Question

A thin uniform angular disc of charge $Q$ has outer radius $4R$ and inner radius $3R$. Find the work done to take a unit positive charge from point $\text{P}$ on its axis to infinity.

• A. $\frac{-2KQ}{7R}[4\sqrt{2}+5]$
• B. $\frac{-2KQ}{7R}[4\sqrt{2}-5]$
• C. $\frac{2KQ}{7R}[4\sqrt{2}-5]$
• D. None of these

Answer 1

The correct option is B

$\frac{-2KQ}{7R}[4\sqrt{2}-5]$

Work done to take a positive charge $q$ from point $\text{P}$ to infinity,

$W=q(V_{\infty }-V_P)....\left(1\right)$

where, Potential at infinity, $V_{\infty }=0$
Potential at point $\text{P}$, $V_P$
Unit positive charge, $q=1\text{C}$

Substituting the values in equation $\left(1\right)$,

$W=-V_P....\left(2\right)$

Now, we need to calculate potential at $\text{P}$.

Let us consider a element of size $dr$ at distance $r$ from the centre of the disc.


Charge on the element,

$dq=\frac{Q}{\pi (16R^2-9R^2)}dA$

$\Rightarrow dq=\frac{Q}{7\pi R^2}2\pi rdr(\because dA=2\pi rdr)$

$\Rightarrow dq=\frac{2Qr}{7R^2}dr$

Potential at point $\text{P}$ due to this elemental ring,

$dV=\frac{Kdq}{\sqrt{r^2+x^2}}$

Substituting the value of $dq$,

$\Rightarrow dV=\frac{2KQ}{7R^2}\frac{rdr}{\sqrt{r^2+x^2}}$

Total potential at point $\text{P}$ due to complete disc,

$V_P=\underset{r=3R}{\overset{r=4R}{\int }}\frac{2KQ}{7R^2}\frac{rdr}{\sqrt{r^2+16R^2}}...\left(3\right)$

Let, $r^2+16R^2=t^2$

$\therefore rdr=tdt$

when, $r=3R,t=5R$

when, $r=4R,t=4\sqrt{2}R$

Now, substituting the values in $\left(3\right)$, we get

$V_P=\frac{2KQ}{7R^2}\underset{t=5R}{\overset{t=4\sqrt{2}R}{\int }}\frac{tdt}{t}$

$V_P=\frac{2KQ}{7R}[4\sqrt{2}-5]....\left(4\right)$

Now, from $\left(3\right)$ and $\left(4\right)$, we get

$W=\frac{-2KQ}{7R}[4\sqrt{2}-5]$

Hence, option (b) is correct answer.