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Electric Potential Due to a Solid Sphere of Charge

Two metal sph...

Question

Two metal spheres (radii $r_{1}, r_{2}$ with $r_{1} < r_{2}$ ) are very far apart but are connected by a thin wire. If their combined charge is $Q$ , then what is their common Potential?

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

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

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

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

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Solution

The correct option is A $\frac{k Q}{(r_{1} + r_{2})}$
Let $q_{1}$ and $q_{2}$ be the charges on the spheres after connecting them. Let $V_{1}$ and $V_{2}$ be the potential on the spheres of radius $r_{1}$ and $r_{2}$ respectively.
After connecting the spheres from wire they will attain some common potential due to flow of charges through the wire.
$∴ V_{1} = V_{2} = V$
$⟹ \frac{k q_{1}}{r_{1}} = \frac{k q_{2}}{r_{2}}$
$⟹ \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ..............(i)
Also $Q = q_{1} + q_{2}$ .........(ii)
Using (i) and (ii) we will get
$q_{2} = \frac{Q r_{2}}{r_{1} + r_{2}}$ ........(iii)
$∴ V = \frac{k q_{2}}{r_{2}} = \frac{k Q}{r_{1} + r_{2}}$

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