Electric field on the surface of an infinite sheet of a perfect electric conductor

Question

let's consider ai infinite and flat sheet of charge distributed on a conductor. Well, it is known that the electric field at steady state on its surface (which is orthogonal to it) is equal to:

$\ E = \frac\sigma {2 \epsilon} $

where $\sigma$ is the surface charge density.

Now let's consider the specific case in which this conductor is a perfect electric conductor. Let's apply the interface conditions for electromagnetic fields, precisely for the normal component of the electric field (let's call 1 the space inside the conductor and 2 the space outside it):

$\ D_{2}-D_{1} = \sigma $

So, since the internal electric field is 0 in a perfect conductor, we get:

$\ D_{2} = \sigma $

that means:

$\ \epsilon \cdot E_{2} = \sigma $

Finally:

$\ E_{2} = \frac \sigma \epsilon$ that is double with respect to the field evaluated with the first method.

Notes: as stated in the reference of the first method, the conductor electric field is double than that of a simply sheet of charge (without speaking of conductors). But I have seen, for instance, people apply the equation $\ E = \frac\sigma {2 \epsilon} $ to find the E fild inside a parallel plate capacitor (there is also this evaluation in the reference). In this case, the plates are conductors, so why do not we use $\ E = \frac\sigma {\epsilon} $?

Answer 1

A surface charge is defined to have zero thickness. It is therefore not possible to talk about a point "inside" it. Because all charge moves to the surface of a conductor, there will be two surface charges on an infinite conducting plane, one on each side.

The boundary condition for a surface charge density $\sigma$ is a difference of $\frac{\sigma}{\epsilon_0} \mathbf{\hat{n}}$, where $\mathbf{\hat{n}}$ is the unit normal vector.

In fact, this boundary condition is derived from the electric field of a surface charge. Suppose you have some external electric field $\mathbf{E}$. Now place a uniform surface charge $\sigma$ in this field. We already know from first principles that the field of the surface charge is $\frac{\sigma}{2\epsilon_0} \mathbf{\hat{n}}$ above and $-\frac{\sigma}{2\epsilon_0} \mathbf{\hat{n}}$ below.

By the principle of superposition, the field immediately above the surface will be $\mathbf{E} +\frac{\sigma}{2\epsilon_0} \mathbf{\hat{n}}$, and the field immediately below will be $\mathbf{E} -\frac{\sigma}{2\epsilon_0} \mathbf{\hat{n}}$,

Thus, the discontinuity across the surface is obtained as $\left(\mathbf{E} +\frac{\sigma}{2\epsilon_0} \mathbf{\hat{n}}\right) - \left(\mathbf{E} -\frac{\sigma}{2\epsilon_0} \mathbf{\hat{n}}\right) = \frac{\sigma}{\epsilon_0} \mathbf{\hat{n}}$, as expected.

Answer 2

On the other side of the sheet there is an electric field pointing in the opposite direction. The total discontinuity is therefore $\Delta \vec E = \sigma/\epsilon_0$, as you indeed find from the interface conditions.