Charge on inner/outer surfaces of two large parallel conducting plates

[zenterix]

Homework Statement

Consider two very large conducting plates which contain different total charges $Q_1$ and $Q_2$. Find the amount of charge on the inner and outer surfaces of the two plates. Neglect all edge effects.

Relevant Equations

I've thought about this for a while and am quite stuck.

I could write a long piece below about all my attempts at using Gauss' Law on different surfaces of the conductors to try to glean some insight.

I will initially try to reason about the simpler case of a capacitor made of two large parallel conducting plates with same charge (but opposite signs).

Let me first think about a simpler case. Suppose we have a capacitor. That is, the two plates have charges of equal magnitude and opposite signs.

Consider the purple rectangle which represents a Gaussian pillbox.

The electric field due to one of the plates individually has field lines perpendicular to the plate.

If the positive plate were a single surface, then by Gauss' Law we'd have $E_+=\frac{\sigma_+}{2\epsilon_0}$.

If instead the positive plate were a conductor with uniform charge distribution then we'd have $E_+=\frac{\sigma_+}{\epsilon_0}$.

Is it possible to have different charge densities on the inner and outer surfaces of such a conductor?

When we have a capacitor formed by two charged surfaces, we use superposition to find that the electric field is zero in the two outer regions and $\frac{2\sigma_+}{\epsilon_0}$ in the inner region.

But suppose we have infinitely large plate conductors, still with the same charge with opposite signs. Each plate has four side surfaces, an inner surface, an outer surface, and an interior.

It appears that we have induced charges on the surfaces from the interaction.

However, by symmetry, it seems we still have zero electric field in the outer regions.

Using a Gaussian pillbox (represented by the purple rectangle above) on the outer surface we find that since the electric field is zero then the flux through the pillbox is zero and so the enclosed charge is zero.

It would appear that all the charge is on the inner surface.

By symmetry, all the negative charge is on the inner surface of the bottom plate in the picture above.

This reasoning is an attempt at the solution of the original problem in the simpler case of a capacitor.

My first question is if this reasoning is correct?

 

[haruspex]

It would appear that all the charge is on the inner surface.

For equal and opposite charges, yes. You could also reason it by considering what would happen to any charge remaining on an outer surface. The net field from all the other charges would attract it to the inner surface.
Maybe you can extend that reasoning to deal with asymmetric charges.

The way I approach such problems is to work with potentials.

 

[Delta2]

Your reasoning is correct for an ideal capacitor, that is a capacitor where the electric field is enclosed between the plates and there is zero electric field fringing out. This would require infinite sized plates.

But in real world capacitors have finite plates and there is e-field outside of the capacitor, hence there is surface charge on the outer surface too, which of course is very little compared to the inner surface charge. I would say at least 90% of the total charge is on the inner surface.

 

[Steve4Physics]

Is it possible to have different charge densities on the inner and outer surfaces of such a conductor?

Yes, providing $Q_1+Q_2 \ne 0$. An (extreme) example would be if $Q_1=Q_2=+1C$.

IMO this is a somewhat unfair problem, so I’ll lay myself open to criticism by saying more than perhaps I should.

There are 2 key steps.

Step1: Show that (using your notation) $Q_{int,1} + Q_{int, 2} = 0$ . (You’ve basically already done that.)

Step 2: Prove that the charges on the outer faces are equal and are given by $Q_{out,1} = Q_{out, 2} = \frac 12 (Q_1+Q_2)$.

I can't see how to prove this just using Gauss's law. It's a 'symmetry thing'. I believe it follows from requiring the system's potential energy to be minimised; you can try that. Or maybe someone knows a better method.

The rest is then simple algebra.

 

[zenterix]

Given that the reasoning in the case of an ideal capacitor is correct. I've spent some time thinking about the problem of asymmetric charges. I am stumped. I am even questioning my understanding of Gauss' Law at this point.

For example, consider the following picture

The different colored rectangles are just different Gaussian pillboxes.

Consider the purple arrow.

The inside surface of the top conductor generates an electric field, call it $\vec{E}_{1,\text{in}}$.

The inside of the bottom conductor likewise has a $\vec{E}_{2,\text{in}}$.

Similarly, we have $\vec{E}_{1,\text{out}}$ and $\vec{E}_{2,\text{out}}$ for the outer surfaces.

Consider the purple Gaussian pillbox.

The enclosed charge is $\frac{\sigma_{1,\text{in}}}{\epsilon_0}$.

When we compute the flux through the pillbox, which electric fields have a non-zero contribution to the flux?

$\vec{E}_{1,\text{in}}$ contributes $E_{1,\text{in}}A$.

It seems all the other electric fields contribute nothing.

Then again

- The charges are distributed on the surface of each conductor in a way that makes the field in the interior of the conductor zero.
- This zero internal field is the result of summing all of the four fields in the interior of the conductors.
- The fields everywhere else (regions 1, 2, and 3 in the picture) are likewise the sum of all four fields in those regions.

So when I apply Gauss law to the purple Gaussian pillbox, can I not consider the field on portion of the pillbox outside of the conductor as the sum of all four fields?

 

[TSny]

Consider the purple Gaussian pillbox.

The enclosed charge is $\frac{\sigma_{1,\text{in}}}{\epsilon_0}$.

The enclosed charge would be $\sigma_{1, \text{in}}\,A$, where $A$ is the area of the inner surface enclosed by the purple box.

Then again

- The charges are distributed on the surface of each conductor in a way that makes the field in the interior of the conductor zero.
- This zero internal field is the result of summing all of the four fields in the interior of the conductors.
- The fields everywhere else (regions 1, 2, and 3 in the picture) are likewise the sum of all four fields in those regions.

Yes.

So when I apply Gauss law to the purple Gaussian pillbox, can I not consider the field on portion of the pillbox outside of the conductor as the sum of all four fields?

Yes. When using Gauss' law for the purple pillbox, the electric field contributing to the flux through the bottom surface of the box is the net electric field due to the charge on all four surfaces.

 

[TSny]

There are 2 key steps.

Step1: Show that (using your notation) $Q_{int,1} + Q_{int, 2} = 0$ . (You’ve basically already done that.)

Step 2: Prove that the charges on the outer faces are equal and are given by $Q_{out,1} = Q_{out, 2} = \frac 12 (Q_1+Q_2)$.

I can't see how to prove this just using Gauss's law. It's a 'symmetry thing'. I believe it follows from requiring the system's potential energy to be minimised; you can try that. Or maybe someone knows a better method.

Yes, symmetry is important for proving Step 2. I don't think you need to use minimization of energy.

You can prove Step 2 by using the result of Step 1 and the fact that the net electric field must be zero inside the material of either plate. (You can treat the surface charge on any one of the inner or outer surfaces as a uniform, infinite sheet of charge.)

For example, pick an arbitrary point inside the material of the upper plate. From the result of Step 1, what can you say about the total electric field at this point due to the charges on the two inner surfaces? Then, what can you deduce about the charges on the two outer surfaces?

 

[Steve4Physics]

When we compute the flux through the pillbox, which electric fields have a non-zero contribution to the flux?

Only the field emanating from charges inside the pillbox have a non-zero contribution to the pillbo's flux.

Field lines from charges outside the pillbox do not contribute to the pillbox's net flux. That's because each field line from an external charge enters and leaves (i.e. passes through) the pillbox; it produces equal + and - flux contributions - net effect is zero.

The flux from the purple pillbox is due only to the charge inside it. (Likewise for all other pillboxes.)

It seems all the other electric fields contribute nothing.

Correct.

- The charges are distributed on the surface of each conductor in a way that makes the field in the interior of the conductor zero.
- This zero internal field is the result of summing all of the four fields in the interior of the conductors.

Correct. Are you getting confused between electric field and flux? For example, there is a field in the gap between the 2 plates - but the net flux through a pillbox in the gap is zero.

- The fields everywhere else (regions 1, 2, and 3 in the picture) are likewise the sum of all four fields in those regions.

So when I apply Gauss law to the purple Gaussian pillbox, can I not consider the field on portion of the pillbox outside of the conductor as the sum of all four fields?

The flux from each of three of the fields passes through (in and out) the pillbox, giving zero net flux contribution.

Gauss's law alone looks insufficent to solve the asymmetric problem. See 'Step 2' in Post #4 (if you haven't already). EDIT -see @TSny's post.

Aha. I see @TSny beat me. But since I've already written it, I'll post anway.

 

[zenterix]

Step 1

$$\frac{Q_{1,\text{in}}+Q_{2,\text{in}}}{\epsilon_0}=\frac{\sigma_{1,\text{in}}A+\sigma_{2,\text{in}}A}{\epsilon_0}=0$$

$$\implies Q_{1,\text{in}}=-Q_{2,\text{in}}$$

 

[Steve4Physics]

Yes, symmetry is important for proving Step 2. I don't think you need to use minimization of energy.

You can prove Step 2 by using the result of Step 1 and the fact that the net electric field must be zero inside the material of either plate. (You can treat the surface charge on any one of the inner or outer surfaces as a uniform, infinite sheet of charge.)

For example, pick an arbitrary point inside the material of the upper plate. From the result of Step 1, what can you say about the total electric field at this point due to the charges on the two inner surfaces? Then, what can you deduce about the charges on the two outer surfaces?

Neat. Works a treat!

 

[zenterix]

If we just consider the two inner surfaces then we have two parallel infinitely large charged surfaces, each with the same magnitude of charge but opposite signs.

The electric field they generate is zero on the outer portions and $\frac{\sigma}{\epsilon_0}$ between them.

If we now go back to the original problem and consider a point inside one of the conductor plates, the electric field is zero. Thus, the electric fields due to the outer surfaces must cancel inside the conductors.

The charge on the outer surface of the top plate is $Q_1-Q_{1,\text{in}}$.

Analogously, the charge on the outer surface of the bottom plate is $Q_2-Q_{2,\text{in}}=Q_2+Q_{1,\text{in}}$.

 

[zenterix]

The pair of outer surfaces also represents two parallel infinitely large charged surfaces.

Since the electric field inside the conductors must be zero (and given that the two inner surfaces contribute a zero electric field inside the conductors) then the electric field due to these two outer surfaces must be zero between them.

For this to happen they must have the same charge density as this will generate electric fields of same magnitude and opposite direction between these surfaces.

$$\frac{\sigma_{1,\text{out}}}{\epsilon_0}=\frac{\sigma_{2,\text{out}}}{\epsilon_0}$$

$$\sigma_{1,\text{out}}=\sigma_{2,\text{out}}$$

$$\frac{Q_{1,\text{out}}}{A}=\frac{Q_{2,\text{out}}}{A}$$

$$Q_{1,\text{out}}=Q_{2,\text{out}}$$

Then

$$Q_1=Q_{1,\text{in}}+Q_{1,\text{out}}$$
$$Q_2=Q_{2,\text{in}}+Q_{2,\text{out}}$$

Summing these equations

$$Q_1+Q_2=2Q_{\text{out}}$$

$$Q_{\text{out}}=\frac{Q_1+Q_2}{2}$$

Subtracting one equation from the other

$$Q_1-Q_2=2Q_{1,\text{in}}$$

$$Q_{1,\text{in}}=\frac{Q_1-Q_2}{2}$$

and

$$Q_2-Q_1=2Q_{2,\text{in}}$$

$$Q_{2,\text{in}}=\frac{Q_2-Q_1}{2}$$