JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 19)
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In the first configuration (1) as shown in the figure, four identical charges $\left(q_0\right)$ 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 $K=\frac{1}{4 \pi \epsilon_0}$, the difference between the potential energies of configuration (2) and (1) is given by :
$\frac{K q_0^2}{a}(4-2 \sqrt{2})$
$\frac{K q_0^2}{a}(4 \sqrt{2}-2)$
$\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3-\sqrt{2})$
$\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)$
Explanation
$$\begin{aligned}
& \mathrm{U}_1=\frac{4 \mathrm{Kq}_0^2}{\mathrm{a}}+\frac{2 \mathrm{Kq}_0^2}{\sqrt{2} \mathrm{a}}=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4+\sqrt{2}) \\
& \mathrm{U}_2=\frac{\mathrm{Kq}_0^2}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}(4+\sqrt{2})=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4 \sqrt{2}+2) \\
& \mathrm{U}_2-\mathrm{U}_1=\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)
\end{aligned}$$
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