A) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(R-r)Q}{({{R}^{2}}+{{r}^{2}})}\]
B) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(R+r)Q}{2({{R}^{2}}+{{r}^{2}})}\]
C) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(R+r)Q}{({{R}^{2}}+{{r}^{2}})}\]
D) \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(R-r)Q}{2\,({{R}^{2}}+{{r}^{2}})}\]
Correct Answer: C
Solution :
[c] 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, \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{1}}}{r}+\frac{{{q}_{2}}}{R} \right]\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{Qr}{{{R}^{2}}+{{r}^{2}}}+\frac{QR}{{{R}^{2}}+{{r}^{2}}} \right]\] \[=\frac{Q(R+r)}{{{R}^{2}}+{{r}^{2}}}\frac{1}{4\pi {{\varepsilon }_{0}}}\]
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