Eight point charges of magnitude $Q$ are arranged to form the corners of a cube of side $L$ . The arrangement is made in manner such that the nearest neighbour of any charge has the opposite sign. Initially, the charges are held at rest. If the system is let free to move, what happens to the arrangement? Does the cube-shape shrink or expand? Calculate the velocity of each charge when the side-length of the cube formation changes from $L$ to $n L$ . Assume that the mass of each point charge is $m$ .

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$\sqrt{\frac{q^{2} (1 - n) (8 \sqrt{6} + \sqrt{2} - 5 \sqrt{3})}{6 n m π L ε_{0} \sqrt{3}}}$

$\sqrt{\frac{Q^{2} (1 - n) (3 \sqrt{6} + \sqrt{2} - 3 \sqrt{3})}{4 n m π L ε_{0} \sqrt{6}}}$

$\sqrt{\frac{Q^{3} (1 - 2 n) (7 \sqrt{3} + \sqrt{2} - 3 \sqrt{2})}{5 n m π L ε_{0} \sqrt{6}}}$

answer is A.

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Detailed Solution

There are total $28$ pairs of charges.

$12$ pairs $\to Q$ and $- Q \to$ distance $L$

$12$ pairs $\to (Q$ and $Q)$ or $(- Q$ and $- Q) \to \sqrt{2} L$

$4$ pairs $\to Q$ and $- Q \to \sqrt{3} L$

$∴ U = 12 (\frac{1}{4 π ε_{0}}) (\frac{- Q^{2}}{L}) + 12 (\frac{1}{4 π ε_{0}} \cdot \frac{Q^{2}}{\sqrt{2} L}) + 4 (\frac{1}{4 π ε_{0}}) (\frac{- Q^{2}}{\sqrt{3} L}) = - \frac{Q^{2}}{π ε_{0} L} (\frac{3 \sqrt{6} + \sqrt{2} - 3 \sqrt{3}}{\sqrt{6}})$

with decrease in $L$ , potential energy will decrease. Therefore, cube should shrink as the conservative forces act in the direction of decreasing potential energy.

Increase in $K E$ of the system = decrease in $P E$

$or 8 (\frac{1}{2} m v^{2}) = U_{i} - U_{f} = \frac{Q^{2}}{π ε_{0}} (\frac{3 \sqrt{6} + \sqrt{2} - 3 \sqrt{3}}{\sqrt{6}}) (\frac{1}{n L} - \frac{1}{L}) or v = \sqrt{\frac{Q^{2} (1 - n) (3 \sqrt{6} + \sqrt{2} - 3 \sqrt{3})}{4 n m π L ε_{0} \sqrt{6}}}$