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Electric Potential Due to a Solid Sphere of Charge

Two metal sph...

Question

Two metal spheres are very far apart but are connected by a thin wire as shown-

If the combined charge of both the sphere is $Q$ , then the common potential will be

A

\frac{k Q}{r_{1} - r_{2}}

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B

\frac{- k Q}{r_{1} + r_{2}}

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C

\frac{k Q}{r_{1} + r_{2}}

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D

\frac{- k Q}{r_{1} r_{2}}

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Solution

The correct option is C $\frac{k Q}{r_{1} + r_{2}}$

Let the charge and potential of sphere having radius $r_{1}$ and $r_{2}$ is $q_{1}, V_{1}$ and $q_{2}, V_{2}$ respectively.

As the metal spheres are connected by a thin wire, the potential on them will become same.

$\Rightarrow V_{1} = V_{2}$
$\Rightarrow \frac{k q_{1}}{r_{1}} = \frac{k q_{2}}{r_{2}}$
$\Rightarrow \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$
$\Rightarrow \frac{q_{1}}{q_{2}} = \frac{r_{1}}{r_{2}} . . . (i)$

Also $q_{1} + q_{2} = Q . . . (i i)$

Dividing equation $(i i)$ by $q_{2}$
$\Rightarrow \frac{q_{1}}{q_{2}} + 1 = \frac{Q}{q_{2}}$

From equation $(i)$
$\Rightarrow \frac{r_{1}}{r_{2}} + 1 = \frac{Q}{q_{2}}$
$\Rightarrow \frac{r_{1} + r_{2}}{r_{2}} = \frac{Q}{q_{2}}$
$\Rightarrow q_{2} = \frac{Q r_{2}}{r_{1} + r_{2}} . . . (i i i)$

Using $(i i i)$ and $(i)$
$q_{1} = q_{2} \times \frac{r_{1}}{r_{2}}$
$\Rightarrow q_{1} = \frac{Q r_{1}}{r_{1} + r_{2}} . . . (i v)$

The common potential is
$V = V_{1} = V_{2}$

So $V = V_{1} = \frac{k q_{1}}{r_{1}} = \frac{k Q}{r_{1} + r_{2}} (From (i v))$

Hence, option (c) is correct.

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