Infinite Plane Of Charge

A Continuous Charge Distribution represented as a 2D plane with infinite area.

Electric Field §

Given a sheet of infinite length and width, we can’t really derive a total charge, $q$, since the area is infinite. However, if the sheet has a uniform Charge Density, $\sigma$, it means there is the same amount of charge for some section of area, which we can use in our calculations. If a point charge is placed at a distance of $z$ away from the sheet, perpendicular to it, the Electric Field of the plane can be represented as:

Formula §

$$\vec{E}=\frac{\sigma }{2{ϵ}_0}\hat{u}$$

• $\sigma$ = Surface Charge Density, assumed to be uniform (in $C/m^2$)
• $\hat{u}$ = Unit vector parallel to plane

Derivation (Without Gauss’ Law) §

Assume we have an infinite sheet $S$, a test charge of $+1C$, $A$.

In order to obtain the Electric Field we’ll use a circular region of the plane. If this was a finite plane, there would be some sections of the plane unaccounted for, but because it’s an infinite plane, it’s allowed. (#tosee )

This circular region can be divided into infinitesimal charged rings, which we can integrate to obtain the total electric field. This means we need to treat $r$, the radius of the rings, as the main independent variable, since that’s what changes as the rings increase.

We can use the definition of charge density and circle equations to obtain the charge in terms of the radius:

#ask Why is it $dr$ and not $r+\text{d}r$?

$$\sigma =\frac{\text{d}q}{\text{d}A}\to dq=\sigma dA$$

$$A=\pi r^2\to \frac{\text{d}A}{\text{d}r}=2\pi r\to \text{d}A=2\pi r\text{d}r$$

I.e. the Area of an infinitesimally small hoop is just the circumference of the hoop times the thickness, $dr$ So now, we have $dq$ in times of $r$

$$\text{d}q=\sigma \times 2\pi r\text{d}r$$

Using the formula for the electric field for a Ring of Charge:

$$|{\vec{E}}_{ring}|=k_e\times \frac{q\times z}{(z^2+r^2{)}^{3/2}}$$

We now need to get multiple of such rings, with varying radii, but more importantly, varying charge. Why? Well if the plane is uniformly charged, then the charge on a larger ring would be greater than that of a smaller ring.

$$\text{d}E=k_e\times \frac{\text{d}q\times z}{(z^2+r^2{)}^{3/2}}=k_e\times \frac{2\pi \sigma rz\text{d}r}{(z^2+r^2{)}^{3/2}}$$

#ask Why is not $\text{d}r$ and $\text{d}q$?

To obtain the total field from all the rings, we need to integrate:

$${\vec{E}}_S=\int k_e\times \frac{2\pi \sigma rz}{(z^2+r^2{)}^{3/2}}dr$$

We can replace $k_e$ with $\frac{1}{4\pi {ϵ}_0}$ to make things easier, and pull out the constants, $\sigma ,z$ :

$${\vec{E}}_S=\frac{\sigma z}{2{ϵ}_0}\int \frac{r}{(z^2+r^2{)}^{3/2}}dr$$

Furthermore we know that, with respect to $r$, the circular section will range from $r\in (0,\infty )$, which we can use as bounds for the integral:

$${\vec{E}}_S=\frac{\sigma z}{2{ϵ}_0}{\int }_0^{\infty }\frac{r}{(z^2+r^2{)}^{3/2}}dr$$

Using a U-Substitution, we can evaluate the integral:

$$\text{ Let }u=z^2+r^2\to \frac{du}{dr}=2r$$

$${\vec{E}}_S=\frac{\sigma z}{2{ϵ}_0}\times \frac{1}{2}{\int }_{z^2}^{\infty }{u}^{-3/2}du$$

$$=\frac{\sigma z}{4{ϵ}_0}{\left[-\frac{2}{\sqrt{u}}\right]}_{z^2}^{\infty }=\frac{\sigma z}{4{ϵ}_0}\times \left(\frac{-2}{\infty }+\frac{2}{\sqrt{z^2}}\right)$$

$$=\frac{\sigma z}{2{ϵ}_0}\times \frac{1}{z}=\frac{\sigma }{2{ϵ}_0}$$

This shows that the electric field is constant, and does not fall off as the distance from the plane increases.

Derivation With Gauss’ Law §

We know that the infinite plane can be translated around the XY plane, or rotated along the Z axis. As such, the electric field symmetry must match that:

Take a cylinder along the Z axis, such that the caps are parallel to the plane. We can use this cylinder as the Gaussian Surface, as the curved part of it will always be perpendicular to the E field, meaning it does not contribute to the total flux:

Using Gauss’ Law:

$$\Phi =\oint \vec{E}\cdot \text{d}\vec{A}=\frac{q_{enclosed}}{{ϵ}_0}$$

We know that the only the flat caps of the cylinder contribute to the total flux. Furthermore, even though the fields point in different directions, they are parallel to the area vectors of the caps, so they add up:

$$\Phi =\oint \vec{E}\cdot \text{d}\vec{A}=\oint |E|\text{d}|A|(A\parallel E)$$

$${\Phi }_{total}=|E||A_{cap}|+-|E|\times -|A_{cap}|=2E{A}_{cap}$$

Finally, we can obtain $q_{enclosed}$ using the surface charge density, $\sigma$:

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

$$2EA=\frac{A\sigma }{{ϵ}_0}\to E=\frac{\sigma }{2{ϵ}_0}$$

Which gets back to our equation!

Resources: §

• https://spinningnumbers.org/a/plane-of-charge.html