A) \[\frac{Q}{4\pi {{\varepsilon }_{0}}r}\]for points outside and \[\frac{Q}{4\pi {{\varepsilon }_{0}}R}\]points inside the shell
B) \[\frac{Q}{4\pi {{\varepsilon }_{0}}r}\]for both points inside and outside the shell
C) zero for points outside and\[\frac{Q}{4\pi {{\varepsilon }_{0}}r}\]for points inside the shell
D) zero for both points inside and outside the shell
Correct Answer: A
Solution :
If charge on a conducting sphere of radius R is Q then potential outside the sphere, \[{{V}_{out}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{r}\] At the surface of sphere \[{{V}_{s}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Q}{R}\]
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