In the first configuration (1) as shown in the figure, four identical charges () are kept at the corners A, B, C and D of suare of side length 'a'. In the second configuration (2), the same charges are shifted to mid points G, E, H and F, of the square.

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Solution

Q.

In the first configuration (1) as shown in the figure, four identical charges ( $q_{0}$ ) are kept at the corners A, B, C and D of square of side length 'a'. In the second configuration (2), the same charges are shifted to mid points G, E, H and F, of the square. If $K = \frac{1}{4 π ε_{0}}$ , the difference between the potential energies of configuration (2) and (1) is given by [2025]

1 $\frac{K q_{0}^{2}}{a} (4 \sqrt{2} - 2)$

2 $\frac{K q_{0}^{2}}{a} (3 - \sqrt{2})$

3 $\frac{K q_{0}^{2}}{a} (4 - 2 \sqrt{2})$

4 $\frac{K q_{0}^{2}}{a} (3 \sqrt{2} - 2)$

Ans.
(4)

$U_{1} = (2 \frac{K q_{0}}{a} + \frac{K q_{0}}{\sqrt{2} a}) q_{0} \times 2$

$U_{2} = (2 \frac{K q_{0} \sqrt{2}}{a} + \frac{K q_{0}}{a}) q_{0} \times 2$

So, $△ U = U_{2} - U_{1} = 2 q_{0} \frac{K q_{0}}{a} [2 \sqrt{2} + 1 - 2 - \frac{1}{\sqrt{2}}]$

$\Rightarrow △ U = \frac{2 q_{0}^{2}}{4 π ε_{0} a} [\frac{4 - \sqrt{2} - 1}{\sqrt{2}}] = \frac{2 q_{0}^{2}}{4 π ε_{0} a} \frac{(3 - \sqrt{2})}{\sqrt{2}}$

$\Rightarrow △ U = \frac{2 K q_{0}^{2}}{a} [\frac{3 - \sqrt{2}}{\sqrt{2}}] = \frac{K q_{0}^{2}}{a} (3 \sqrt{2} - 2)$

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