question_answer

A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is     AIEEE  Solved  Paper-2003

A) \[\frac{2Q}{4\pi {{\varepsilon }_{0}}R}\]       

B)                       \[\frac{2Q}{4\pi {{\varepsilon }_{0}}R}-\frac{2q}{4\pi {{\varepsilon }_{0}}R}\]     

C) \[\frac{2Q}{4\pi {{\varepsilon }_{0}}R}+\frac{q}{4\pi {{\varepsilon }_{0}}R}\] 

D)       \[\frac{(q+Q)}{4\pi {{\varepsilon }_{0}}R}\frac{2}{R}\]

Correct Answer: C

Solution :

At P, potential due to shell                     \[{{V}_{1}}=\frac{q}{4\pi \,{{\varepsilon }_{0}}R}\] At P, due to charge                                     \[{{V}_{2}}=\frac{Q}{4\pi {{\varepsilon }_{0}}\frac{R}{2}}\]                 \[=\frac{2Q}{4\pi {{\varepsilon }_{0}}R}\] \[\therefore \] Net potential at P,                     \[V={{V}_{1}}+{{V}_{2}}=\frac{q}{4\pi {{\varepsilon }_{0}}R}+\frac{2Q}{4\pi {{\varepsilon }_{0}}R}\]                     (\[\because \] v is scalar quantity)