A) \[\frac{Q(R+r)}{4\pi {{\varepsilon }_{0}}({{R}^{2}}+{{r}^{2}})}\]
B) \[\frac{Q({{R}^{2}}+{{r}^{2}})}{4\pi {{\varepsilon }_{0}}(R+r)}\]
C) \[\frac{Q}{(R+r)}\]
D) zero
Correct Answer: A
Solution :
We know that the surface charge densities \[(\sigma )\] \[=\frac{Q}{4\pi ({{r}^{2}}+{{R}^{2}})}\] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\left[ \frac{\sigma \times 4\pi {{r}^{2}}}{r}+\frac{\sigma \times 4\pi {{R}^{2}}}{R} \right]\] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\left[ \frac{{{r}^{2}}}{r}+\frac{{{R}^{2}}}{R} \right]\,=\frac{\sigma }{{{\varepsilon }_{0}}}(r+R)\] And \[V=\frac{1}{{{\varepsilon }_{0}}.4\pi ({{r}^{2}}+{{R}^{2}})}.(r+R)\] \[=\frac{Q}{4\pi {{\varepsilon }_{0}}}\,.\frac{(R+r)}{({{R}^{2}}+{{r}^{2}})}\]
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