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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)

Detailed Solution
$W=q\Delta V=q(V_B-V_A)=1\left(\frac{kQ}{3R}-\frac{kQ}{2R}\right)=-\frac{kQ}{6R}$. With $Q=\rho\times\frac{4}{3}\pi R^3$: $|W|=\frac{1}{4\pi\varepsilon_0}\cdot\frac{\rho\frac{4}{3}\pi R^3}{6R}=\frac{\rho R^2}{18\varepsilon_0}$, so n = 18.
