Question
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 distance $2R$ and $3R$ respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is $\frac{\rho R^{2}}{n\varepsilon_{0}}$. The value of $n$ is : ( $\varepsilon_{0}$ is the permittivity of vacuum)
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Solution
Sol. [Key Concept: Work = q×potential (charge moved through a charged sphere)]
Work done $W = q\left( V_{B} - V_{A} \right)$
$$\begin{matrix} & \ = q\left\lbrack \frac{KQ}{3R} - \frac{KQ}{2R} \right\rbrack = \frac{Kq}{R}\left\lbrack \frac{- Q}{6} \right\rbrack \\ & \ |W| = \frac{KqQ}{6R} = \frac{1}{4\pi q_{0}} \times \frac{q}{R} \times \frac{1}{6} \times \rho \times \frac{4}{3}\pi R^{3} \\ & \ |W| = \frac{\rho R^{2}}{18\varepsilon_{0}}\lbrack\because\ q = 1\rbrack \\ & \ \therefore\ n = 18 \end{matrix}$$
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