Question

Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

Answer 1

Step 1: Draw a rough diagram.



Step 2: Find potential on the axis of a disc.

Given,

Radius of disk =$R$

Magnitude of charge distributed on the disk = $Q$

Suppose an elementary part of ring of radius r of thickness dr on disc of radius R.

Charge on the ring is dq, then potential $dV$ due to ring at P, will be,

$dV=\frac{kdq}{r^{\prime }}$ $\therefore r^{\prime }=\sqrt{r^2}+x^2$

dq is the charge on the ring =$\sigma \times areaofthering$

$dq=\sigma \left(2\pi rdr\right)=2\pi r\sigma \cdot dr$

Now apply potential at P due to element,

$dV=\frac{Kdq}{r^{\prime }}$

$dV=\frac{K\cdot 2\pi r\sigma dr}{\sqrt{r^2+x^2}}$

Therefore, the total electric potential due to the disk is then obtained by summing or integrating the potentials due to all the elemental rings at point P,

${\int }_0^{V}dV={\int }_0^R\frac{K2\mu r\sigma dr}{\sqrt{r^2+x^2}}$

$V=\frac{1}{4\pi {ϵ}_o}.2\pi \sigma {\int }_0^R\frac{rdr}{(r^2+x^2{)}^{1/2}}$

$V=\frac{\sigma }{2{ϵ}_o}{\int }_0^R\frac{r}{(r^2+x^2{)}^{1/2}}dr$

$V=\frac{\sigma }{2{ϵ}_o}[\sqrt{r^2+x^2}{]}_0^R$

$V=\frac{\sigma }{2{ϵ}_o}\left[\sqrt{R^2+x^2-x}\right]$

[ we know that $\pi R^2\sigma =Q$ (charge on disc), So,$\sigma =\frac{Q}{R^2}$]

$V=\frac{2\pi R^2\sigma }{4\pi {ϵ}_0{R}^2}[\sqrt{R^2+x^2}-x]$

$V=\frac{2Q}{4\pi {ϵ}_0{R}^2}[\sqrt{R^2+x^2}-x]$

Final Answer: $V=\frac{2Q}{4\pi {ϵ}_0{R}^2}[\sqrt{R^2+x^2}-x]$