A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( R-r \right)Q}{\left( {{R}^{2}}+{{r}^{2}} \right)}\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( R+r \right)Q}{2\left( {{R}^{2}}+{{r}^{2}} \right)}\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( R+r \right)Q}{\left( {{R}^{2}}+{{r}^{2}} \right)}\]
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\left( R-r \right)Q}{\left( {{R}^{2}}+{{r}^{2}} \right)}\]
Correct Answer: C
Solution :
Let \[{{q}_{1}}\]and \[{{q}_{2}}\]be charge on two spheres of radius 'r' and ?R respectively As,\[{{q}_{1}}+{{q}_{2}}=Q\]and \[{{\sigma }_{1}}={{\sigma }_{2}}\][Surface charge density are equal] \[\therefore \]\[\frac{{{q}_{1}}}{r\pi {{r}^{2}}}=\frac{{{q}_{2}}}{4\pi {{R}^{2}}}\] So,\[{{q}_{1}}=\frac{Q{{r}^{2}}}{{{R}^{2}}+{{r}^{2}}}\]and\[{{q}_{2}}=\frac{Q{{R}^{2}}}{{{R}^{2}}+{{r}^{2}}}\] Now, potential,\[V=\frac{Q{{R}^{2}}}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{1}}}{r}+\frac{{{q}_{2}}}{R} \right]\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Qr}{{{R}^{2}}+{{r}^{2}}}+\frac{QR}{{{R}^{2}}+{{r}^{2}}} \right]\] \[=\frac{Q(R+r)}{{{R}^{2}}+{{r}^{2}}}\frac{1}{4\pi {{\varepsilon }_{0}}}\]
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