Question

An electric dipole is placed at a distance x from center $O$ on the axis of a charged ring of radius $R$ and charge $Q$ uniformly distributed over it. What is the work done in rotating the dipole through ${180}^{\circ }$?

• A. $\frac{aqQx}{2\pi {\epsilon }_0(R^2+x^2{)}^{\frac{3}{2}}}$
• B. $\frac{aqQx}{\pi {\epsilon }_0(R^2+x^2{)}^{\frac{3}{2}}}$
• C. $aqQx(R^2+x^2)$
• D. zero

Answer 1

The correct option is B $\frac{aqQx}{\pi {\epsilon }_0(R^2+x^2{)}^{\frac{3}{2}}}$

Here dipole is placed in the electric field of the ring $$E_x=\frac{kQx}{(R^2+x^2{)}^{\frac{3}{2}}}$$and dipole moment $p=q\times \left(2a\right)$ , the initial angle between $\overrightarrow{p}$ and field is ${\theta }_i=0^0$ and when rotated by ${180}^{\circ }$ is ${\theta }_f={180}^{\circ }$

Workdone in rotating from ${\theta }_i\to {\theta }_f$ is given by

$W=pE(\cos {\theta }_i-\cos {\theta }_f)$

$\Rightarrow W=p{E}_x(\cos 0^0-\cos {180}^0)$

$\Rightarrow W=q\left(2a\right)\times \frac{kQx}{(R^2+x^2{)}^{\frac{3}{2}}}\times (1+1)$

$\therefore W=\frac{4aqkQx}{(R^2+x^2{)}^{\frac{3}{2}}}\Rightarrow W=\frac{1}{\pi {\epsilon }_0}\frac{aqQx}{(R^2+x^2{)}^{\frac{3}{2}}}$

Hence, option (b) is the correct answer.