Question
A non-conducting solid sphere, of radius R, with its center at A, has a spherical cavity of diameter R with center at B as shown. There is no charge in the cavity while the solid part has a uniform volume charge density ρ. Electric potential at the center of the sphere (at point A) is V = kρR2/(12ε0) (in SI units) where the value of k is :-
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Solution
[Key Concept: A cavity is a superposition of the full sphere and a sphere of opposite charge density filling the cavity.]
$$Q_{\text{total~}} = \rho \times \frac{4}{3}\pi R^{3}$$
$$\begin{matrix} & Q_{\text{removed~}} = \rho \times \frac{4}{3}\pi\left( \frac{R}{2} \right)^{3} \\ & \begin{matrix} V_{A} & \ = \frac{3}{2} \times \frac{1}{4\pi\varepsilon_{0}}\frac{Q_{T}}{R} - \frac{1}{4\pi\varepsilon_{0}}\frac{Q_{R}}{R/2} \\ & \ = \frac{1}{4\pi\varepsilon_{0}R}\left\lbrack \frac{3}{2}\rho \times \frac{4}{3}\pi R^{3} - 2\rho \times \frac{4}{3}\pi\frac{R^{3}}{8} \right\rbrack \\ & \ = \frac{5\rho R^{2}}{12\varepsilon_{0}} \\ K & \ = 5 \end{matrix} \end{matrix}$$
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