Question

An annular disc of inner radius $a$ and outer radius $2a$ is uniformly charged with uniform surface charge density $\sigma$. Find the potential at a distance $a$ from the centre at a point $P$ lying on the axis which is perpendicular to the plane containing the disc.

• A. $\frac{\sigma }{{ϵ}_0}a(\sqrt{5}-\sqrt{2})$
  
• B. $\frac{\sigma }{2{ϵ}_0}a(\sqrt{5}-\sqrt{2})$
• C. $\frac{\sigma }{2{ϵ}_0}(\sqrt{3}-\sqrt{2})$
  
• D. $$\frac{\sigma }{2{ϵ}_0}a$$
  

Answer 1

The correct option is B $\frac{\sigma }{2{ϵ}_0}a(\sqrt{5}-\sqrt{2})$

Given that,
Inner radius of the disc $=a$
Outer radius of the disc $=2a$
Uniform charge density $=\sigma$
Consider an elementary ring as shown in figure. The charge contained in this elementary ring is $dq$.
$\Rightarrow dq=2\sigma \pi rdr$

The potential due to this ring at a point $P$ is
$dV=\frac{dq}{4\pi {ϵ}_0{r}^{\prime }}$

Total potential due to the given disc.
$V=\frac{1}{4\pi {ϵ}_0}\int \frac{dq}{r^{\prime }}$
Now $r^{{}^{\prime }}=\sqrt{a^2+r^2}$
where $r$ is changing from $a$ to $2a$
$V=\frac{\sigma }{2{ϵ}_0}{\int }_a^{2a}\frac{rdr}{\sqrt{a^2+r^2}}$
$V=\frac{\sigma }{2{ϵ}_0}{\left\{\sqrt{a^2+r^2}\right\}}_a^{2a}$
$V=$ $\frac{\sigma a}{2{ϵ}_0}(\sqrt{5}-\sqrt{2})$

Why this question? Concept: This question helps in understanding that the element in a disc is a ring.