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 $\mathcal{Q}$ and $q\ (\mathcal{Q} q)\ $and are insulated from each other. The potential difference between the spheres will be

Question

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 $\mathcal{Q}$ and $q\ (\mathcal{Q} > q)\ $and are insulated from each other. The potential difference between the spheres will be

Dr. Abubacker · Accepted Answer

(a) The potential $V_{1}$of smaller sphere is given by
$V_{1} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{r} + \frac{1}{4\pi\varepsilon_{0}}.\frac{\mathcal{Q}}{R}\ \ \ \ldots\ldots(i)$
The potential $V_{2}$of bigger sphere is given by
$V_{2} = \frac{1}{4\pi\varepsilon_{0}}.\frac{\mathcal{Q}}{R} + \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{R}$
So, the potential difference between the plates
$V = V_{1} - V_{2}$
Or
$V = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{r} + \frac{1}{4\pi\varepsilon_{0}}.\frac{\mathcal{Q}}{R} - \frac{1}{4\pi\varepsilon_{0}}.\frac{\mathcal{Q}}{R} - \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{R}$
$= \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r} - \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{R} = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{q}{r} - \frac{q}{R} \right)$
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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 $\mathcal{Q}$ and $q\ (\mathcal{Q} > q)\ $and are insulated from each other. The potential difference between the spheres will be

Explanation

Reviewed by Dr. Abubacker

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