Two equal point charges are kept at a separation of r = 8 cm from each other and potential energy of the system is found to be U. Now 50 % of charge is transferred from one charge to another. What will be the new separation between the charges for which the electrostatic potential energy of the system will be U?

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12 cm

7.5 cm

4 cm

6 cm

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{k \cdot Q \cdot Q}{r} = U and \frac{k \cdot \frac{3 Q}{2} \cdot \frac{q}{2}}{r^{'}} = U \\ \frac{3}{4} \cdot \frac{{KQ}^{2}}{r^{'}} = \frac{{KQ}^{2}}{r} \Rightarrow r^{'} = \frac{3}{4} r = 6 cm \end{array}$

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