A point charge 'q' is placed at the centre of an uncharged conducting shell of inner radius R and outer radius 2R. If the point charge is removed from the centre of the shell, to infinity (without any electric contact with the shell), work done by electric force is:

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$\frac{- q^{2}}{16 π \in_{0} R}$

$\frac{- 3 q^{2}}{8 π \in_{0} R}$

$\frac{3 q^{2}}{8 π \in_{0} R}$

$\frac{q^{2}}{16 π \in_{0} R}$

answer is B.

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Detailed Solution

$U_{in} = U_{self} = (U_{due to outer} - U_{due to inner})$

$= \frac{q^{2}}{16 {πε}_{0} R} - \frac{q^{2}}{8 {πε}_{0} R} = - \frac{q^{2}}{16 {πε}_{0} R}$

And finally the potential energy of the system is zero. $\Rightarrow U_{f} = 0$

$W = - (dU) \Rightarrow W = [0 - (\frac{q^{2}}{16 {πε}_{0} R})] \Rightarrow W = \frac{- q^{2}}{16 {πε}_{0} R}$

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