A) \[\frac{qRC}{(4\pi {{\varepsilon }_{0}}R+C)}\left( \frac{2}{x}+\frac{1}{y} \right)\]
B) \[\frac{qRC}{(4\pi {{\varepsilon }_{0}}R+C)}\left( \frac{1}{x}+\frac{2}{y} \right)\]
C) \[\frac{2qRC}{(4\pi {{\varepsilon }_{0}}R+C)}\left( \frac{1}{x}+\frac{1}{y} \right)\]
D) \[\frac{qRC}{(4\pi {{\varepsilon }_{0}}R+C)}\left( \frac{1}{x}+\frac{1}{y} \right)\]
Correct Answer: D
Solution :
[d] Let a charge Q be induced on capacitor plate connected to the ball. Induced charge on the ball is \[-Q\]. \[\therefore \] Potential at the centre of the ball will be \[V=K\frac{q}{x}+K\frac{q}{y}+K\frac{-Q}{R}\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{x}+\frac{1}{y} \right)-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}\] This is the potential of the entire ball and this is the potential of the capacitor plate connected to the sphere. The other plate of the capacitor is at zero potential. \[\therefore \] Potential difference across capacitor plates = V \[V=\frac{Q}{C}=\frac{q}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{x}+\frac{1}{y} \right)-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,Q\left( \frac{1}{C}+\frac{1}{4\pi {{\varepsilon }_{0}}R} \right)=\frac{q}{4\pi {{\varepsilon }_{0}}}\left( \frac{1}{x}+\frac{1}{y} \right)\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,Q=\frac{qRC}{(4\pi {{\varepsilon }_{0}}R+C)}\left( \frac{1}{x}+\frac{1}{y} \right)\]
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