Consider two conducting spheres with radii $R_{1}$ and $R_{2}$ separated by a distance much greater than radius of either. A total charge $Q$ is shared between the spheres, subject to the condition that the electric potential energy of the system has the smallest possible value. The spheres being very far apart, you can assume that the charge of each is uniformly distributed over the surface of each.
Find the potential difference between the surfaces of the spheres.

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$3$

$0$

$2$

$1$

answer is D.

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Detailed Solution

$I f c h a r g r o n t h e s p h e r e o f r a d i u s R_{1} b e q, t h e n U = \frac{q^{2}}{8 {πε}_{0} R_{1}} + \frac{{(Q - q)}^{2}}{8 {πε}_{0} R_{2}} F o r m i n i m u m U, \frac{d U}{d q} = 0 \Rightarrow q = \frac{Q R_{1}}{R_{1} + R_{2}} V_{1} = \frac{1}{4 {πε}_{0}} \frac{q_{1}}{R_{1}} = \frac{1}{4 {πε}_{0}} \frac{R_{1} Q}{R_{1} (R_{1} + R_{2})} \Rightarrow V_{1} = \frac{1}{4 {πε}_{0}} \frac{Q}{R_{1} + R_{2}} V_{2} = \frac{1}{4 {πε}_{0}} \frac{q_{2}}{R_{2}} = \frac{1}{4 {πε}_{0}} \frac{R_{2} Q}{R_{2} (R_{1} + R_{2})} \Rightarrow V_{2} = \frac{1}{4 {πε}_{0}} \frac{Q}{R_{1} + R_{2}} \Rightarrow V_{1} - V_{2} = 0$