Class 12 NEET Physics Electrostatics Questions Part 2

Concept:Work done on a charge moving slowly between two points equals the charge times the potential difference.For points outside a uniformly charged sphere, the potential is $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$.Explanation:The point charge $q = 1$ moves from $A$ at distance $2R$ to $B$ at distance $3R$ from the centre.Both points are outside the sphere.Potential at $A$: $V_A = \frac{1}{4\pi\varepsilon_0} \frac{Q}{2R}$.At $B$: $V_B = \frac{1}{4\pi\varepsilon_0} \frac{Q}{3R}$.Work done $W = q(V_B - V_A) = \frac{Q}{4\pi\varepsilon_0} \left( \frac{1}{3R} - \frac{1}{2R} \right) = \frac{Q}{4\pi\varepsilon_0} \left( -\frac{1}{6R} \right)$.Magnitude $|W| = \frac{Q}{24\pi\varepsilon_0 R}$.Total charge $Q = \rho \cdot \frac{4}{3}\pi R^3$.Substitute: $|W| = \frac{1}{4\pi\varepsilon_0} \cdot \frac{\rho \frac{4}{3}\pi R^3}{6R} = \frac{\rho R^2}{18\varepsilon_0}$.Thus $n = 18$.Answer:$\boxed{18}$ (Option A)