question_answer

Charge q is distributed uniformly throughout a non-conducting spherical volume of radius R Energy of charge distribution is (k = dielectric constant)

A) \[\frac{8{{q}^{2}}}{4\pi {{\in }_{0}}kR}\]                   

B) \[\frac{{{q}^{2}}}{8\pi {{\in }_{0}}kR}\]

C) \[\frac{30{{q}^{2}}}{\pi {{\in }_{0}}kR}\]                   

D) \[\frac{3{{q}^{2}}}{20\pi {{\in }_{0}}kR}\]

Correct Answer: D

Solution :

Consider a sphere of radius z charge \[\rho \cdot V,\] where p = charge density \[(c{{m}^{-\,3}})\]and V = volume \[({{m}^{3}}).\]

Potential of sphere \[=\frac{kq}{r}=\frac{\frac{4}{3}\pi {{z}^{3}}\cdot \rho }{4\pi \in z}=\frac{\rho {{z}^{2}}}{3\in }\]

To increase charge layer by thickness dz.

Work done, \[dW=qV=\frac{{{z}^{2}}\rho }{3\in }4\pi {{z}^{2}}dz.\rho \]

or         \[dW=qV=\frac{{{z}^{2}}\rho }{3\in }\cdot {{z}^{4}}\cdot dz\]

Work done in assembling complete sphere\[\int_{0}^{R}{dW=\int_{0}^{R}{\frac{4\pi {{\rho }^{2}}}{3\in }}}{{z}^{4}}\cdot dz\]

\[=\frac{4\pi {{\sigma }^{2}}}{3\times 5\times \in }\cdot ({{z}^{5}})/_{0}^{R}=\frac{4\pi {{\sigma }^{2}}{{R}^{5}}}{3\times 5\in }\]

\[=\frac{4\pi \times {{\left( \frac{3q}{4\pi {{R}^{3}}} \right)}^{2}}\times {{R}^{5}}}{3\times 5\times \in }\] \[=\frac{3{{q}^{2}}}{20\pi \in R}=\frac{3{{q}^{2}}}{20\pi {{\in }_{0}}kR}\]