question_answer

A small conducting sphere of radius r is lying concentrically inside a bigger hollow conducting sphere of radius R. The bigger and smaller spheres are charged with Q and \[q\,(Q>q)\] and are insulated from each other. The  potential difference between the spheres will be

A)  \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{q}{r}-\frac{q}{R} \right)\]              

B)  \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{q}{R}-\frac{Q}{r} \right)\]

C)  \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{q}{r}-\frac{Q}{R} \right)\]              

D)  \[\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{Q}{R}+\frac{q}{r} \right)\]

Correct Answer: A

Solution :

The potential V[ of smaller sphere is given by                 \[{{V}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}\]                                ?. (i) The potential \[{{V}_{2}}\] of bigger sphere is given by                 \[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{R}\] So, the potential difference between the plates                 \[V={{V}_{1}}-{{V}_{2}}\] or\[{{V}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{r}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{R}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{R}\]                 \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q}{r}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{R}\]                 \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left( \frac{q}{r}-\frac{q}{R} \right)\]