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Two particles having positive charges +Q and +2Q are fixed at equal distance x from centre of a conducting sphere having zero net charge and radius r as shown. Initially the switch S is open. After the switch S is closed, the net charge flowing out of sphere is $\frac{(5 - P) Qr}{x}$ find P

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Detailed Solution

After earthing

Potential at centre of sphere = 0

$\frac{KQ}{x} + \frac{K 2 Q}{x} + \frac{{KQ}_{s}}{r} = 0$

$\Rightarrow Q_{s} = \frac{- 3 Qr}{x}$

So the charge flowing out is $\frac{3 Qr}{x}$ .

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