Electric Field of a Charged Disk

Problem

Figure 1. Disc of radius R

Find the electric field a distance $x$ along the axis from a disc of radius $R$ and uniform charge density $\sigma$.

Hint

(1) Think of this disk as a bunch of concentric rings. Begin by finding the electric field a distance $x$ along the axis up from a thin ring of charge $dQ$ and radius $a$.

(2) You may use the integral

$$\int \frac{xdx}{(\sqrt{x^2 + a^2})^3} = -\frac{1}{\sqrt{x^2 + a^2}} + C.$$

Solution

Figure 2. Electric field at point P

Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is

$$r = \sqrt{x^2 + a^2}.$$

Hence, the differential element of electric field (along the x-axis) is

$$dE_x = dE \cos{(\theta)}$$

$$dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{dQ}{x^2 + a^2} \frac{x}{\sqrt{x^2 + a^2}}.$$

Integrating the differential yields:

$$E_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{x}{(\sqrt{x^2 + a^2})^3} \int_{0}^{Q} dQ :$$

$$E = \frac{1}{4\pi {\epsilon}_{0}} \frac{xQ}{(\sqrt{x^2 + a^2})^3} \boldsymbol{i}.$$

For the disk problem, replace the differential charge with $dQ = \sigma 2\pi r dr$ because the areal charge density is defined as $\sigma = \frac{dQ}{dA}$ and $A = \pi r^2$ for a disk.

Since we are integrating with respect to the variable $r$ now, the variable radii $a$ is replaced with $r$. Therefore, the electric field is modified to

$$dE_x = \frac{1}{4\pi {\epsilon}_{0}} \frac{\sigma 2\pi rdr}{x^2 + r^2} \frac{x}{\sqrt{x^2 + r^2}}.$$

Using the given integral yields,

$$E_x = \frac{1}{4\pi {\epsilon}_{0}} \int_{0}^{R} \frac{(\sigma 2\pi rdr) x}{(\sqrt{x^2 + r^2})^3}$$

$$E_x = \frac{\sigma x}{2 {\epsilon}_{0}} \left( -\frac{1}{\sqrt{x^2 + R^2}} + \frac{1}{x} \right).$$

Therefore, the electric field a distance $x$ along the axis from a disc of radius $R$ and uniform charge density $\sigma$is

$$E = \frac{\sigma}{2 {\epsilon}_{0}} \left(1 -\frac{x}{\sqrt{x^2 + R^2}} \right) \boldsymbol{i}.$$