question_answer

The concentric spheres of radii R and r have similar charges with equal surface densities (\[\sigma \]) what is the electric potential at their common centre?

A) \[\frac{\sigma }{{{\varepsilon }_{0}}}(R-r)\]                        

B) \[\frac{\sigma }{{{\varepsilon }_{0}}}(R+r)\]

C) \[\frac{R\sigma }{{{\varepsilon }_{0}}}\]                                               

D) \[\frac{\sigma }{{{\varepsilon }_{0}}}\]

Correct Answer: B

Solution :

Key Idea: We have given, \[\sigma =\frac{Q}{4\pi {{R}^{2}}}=\frac{q}{4\pi {{r}^{2}}}\] The surface of sphere is spherical and distribution of charge is uniform. Therefore at every point on the surface the potential will be same. Hence, lines of force emerging will be. perpendicular, to the surface . The electric potential will be sum of potential due to the two surfaces. \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}\] and \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] \[\therefore \]   \[V={{V}_{1}}+{{V}_{2}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}\] \[=\frac{1}{{{\varepsilon }_{0}}}\left[ \frac{Q\times R}{4\pi {{R}^{2}}}+\frac{q\times r}{4\pi {{r}^{2}}} \right]\] Given,      \[\frac{Q}{4\pi {{R}^{2}}}=\frac{q}{4\pi {{r}^{2}}}=\sigma \] \[\therefore \]           \[V=\frac{\sigma }{{{\varepsilon }_{0}}}(r+R)\]