• question_answer Four point charges q, q, q and $-q$are placed at the vertices of a square of side length a. The configuration is changed and the charge are positioned at the vertices of a rhombus of side length a with $-q$ charge at the vertex where angle is$120{}^\circ$. Find the work done by the external agent in changing the configuration.   A) $-\frac{K{{q}^{2}}}{a}\left( 1-\frac{1}{\sqrt{3}} \right)$            B)  $\frac{K{{q}^{2}}}{a}\left( 1+\frac{1}{\sqrt{3}} \right)$ C) $-\frac{K{{q}^{2}}}{2a}\left( 1-\frac{1}{\sqrt{3}} \right)$         D) $\frac{K{{q}^{2}}}{2a}\left( 1+\frac{1}{\sqrt{2}} \right)$

 [a] On square: There will be six terms in the expression of potential energy for the system, as there are six pairs of interactions. ${{U}_{AB}}$ and ${{U}_{AD}}$ are negative whereas ${{U}_{BC}}$ and ${{U}_{CD}}$positive terms of same magnitude. Similarly ${{U}_{BD}}$ and ${{U}_{AC}}$ cancel out. Hence PE is zero. On Rhombus: ${{U}_{AB}}$ and ${{U}_{AD}}$ cancel out with ${{U}_{BC}}$and ${{U}_{CD}}$. Length $BD=2a\cos 30{}^\circ =\sqrt{3}a$ $AC=2a\,\cos 60{}^\circ =a$ $\therefore \,\,\,\,U={{U}_{BD}}+{{U}_{AC}}$ $=K\frac{{{q}^{2}}}{\sqrt{3}a}-K\frac{{{q}^{2}}}{a}=-\frac{K{{q}^{2}}}{a}\left( 1-\frac{1}{\sqrt{3}} \right)$ $\therefore \,\,\,{{W}_{ext\,\,Agt}}={{U}_{\text{Rhombus}}}-{{U}_{square}}=-\frac{K{{q}^{2}}}{a}\left( 1=\frac{1}{\sqrt{3}} \right)$