A) \[\frac{Q({{R}^{2}}+{{r}^{2}})}{4\pi {{\varepsilon }_{0}}(R+r)}\]
B) \[\frac{QR}{R+r}\]
C) Zero
D) \[\frac{Q(R+r)}{4\pi {{\varepsilon }_{0}}({{R}^{2}}+{{r}^{2}})}\]
Correct Answer: D
Solution :
\[{{q}_{1}}+{{q}_{2}}=Q\] and \[\frac{{{q}_{1}}}{4\pi {{r}^{2}}}=\frac{{{q}_{2}}}{4\pi {{R}^{2}}}\] (given) \[{{q}_{1}}=\frac{Q{{r}^{2}}}{{{R}^{2}}+{{r}^{2}}}\] and \[{{q}_{2}}=\frac{Q{{R}^{2}}}{{{R}^{2}}+{{r}^{2}}}\] Potential at common centre \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Q{{r}^{2}}}{({{R}^{2}}+{{r}^{2}})r}+\frac{Q{{R}^{2}}}{({{R}^{2}}+{{r}^{2}})R} \right]=\frac{Q(R+r)}{4\pi {{\varepsilon }_{0}}({{R}^{2}}+{{r}^{2}})}\]You need to login to perform this action.
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