A unit positive point charge is taken slowly | vedclass

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(D) The work done by an external agent in moving a charge $q$ slowly from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.Since the charge is a

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$18$

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(D) The work done by an external agent in moving a charge $q$ slowly from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.Since the charge is a unit positive point charge, $q = 1$, so $W = V_B - V_A$.The potential at a distance $r$ from the center of a uniformly charged dielectric sphere of radius $R$ is:For $r \ge R$, $V(r) = \frac{Q}{4\pi\epsilon_0 r}$, where $Q = \rho \cdot \frac{4}{3}\pi R^3$.Thus, $V(r) = \frac{\rho \cdot \frac{4}{3}\pi R^3}{4\pi\epsilon_0 r} = \frac{\rho R^3}{3\epsilon_0 r}$.Given $r_A = 2R$ and $r_B = 3R$, we have:$V_A = \frac{\rho R^3}{3\epsilon_0 (2R)} = \frac{\rho R^2}{6\epsilon_0}$$V_B = \frac{\rho R^3}{3\epsilon_0 (3R)} = \frac{\rho R^2}{9\epsilon_0}$The work done is $W = |V_B - V_A| = |\frac{\rho R^2}{9\epsilon_0} - \frac{\rho R^2}{6\epsilon_0}| = |\frac{2\rho R^2 - 3\rho R^2}{18\epsilon_0}| = \frac{\rho R^2}{18\epsilon_0}$.Comparing this with $\frac{\rho R^2}{n \epsilon_0}$, we get $n = 18$.

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