Infinite Conducting Charged Slab

Here, an infinite conducting charged plate means a 3D dimensional cuboid that has 2 faces much larger than the remaining 4, and is also a conductor. The derivation and shape are very similar to that of Infinite Parallel Planes Of Charge, with the key exception being it can only contain like charges

We assume the plate to be in Electrostatic Equilibrium due to being a conductor. Furthermore, due to it being infinitely large, we can assume the fields on the smaller 4 faces do not contribute to the total field.

Electric Field §

Assume the test charge is placed along the axis of the larger faces of the shape. Then the Electric Field can be given by:

$$\vec{E}=\frac{\sigma }{{ϵ}_0}\hat{u}$$

• $\sigma$ =Surface charge density (in $C/m^2$)
• $\hat{u}$ = Unit vector perpendicular to sheet

Why not volume charge density? Is it not a 3D shape?

Well, yes. But because we’ve specified that this object is a conductor, we use the property of conductors, specifically, Electrostatic Equilibrium of conductors, which causes their charges to accumulate on the surface of the object. Hence we use surface charge density, $\sigma$ instead of $\rho$.

Note how it is double that of an Plane of Charge. This is because there is, essentially, a double up of charges. But it’s important to note that, for the same amount of charge, $q$:

• A plane of charge with area $A$ will have $\sigma =\frac{q}{A}$
• A slab of charge with one face having area $A$ will have double the surface area, so $2A$. Implying the surface charge density is half: $\sigma =\frac{q}{2A}=\frac{1}{2}{\sigma }_{plane}$

So they’ll have the same electric field for the same amount of charge.

Derivation With Gauss’ Law §

We can deduce that the Electric Field Symmetry of this shape is such that the field is perpendicular to the surfaces and is evenly spread.

Interestingly enough, there are 2 configurations of Gauss’ Law we can use, which both give the same equation.

Asymmetric Version §

Let use a similar approach to that of a plane of charge, with a cylinder acting as our Gaussian Surface. However, only a part of the cylinder goes inside the sheet. We can use the property of Electrostatic Equilibrium to aid us in the derivation.

Let’s recall Gauss’ Law:

Firstly, as was the case with the plane of charge, the curved surface of the cylinder can be ignored since it is perpendicular to the electric field.

$$\Phi =\oint \vec{E}\cdot \text{d}\vec{A}=|E||A{|}_{cap}+|E||A{|}_{cap}$$

The key difference here is that the cap inside the conductor experiences 0 electric field, and so, zero flux:

$$\Phi =|E||A{|}_{cap}+0=E{A}_{cap}$$

We do the same thing we did for the sheet of charge, in regards to $q_{enclosed}$

$$\sigma =\frac{q_{enclosed}}{A}\to q_{enclosed}=\sigma A=\sigma A_{cap}$$

Finally:

$$E{A}_{cap}=\frac{\sigma A_{cap}}{{ϵ}_0}\to E=\frac{\sigma }{{ϵ}_0}$$

Symmetric Version §

We can use the same cylinder in a symmetric manner and still obtain the same result:

Note that now both caps of the cylinder experience flux:

$$\Phi =\oint \vec{E}\cdot \text{d}\vec{A}=E{A}_{cap}+-E\times -A_{cap}=2E{A}_{cap}$$

However, the charge enclosed is also doubled!

$$q_{enclosed}=\sigma A_{cap}\times 2$$

$$2E{A}_{cap}=\frac{2\sigma A_{cap}}{{ϵ}_0}$$

$$E=\frac{\sigma }{{ϵ}_0}$$