Q. 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) $\frac{1}{4\pi{\varepsilon r}_{0}}\left( \frac{q}{R} - \frac{Q}{r} \right)$
Explanation

(c) The potential of $Q$ at the surface
$A = \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{R}$; The potential of $q$
At the surface $A = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{R}$
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The potential at $B$ is due to $Q$
Inside $= \frac{1}{4\pi\varepsilon_{0}}.\frac{Q}{R}$
The potential at $B$ due to $q = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{r}$
$\therefore$ Potential at $A,\ V_{A} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{Q}{R} + \frac{q}{R} \right)$
Potential at $B,\ V_{B} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{Q}{R} + \frac{q}{R} \right)$
$\therefore V_{B} - V_{A} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q}{r} - \frac{q}{R} \right)$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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