Question:

A conducting sphere of radius R carrying charge +Q is connected to an uncharged conducting sphere of radius 2R. The charge that flows between them is

Updated On: Jun 23, 2023
  • $Q / 2$
  • $Q / 3$
  • $Q / 4$
  • $2Q / 3$
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The Correct Option is D

Solution and Explanation

Charge will not flow if both the sphere get the same potential i.e.
$V_{1} =V_{2} $ or, $\frac{Q_{1}}{C_{1}} = \frac{Q_{2}}{C_{2}}$
Here, $ C_{1} = 4 \pi \varepsilon_{0} R, C_{2} = 4 \pi \varepsilon_{0}\left(2R\right) $
$ Q_{1 } +Q_{2} = Q $
$\therefore \, \frac{Q_{1}}{4\pi\in_{0}R} = \frac{Q_{2}}{4\pi\in _{0}\left(2R\right)} or ,Q_{1} =\frac{Q_{2}}{2} $
From equation (i),
$\frac{Q_{2}}{2} + Q_{2} = Q or \frac{3}{2}Q_{2} = Q $
or , $ Q_{2} = \frac{2}{3} Q$
And $Q_{1} = \frac{1}{3}Q $
Flow of charge $= Q -Q_{1} = Q_{2} =\frac{2}{3} Q$
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