Question

Two identical rings of radii 0.1 m are placed co-axially at a distance 0.5 m apart. The charges on the rings are 2$\mu C$ and 4$\mu C$ respectively. The work done in transferring 5$\mu C$ charge from the centre of one ring to that of the other will be nearest to :

• A. 0.50 J
• B. 0.75 J
• C. 1.00 J
• D. 1.50 J

Answer 1

Answer: (B)

0.75 J

We know that the potential at a point on the axis of the ring due to the ring is $V=\frac{1}{4\pi {ϵ}_0}\frac{q}{\sqrt{x^2+R^2}}$
When charge $Q=5\mu C$ at the center of 1st ring, the potential energy is
$U_i=Q($ potential due to 1st ring $+$ potential due to 2nd ring $)=Q[\frac{1}{4\pi {ϵ}_0}\frac{q_1}{\sqrt{0^2+R^2}}+\frac{1}{4\pi {ϵ}_0}\frac{q_2}{\sqrt{x^2+R^2}}]$
When charge $Q=5\mu C$ at the center of 2nd ring, the potential energy is
$U_f=Q($ potential due to 2 nd ring $+$ potential due to 1st ring $)=Q[\frac{1}{4\pi {ϵ}_0}\frac{q_2}{\sqrt{0^2+R^2}}+\frac{1}{4\pi {ϵ}_0}\frac{q_1}{\sqrt{x^2+R^2}}]$
Work done , $W=\Delta U=U_f-U_i=\frac{Q}{4\pi {ϵ}_0}[\frac{q_2}{R}-\frac{q_1}{R}+\frac{q_1}{\sqrt{x^2+R^2}}-\frac{q_2}{\sqrt{x^2+R^2}}]$
or $W=(9\times {10}^9)\times 5\times {10}^{-6}[\frac{4\times {10}^{-6}}{0.1}-\frac{2\times {10}^{-6}}{0.1}+\frac{2\times {10}^{-6}}{\sqrt{{0.5}^2+{0.1}^2}}-\frac{4\times {10}^{-6}}{\sqrt{{0.5}^2+{0.1}^2}}]$
$=0.72J$