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 \[\Delta =E=\int\limits_{R}^{2R}{\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}dV=\frac{1}{2}{{\varepsilon }_{0}}\int\limits_{R}^{2R}{\left( \frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{{{r}^{2}}} \right)4\pi {{r}^{2}}dr}}\] \[=\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}}\int\limits_{R}^{2R}{\frac{dr}{{{r}^{2}}}}=-\frac{{{Q}^{2}}}{8\pi {{\varepsilon }_{0}}}\left[ \frac{1}{2R}-\frac{1}{R} \right]=\frac{{{Q}^{2}}}{16\pi {{\varepsilon }_{0}}R}\] \[{{W}_{ext}}={{U}_{f}}-{{U}_{i}}=-\frac{{{Q}^{2}}}{16\pi {{\varepsilon }_{0}}R}\]You need to login to perform this action.
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