JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

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 \(\mathrm{G}, \mathrm{E}, \mathrm{H}\) and F, of the square, If \(\mathrm{K}=\frac{1}{4 \pi \varepsilon_0}\), the difference between the potential energies of configuration (2) and (1) is given by :
- A \(\frac{K q_0^2}{\mathrm{a}}(4-2 \sqrt{2})\)
- B \(\frac{K q_0^2}{\mathrm{a}}(3-\sqrt{2})\)
- C \(\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(4 \sqrt{2}-2)\)
- D \(\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & u_{\oplus}=\left(2 \frac{K q_0}{a}+\frac{K q_0}{\sqrt{2} a}\right) q_0 \times 2 \\ & u_0=\left(2 \frac{K q_0 \sqrt{2}}{a}+\frac{K q_0}{a}\right) q_0 \times 2 \end{aligned}\)…
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