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In most pictures I've seen of parallel plate capacitors, charges are drawn so that they're entirely on the inner surface of the plates.

enter image description here

I accept that there can't be any net charge within the conducting plates, as that would lead to a non-zero electric field within the metal, and charges would move to the surface.

However, I see no reason why charge can't reside on both sides of the same plate. I feel like there should be a way to have charge accumulate on both sides of the same plate, and still get a zero electric field within the metal. How can we show that there isn't?

Or...am I incorrect to assume that charge only distributes itself on the inner surfaces? I'd like to understand this both for the infinite plate idealization, and for the finite plate reality.

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2 Answers 2

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This is a good question and let us analyse a general situation and then come to your question.

Let us consider the following diagram:

Diagram 1

Two conducting plates $A$ and $B$ are placed parallel to each other. Let us assume that we supply a charge $Q_1$ to the left plate and $Q_2$ to the right plate. The charge distribution in the surfaces $1$, $2$, $3$ and $4$ is shown above.

We know that the electric field due to a surface charge density is given by $\frac{\sigma}{2\epsilon_0}$ where $\sigma$ is the surface charge density ($Q/A$ where $Q$ is the charge and $A$ is the surface area of the plate). Using this and the fact - net electric field inside a conductor in electrostatics is zero, let us write an expression for electric field inside the plate $A$ due to the charge densities on the four surfaces:

$$\begin{align} \frac{Q_1-q}{2A\epsilon_0}-\frac{q}{2A\epsilon_0}+\frac{q}{2A\epsilon_0}-\frac{Q_2+q}{2A\epsilon_0} &=0 \\ Q_1-q-Q_2-q &=0 \\ Q_1-Q_2 &=2q \\ \end{align}$$

$$q=\frac{Q_1-Q_2}{2}$$

Using this value of $q$, the charge distribution in the four surfaces looks something like the image below:

Diagram 2

Now let us turn our attention to your question.

Usually, the plates of a capacitor are not charged initially. When we connect the two plates of a parallel plate capacitor to the two terminals of a battery, the battery just acts as an electron pump and moves negative charge from the positive plate of the capacitor and pushes it to the negative terminal of the capacitor. Or in short, the battery doesn't supply a net charge just moves it from one plate to the other.

So the two plates have the same magnitude of charge but of opposite signs. Or on the lines of the example we considered earlier $Q_1=-Q_2$. Let the magnitude of this common charge be $Q$. Now the charge distribution looks like the following:

Diagram 3

Thus, the charges on the outer surface of an initially uncharged capacitor when connected to a battery is zero.


Please click on the images to view them in higher resolution.
Image courtesy: My own work :)

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That is not correct that if you had charge on both sides, that the electric field inside the metal would still be zero. Consider a situation similar to the picture you have shown, except that each plate has a charge density of $\sigma_1$ on the outside surface and $\sigma_2$ on the inside surface, with the signs flipped for the opposite plate. The electric field outside the capacitor will still be zero as before, since a Gaussian surface enclosing both plates will still contain zero net charge. The electric field inside will still be $\frac{\sigma}{\epsilon}$ as before, where $\sigma = \sigma_1 + \sigma_2$, because a Gaussian surface enclosing a single plate will still contain a net charge of $\sigma A$.

To find in electric field inside a plate, we can apply the boundary condition that $E_\text{outside} - E_\text{inside} = \frac{\sigma_1}{\epsilon_0}$, because translational symmetry tells us that the electric field is only in the perpendicular direction, assuming large area. Since we know that the electric field outside is zero, we conclude that in order for the electric field inside the plate to be zero, the charge density on the outside surface of the plate must be zero, meaning that all charge will lie in the inside surface of the plate.

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