Question

Two identical rings $P$ and $Q$ of radius $0.1m$ are mounted coaxially at a distance $0.5m$ apart. The charges on the two rings are 2$\mu$C and 4$\mu$C, respectively. The work done in transferring a charge of 2$\mu$C from the center of $Q$ to that of $P$ is :

• A. $1.28J$
• B. $0.289J$
• C. $0.144J$
• D. $2.24J$

Answer 1

The correct option is B $0.289J$

Potential at the centre of the ring is given by $V_c=\frac{Kq}{R}$
where $R=0.1m$ is the radius of ring.
Potential at a distance $r=0.5m$ from the ring is given by $V_r=\frac{Kq}{\sqrt{R^2+r^2}}$
Thus potential at centre of P, $V_P=\frac{K{q}_p}{R}+\frac{K{q}_Q}{\sqrt{R^2+r^2}}$
$\therefore$ $V_P=\frac{(9\times {10}^9)(2\times {10}^{-6})}{0.1}+\frac{(9\times {10}^9)(4\times {10}^{-6})}{\sqrt{(0.1{)}^2+(0.5{)}^2}}=0.251\times {10}^{6}V$
So, potential at centre of Q, $V_Q=\frac{K{q}_Q}{R}+\frac{K{q}_P}{\sqrt{R^2+r^2}}$
$\therefore$ $V_Q=\frac{(9\times {10}^9)(4\times {10}^{-6})}{0.1}+\frac{(9\times {10}^9)(2\times {10}^{-6})}{\sqrt{(0.1{)}^2+(0.5{)}^2}}=0.395\times {10}^{6}V$
Potential difference $\Delta V=0.395-0.251=0.1447\times {10}^{6}V$
THus work done $W=q\Delta V=2\times {10}^{-6}\times 0.1447\times {10}^6=0.289J$