question_answer

A conducting sphere of radius R carries a charge Q. It is enclosed by another concentric spherical shell of radius 2R. Charge from the inner sphere is transferred in infinitesimally small installments to the outer sphere. The work done in transferring the entire charge from the inner sphere to the outer one will be

A) \[-\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}R}\]

B) \[-\frac{{{Q}^{2}}}{2\sqrt{2}\pi {{\varepsilon }_{0}}R}\]

C) \[-\frac{{{Q}^{2}}}{16\pi {{\varepsilon }_{0}}R}\]       

D) \[-\frac{{{Q}^{2}}}{2\sqrt{2}\pi {{\varepsilon }_{0}}R}\]

Correct Answer: C

Solution :

[c] The electric field in initial and final configuration is as shown below: In initial configuration there is electric field in all region\[r>R\]. The field varies as \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}}\]. In final configuration there is no field in the region\[R<r<2R\]. Elsewhere the field is similar to that in initial configuration. Thus initial configuration has more energy stored in electric field given by \[\Delta =E=\int\limits_{R}^{2R}{\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}dV=\frac{1}{2}{{\varepsilon }_{0}}\int\limits_{R}^{2R}{\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}} \right)4\pi {{r}^{2}}dr}}\] \[=\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}}\int\limits_{R}^{2R}{\frac{dr}{{{r}^{2}}}}=-\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}}\left[ \frac{1}{2R}-\frac{1}{R} \right]=\frac{{{Q}^{2}}}{16\pi {{\varepsilon }_{0}}R}\] \[{{W}_{ext}}={{U}_{f}}-{{U}_{i}}=-\frac{{{Q}^{2}}}{16\pi {{\varepsilon }_{0}}R}\]