A) \[\frac{\sigma }{{{\varepsilon }_{0}}}(R+r)\]
B) \[\frac{\sigma }{{{\varepsilon }_{0}}}(R-r)\]
C) \[\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{1}{R}+\frac{1}{r} \right)\]
D) \[\frac{\sigma }{{{\varepsilon }_{0}}}\left( \frac{R}{r} \right)\]
Correct Answer: A
Solution :
: The situation is as shown in the figure. According to the problem \[{{\sigma }_{1}}={{\sigma }_{2}}=\sigma \] \[\therefore \] \[\frac{{{q}_{1}}}{4\pi {{R}^{2}}}=\sigma \] or \[{{q}_{1}}=\sigma 4\pi {{R}^{2}}\] and \[\frac{{{q}_{2}}}{4\pi {{r}^{2}}}=\sigma \] or \[{{q}_{2}}=\sigma 4\pi {{r}^{2}}\] Electric potential at the common centre O is \[V={{V}_{1}}+{{V}_{2}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}}{R}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{2}}}{r}=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{1}}}{R}+\frac{{{q}_{2}}}{r} \right]\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{\sigma 4\pi {{R}^{2}}}{R}+\frac{\sigma 4\pi {{R}^{2}}}{r} \right]\] \[=\frac{4\pi \sigma }{4\pi \in }[(R+r)]=\frac{\sigma }{{{\in }_{0}}}(R+r)\]You need to login to perform this action.
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