Why are the two outer charge densities on a system of parallel charged plates identical?

Question

One of the ways examiners torture students is by asking them to calculate charge distributions and potentials for systems of charged parallel plates like this:

the ellipsis is meant to indicate any number of additional plates could be inserted where I've placed the ellipsis. The plates are assumed to be large enough that edge effects can be neglected i.e. they are effectively infinite.

In practice these problems are solved by the (frequently tedious) application of some simple rules, one of which is:

The charge densities on the two outermost surfaces (labelled $Q_{ext}$ in my diagram) are identical.

My question is how we can prove the statement that the two charge densities, $Q_{ext}$, are equal?

The problem is that our system of plates can have an arbitrary number of plates with arbitrary charges and spacings. Presumably one could with enough effort grind through a massive calculation where the plate charges and positions are all variables, but this seems an awfully complicated way of approaching an exceedingly simple observation. Surely there has to be a better way?

I am guessing that there must be some general argument one can make that means we get the same external charges regardless of the details of plates charges and spacings. Can anyone describe such an approach?

For bonus marks another of the starting points students are taught is that if any plate in the setup is earthed then the exterior charges are zero. If this doesn't emerge naturally from whatever approach is used to answer my first question I would also be curious to see how this could be proved.

Answer 1

The proof turns out to be very simple. Both Farcher and npojo have essentially answered the question but I’m going to present a detailed explanation, if only for my own satisfaction. This also captures elements of a discussion on the subject in the physics problem solving chat room.

There are a couple of preliminaries to get out of the way. Start by considering a single plate with a charge per unit area of $Q$.

The plate is a conductor so we know that:
a) the charge goes to the surfaces
b) the field inside the plate is zero

Since the charge resides at the surfaces we effectively have two sheet charges that I've shown in red and blue. The fields produced by these two sheets of charge are going to be approximately the infinite charged sheet field $E = Q/(2\epsilon)$. If the charge on the left side is $x$ then the charge on the right side has to be $Q-x$ so:

$$\begin{align} E_a &= \frac{x}{2\epsilon} \\ E_b &= \frac{Q-x}{2\epsilon} \end{align}$$

At the point $P$ inside the plate the field is zero, because we are inside a conductor, so the two fields $E_a$ and $E_b$ must be equal and opposite. Equating the fields immediately tells us that $x=Q/2$ and therefore that the charges on the two surfaces are equal to $Q/2$ (which should surprise no-one).

One more preliminary: consider any pair of adjacent plates in our system:

The blue line shows a Gaussian surface. The field inside the (conducting) plates is zero, so assuming the plates are large enough for end corrections to be negligible the flux through the surface is zero and hence the total charge inside the surface must be zero. That means $q_1 = -q_2$ i.e. the charges on the adjacent surfaces are equal and opposite. Again this is kind of obvious but it’s worth stating explicitly as we’ll need it for the proof.

Now to business. To see how the proof works start with a system of two plates with charges $Q_1$ and $Q_2$:

Again set the charge on the left surface of the first plate to $x$, and then the charges on the rest of the surfaces follow. As with the single plate we started with the field at $P$ is zero because it’s inside a conductor, and that means $E_{1a}$ must be equal and opposite to the sum of all the other fields:

$$ E_{1a} = E_{1b} + E_{2a} + E_{2b} $$

And substituting for the fields:

$$ \frac{x}{2\epsilon} = \left(\frac{Q_1-x}{2\epsilon} - \frac{Q_1-x}{2\epsilon}\right) + \frac{Q_2 + Q_1 - x}{2\epsilon} $$

I've bracketed the two terms that cancel. They cancel because the charges on adjacent surfaces are equal and opposite to their contributions to the total field at $P$ are equal and opposite. We end up with:

$$ x = \frac{ Q_2 + Q_1}{2} $$

And therefore the charges on the leftmost surface and the rightmost surface are both $(Q_1+Q_2)/2$ and we’ve proved the outermost charges are equal.

And the extension to $n$ plates is straightforward. The key point is that the fields on adjacent surfaces are all going to cancel so only the fields due to the leftmost and rightmost fields matter. The charges on the plates are:

Starting with a charge $x$ on the leftmost plates the charges on the right facing (blue) sides are $Q_1-x$, $Q_1+Q_2-x$, $Q_1+Q_2+Q_3-x$ and so on up to the rightmost surface and that has the charge $Q_1 + … + Q_n – x$. When we add up all the fields at the point $P$ the fields due to each adjacent sheet of charge cancel and we end up with:

$$ x = \tfrac{1}{2}\sum_n Q_i $$

So the charges on the outermost plates both have the same value of $\tfrac{1}{2}\sum_n Q_i$ and we've proved the result we wanted.

Answer 2

I think it is to do with minimising the electric potential energy stored in the electric fields produced by the arrangement of plates and charges?

Gauss's law across the whole arrangement with the end caps of the Gaussian surface outside the arrangement of plates tells us that the total charge stored on the outside of the outer plates is the same as the net charge stored on the plates.
That charge on the outside of the outer plates dictates what the electric field is outside the outer plates.
In other words if the net charge of the arrangement is $q$ then charge $q$ must be the total charge which resides on the outside of the outer plates.

So let the charges on the outside of the two plates be $x$ and $y$ with $x+y =q$.

The electric potential energy stored in the field outside the plates is proportional to $x^2+y^2$ and so $z=x^2+y^2$ is the quantity that needs to be minimised?

From this $\frac{dz}{dy} = 4x - 2q \Rightarrow x = \frac q 2 = y$ for a minimum, thus the equality of charge on the two plates.

Once a plate is earthed the total enclosed charge on the plates can become zero.
The charge on the outside of the plates is now zero and there is no electric field outside the plates.

Answer 3

Count the surfaces from left to right: S1,S2,S3...,Sn. Each surface generates a symmetric field (left and right). Hence total field left of S1 (E0) is equal to total field right of Sn. Field between S1 and S2 is zero (within the first plate). Field within the rightmost plate is also zero. So Qext is determined solely by E0 at both sides (narrow Gauss box around S1 and another box around Sn) Qext=E0*$\epsilon_0$.

If a plate is grounded, its right surface will accumulate charges that nullify charges on all the plates to its right. Same argument for the left surface. Hence E0=0 and continue as before.

Answer 4

Disclaimer: This answer is just an "attempt". I myself am not quite sure of this explanation. I write this as a high-schooler, with this being a problem being one for which I could not find any literature anywhere.

For bonus marks another of the starting points students are taught is that if any plate in the setup is earthed then the exterior charges are zero. If this doesn't emerge naturally from whatever approach is used to answer my first question I would also be curious to see how this could be proved.

This is what I attempt to prove.

Prerequisite:

When we connect two objects together with a wire, at equilibrium, the potential of the two objects is the same.

This what we use for problems involving earthing/grounding too. The "earth" is equivalent to an object at $\infty$. As per our "assignment" of potential as $0$ at $\infty$, the potential of the earth is thus $0$, and thus, upon "earthing" the potential of the object must also be zero.

This is what the general notion is for earthing a system. We mostly say "potential of the system is zero" although actually, what really happens is, "potential difference between the system and $\infty$ is $0$".

The latter statement can be expressed in the form of an integral:

$$\int_{r_{0}}^{\infty}\vec{E}\cdot d\vec{r}=0$$

This interpretation turns out to be particularly useful for our problem.

Consider this system of three plates.

The field $E_{3}=\dfrac{z}{\epsilon_{0}A}$ exists between the system and $\infty$. Thus, $\int_{r_{0}}^{\infty}\dfrac{z}{\epsilon_{0}A}dr=0$, which is only possible when $z=0$.

An equivalent statement (note: the field here is uniform) is that field lines cannot escape to $\infty$, otherwise there will exist a potential difference between the system and $\infty$. Clearly, the only way for this to happen is for $z$ to become zero.