A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}+\frac{{{Q}_{2}}}{{{R}_{2}}} \right)\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}-\frac{{{Q}_{2}}}{{{R}_{2}}} \right)\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}} \right)\]
D) 0
Correct Answer: A
Solution :
Potential at the surface of the inner sphere due to the charges of the outer sphere \[={{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{2}}}{{{R}_{2}}}\] Potential at the surface of the inner sphere due to its own charges \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{Q}_{1}}}{{{R}_{1}}}\] .'. Net potential on the surface of the inner sphere \[=V={{V}_{1}}+{{V}_{2}}\] or, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{{{Q}_{1}}}{{{R}_{1}}}+\frac{{{Q}_{2}}}{{{R}_{2}}} \right)\] Hence, the correction option is [a].
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