A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is

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$\frac{2 Q}{4 {πε}_{0} R} - \frac{2 q}{4 {πε}_{0} R}$

$\frac{(q + Q)}{4 {πε}_{o}} \frac{2}{R}$

$\frac{2 Q}{4 {πε}_{0} R}$

$\frac{2 Q}{4 {πε}_{0} R} + \frac{q}{4 {πε}_{0} R}$

answer is C.

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Detailed Solution

$\begin{array}{l} V_{1} = \frac{1}{4 {πε}_{0}} \frac{q}{R}; V_{2} = \frac{1}{4 {πε}_{0}} \frac{Q}{R / 2} = \frac{2 Q}{4 {πε}_{0} R} \\ V_{atP} = V_{1} + V_{2} = \frac{q}{4 {πε}_{0} R} + \frac{2 Q}{4 {πε}_{0} R} \end{array}$

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