A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius $R$, having uniform positive charge density $\rho$, as shown in the figure. The initial and final positions of the charge are marked by $A$ and $B$ at distances $2R$ and $3R$ respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is $\dfrac{\rho R^2}{n\epsilon_0}$. The value of $n$ is :
($\epsilon_0$ is the permittivity of vacuum)
Answer: (D) $18$
Concept. The field is conservative, so the work depends only on the end points; outside the sphere $V = \dfrac{\rho R^3}{3\epsilon_0r}$.
Step 1. Potentials. $V_A = \dfrac{\rho R^2}{6\epsilon_0}$, $V_B = \dfrac{\rho R^2}{9\epsilon_0}$.
Step 2. Work. $|W| = \dfrac{\rho R^2}{\epsilon_0}\left(\dfrac16 - \dfrac19\right) = \dfrac{\rho R^2}{18\epsilon_0}$: $n = 18$.
Answer. $18$ (Option D).
Solution by Sreeraj P, M.Sc Physics