capacitor. This is because the electric field between the plates is strongest between the two plates, so the charges are attracted to the inner surface. The outer surface of the plates has a weaker electric field, so charges are not able to accumulate there.

## 5. What happens if the dielectric material between the plates of a capacitor is removed?

If the dielectric material between the plates of a capacitor is removed, the electric field between the plates will increase significantly. This will cause the charges on the plates to move closer together, increasing the capacitance of the capacitor. However, without the dielectric, the capacitor will have a lower breakdown voltage and may not be able to hold as much charge before discharging.

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