A small conducting sphere of radius r is lying concentrically inside a bigger hollow conducting sphe

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2. Electric Potential and Capacitance

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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

$\frac{1}{{4\pi { \in _0}}}\left( {\frac{q}{r} - \frac{Q}{R}} \right)$

$\frac{1}{{4\pi { \in _0}}}\left( {\frac{Q}{R} + \frac{q}{r}} \right)$

$\frac{1}{{4\pi { \in _0}}}\left( {\frac{q}{r} - \frac{q}{R}} \right)$

$\frac{1}{{4\pi { \in _0}}}\left( {\frac{q}{R} - \frac{Q}{r}} \right)$

Solution

The potential of $\mathrm{Q}$ at the surface

${\rm{A}} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\rm{Q}}}{{\rm{R}}};$ The potential of $\mathrm{q}$ at the surface

$A = \frac{1}{{4\pi { \in _0}}} \cdot \frac{q}{R}$

The potential at $\mathrm{B}$ is due to $\mathrm{Q}$ inside $ = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\rm{Q}}}{{\rm{R}}}$

The potential at $\mathrm{B}$ due to ${\rm{q}} = \frac{1}{{4\pi { \in _0}}} \cdot \frac{{\rm{q}}}{{\rm{r}}}$

$\therefore $ Potential at ${\rm{A}},{{\rm{V}}_{\rm{A}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{\rm{Q}}}{{\rm{R}}} + \frac{{\rm{q}}}{{\rm{R}}}} \right)$

Potential at ${\rm{B}},{{\rm{V}}_{\rm{B}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{\rm{Q}}}{{\rm{R}}} + \frac{{\rm{q}}}{{\rm{r}}}} \right)$

$\therefore {{\rm{V}}_{\rm{B}}} – {{\rm{V}}_{\rm{A}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{\rm{q}}}{{\rm{r}}} – \frac{{\rm{q}}}{{\rm{R}}}} \right)$

Standard 12

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Solution

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