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(C) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \s

Vedclass · Accepted Answer

$\frac{kQ^2}{a} \left( 24 + \frac{11}{\sqrt{2}} \right)$

Vedclass · Answer

(C) The electrostatic potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.For a square with side $a$,there are $4$ pairs of charges at distance $a$ (sides) and $2$ pairs of charges at distance $a\sqrt{2}$ (diagonals).Let the charges be $q_1 = Q$,$q_2 = 2Q$,$q_3 = 3Q$,and $q_4 = 4Q$ in counter-clockwise order starting from the bottom-left.The pairs at distance $a$ are:$(q_1, q_2) = (Q, 2Q) \implies U_1 = \frac{k(Q)(2Q)}{a} = 2 \frac{kQ^2}{a}$$(q_2, q_3) = (2Q, 3Q) \implies U_2 = \frac{k(2Q)(3Q)}{a} = 6 \frac{kQ^2}{a}$$(q_3, q_4) = (3Q, 4Q) \implies U_3 = \frac{k(3Q)(4Q)}{a} = 12 \frac{kQ^2}{a}$$(q_4, q_1) = (4Q, Q) \implies U_4 = \frac{k(4Q)(Q)}{a} = 4 \frac{kQ^2}{a}$The pairs at distance $a\sqrt{2}$ are:$(q_1, q_3) = (Q, 3Q) \implies U_5 = \frac{k(Q)(3Q)}{a\sqrt{2}} = 3 \frac{kQ^2}{a\sqrt{2}}$$(q_2, q_4) = (2Q, 4Q) \implies U_6 = \frac{k(2Q)(4Q)}{a\sqrt{2}} = 8 \frac{kQ^2}{a\sqrt{2}}$Total potential energy $U = U_1 + U_2 + U_3 + U_4 + U_5 + U_6 = (2 + 6 + 12 + 4) \frac{kQ^2}{a} + (3 + 8) \frac{kQ^2}{a\sqrt{2}} = 24 \frac{kQ^2}{a} + 11 \frac{kQ^2}{a\sqrt{2}} = \frac{kQ^2}{a} \left( 24 + \frac{11}{\sqrt{2}} \right)$.

The electrostatic potential energy of the given system of charges placed at the corners of a square of side $a$ is:

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