Question

Obtain the expression for potential at a point due to uniformly charged disc?

Answer 1

Question 32548

Let $z$ be the distance from center of the disc to the given point. Also, let the disc have radius $r$ and surface charge density $\sigma$ .

An infinitesimal potential created by charge $dq$ is $d\varphi = \frac{k dq}{R}$ , where $k = \frac{1}{4\pi \epsilon_0}$ is the

Coulomb's constant and $R$ is the distance from point of disc to the given point. In order to find the whole potential, one needs to integrate over the disc.

Elementary charge $dq$ in terms of charge density is $dq = \sigma dS = \sigma 2\pi R' dR'$ , where $R'$ is the distance from the center of disc to the circle at disc which creates elementary potential (in the plane of disc). Distance $R$ according to Pythagoras theorem is $R = \sqrt{R'^2 + z^2}$ . Hence:

$$\varphi = 2 \pi \sigma k \int_ {0} ^ {r} \frac {R ^ {\prime} d R ^ {\prime}}{\sqrt {R ^ {\prime 2} + z ^ {2}}} = 2 \pi k \sigma \sqrt {R ^ {\prime 2} + z ^ {2}} \zeta = 2 \pi \sigma k \left(\sqrt {r ^ {2} + z ^ {2}} - z\right).$$