A) \[\frac{Q({{R}^{2}}+{{r}^{2}})}{4\pi {{\varepsilon }_{0}}(R+r)}\]
B) \[\frac{Q}{R+r}\]
C) Zero
D) \[\frac{Q(R+r)}{4\pi {{\varepsilon }_{0}}({{R}^{2}}+{{r}^{2}})}\]
Correct Answer: D
Solution :
Let the charges on spheres of radii r and R be \[{{q}_{1}}\] and\[{{q}_{2}}\]respectively. Then, \[Q={{q}_{1}}+{{q}_{2}}.\]The potential at the centre will be: \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}(\frac{{{q}_{1}}}{r}+\frac{{{q}_{2}}}{R})\] As surface densities are equal, hence \[\sigma =\frac{{{q}_{1}}}{4\pi {{r}^{2}}}\frac{{{q}_{2}}}{4\pi {{R}^{2}}}\]or\[{{q}_{2}}={{q}_{1}}\left( \frac{{{R}^{2}}}{{{r}^{2}}} \right)\] \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}(\frac{{{q}_{1}}}{r}+{{q}_{1}}\frac{R}{{{r}^{2}}})=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{{{q}_{1}}}{{{r}^{2}}}(r+R)\] Now, \[{{q}_{1}}+{{q}_{2}}+Q\]or\[{{q}_{1}}+{{q}_{1}}\frac{{{R}^{2}}}{{{r}^{2}}}=Q\] \[\Rightarrow \]\[\frac{{{q}_{1}}}{{{r}^{2}}}=\frac{Q}{{{R}^{2}}+{{r}^{2}}};V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\left( \frac{R+r}{{{R}^{2}}+{{r}^{2}}} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
