A metal sphere $A$ of radius $r_{1}$ charged to a potential $ϕ_{1}$ is enveloped by a thin walled conducting spherical shell $B$ of radius $r_{2}$ . Then potential $ϕ_{2}$ of the sphere $A$ after it is connected to the shell $B$ by a thin conducting wire will be:

ϕ_{1} (\frac{r_{1}}{r_{2}})

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ϕ_{1} (\frac{r_{2}}{r_{1}})

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ϕ_{1} (1 - \frac{r_{2}}{r_{1}})

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ϕ_{1} (\frac{r_{1} r_{2}}{r_{1} + r_{2}})

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Solution

The correct option is A $ϕ_{1} (\frac{r_{1}}{r_{2}})$
Initially potential at surface of $A$ = $ϕ_{1}$

Let charge on $^{'} A^{'} = q_{0}$

So, $\frac{k q_{0}}{r_{1}} = ϕ_{1} . . . . . (1)$

$\Rightarrow q_{0} = \frac{ϕ_{1} r_{1}}{k}$

When it is connected to enveloped shell $B$ by a conducting wire.
All the charge $q_{0}$ will flow on outer surface in order to make surface of $A$ & $B$ equipotential.

So, finally charge at surface $B = q_{0}$
Charge at surface $A = 0$ .

Now, surface potential at $A$ will be due to charge present at surface $B$ .

So,

$ϕ_{2} = V_{A}^{f i n a l} = V_{i n s i d e f o r B} = \frac{k q_{0}}{r_{2}}$

Substituting the value of $q_{0}$ ,

$ϕ_{2} = \frac{k}{r_{2}} \frac{ϕ_{1} r_{1}}{k} = \frac{ϕ_{1} r_{1}}{r_{2}}$

Hence, option (a) is correct answer.

Why this quesstion?
This problem gives insight of charge distribution when we connect spherical conducting surfaces.

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A metal sphere A of radius r1 charged to a potential ϕ1 is enveloped by a thin walled conducting spherical shell B of radius r2. Then potential ϕ2 of the sphere A after it is connected to the shell B by a thin conducting wire will be:

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