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A charge Q is distributed over the surfaces of two concentric hollow spheres of radii r and R (R>>r), such that their surface charge densities are equal. Derive the expression for the potential at the common centre.

Answers (1)

Let us consider the two concentric hollow spheres of radius r and R respectively and having charges $q_{1}and q_{2}$ .

The total charge (Q) = $q_{1}+ q_{2}$

We know that; $\sigma = \frac{q_{1}}{A}$

where, $\sigma$ = surface charge density and area of sphere (A) = $4\pi r^{2}$

then,

$\sigma = \frac{q_{1}}{4\pi r^{2}}or q_{1}= \sigma \times 4\pi r^{2}$

similarly for charge $q_{2}$ ,

$\sigma = \frac{q_{2}}{4\pi r^{2}}or q_{2}= \sigma \times 4\pi R^{2}$

Hence, potential (V) at the common centre;

$V= \frac{1}{4\pi \epsilon _{0}}\left [ \frac{q_{1}}{r}+\frac{q_{2}}{R} \right ]$

$V= \frac{1}{4\pi \epsilon _{0}} \left [ \frac{\sigma \times 4\pi r^{2}}{r}+\frac{\sigma \times 4\pi R^{2}}{R} \right ]$

$V= \frac{1}{ \epsilon _{0}} \sigma \left [ r+R \right ]$

$V= \frac{\sigma \left ( r+R \right )}{\epsilon _{0}}$

and, we know that; $\sigma = \frac{Q}{A}= \frac{Q}{4\pi (r+R)^{2}}$ (for two shells)

then,

$V= \frac{(r+R)}{\epsilon _{0}} \times \frac{Q}{4\pi (r^{2}+R^{2})}$

Hence,

$V= \frac{1}{4\pi \epsilon _{0}}\frac{Q(r+R)}{(r^{2}+R^{2})}$

Posted by

Safeer PP

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