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

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

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.

• So, in case of a parallel plate capacitor, do we use $\ E = \frac\sigma {2 \epsilon}$ since we assume its plates very thin? – Kinka-Byo May 9 at 6:39
• @Kinka-Byo No. For a parallel-plate capacitor, there is a positive charge on one plate and an equal amount of negative charge on the other. So there are 2 surface charges, $+\sigma$ and $-\sigma$. Thus, between the plates, we have two copies of $\frac{\sigma}{2\epsilon_0}$, which gives $\frac{\sigma}{\epsilon_0}$. – user7777777 May 9 at 7:31
• and why do not we use $\frac{\sigma}{\epsilon_0}$ for each plate? – Kinka-Byo May 9 at 8:25
• @Kinka-Byo As I have already said, the electric field of each plate is $\frac{\sigma}{2\epsilon_0}$, and a capacitor has 2 such oppositely-charged plates. – user7777777 May 9 at 8:30
• @Kinka-Byo Yes, if you choose to assume that the plates are not thin, then it will be $\frac{2\sigma}{\epsilon_0}$. This is because there are now 2 layers of surface charge on each plate (see the first part of my answer). So the value of $\sigma$ will be half of what it is supposed to be, because the same amount of charge on the plate is now evenly distributed over 2 layers instead of 1. So the capacitor equation $Q=CV$ still holds. – user7777777 May 9 at 14:16

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.