$\begin{aligned}
& u_{\oplus}=\left(2 \frac{K q_0}{a}+\frac{K q_0}{\sqrt{2} a}\right) q_0 \times 2 \
& u_0=\left(2 \frac{K q_0 \sqrt{2}}{a}+\frac{K q_0}{a}\right) q_0 \times 2
\end{aligned}\begin{aligned}
& \text { So, } \quad \Delta u=u_2-u_1=2 q_0 \frac{k q_0}{a}\left[2 \sqrt{2}+1-2-\frac{1}{\sqrt{2}}\right] \
& \Rightarrow \quad \Delta u=\frac{2 q_0^2}{4 \pi \varepsilon_0 a}\left[\frac{4-\sqrt{2}-1}{\sqrt{2}}\right]=\frac{2 q_0^2}{4 \pi \varepsilon_0 a} \frac{(3-\sqrt{2})}{\sqrt{2}} \
& \Rightarrow \quad \Delta u=\frac{2 k q_0^2}{a}\left[\frac{3-\sqrt{2}}{\sqrt{2}}\right]=\frac{k q_0^2}{a}(3 \sqrt{2}-2)
\end{aligned}$
JEE Main 2025 — Physics Electromagnetism
2025mcqhard
In the first configuration (1) as shown in the figure, four identical charges (q0) 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=4πε01, the difference between the potential energies of configuration (2) and (1) is given by :
Official previous-year question
Held on 24 Jan 2025 · Verified 6 Jul 2026.
Options
- A
aKq02(4−22)
- B
aKq02(3−2)
- C
aKq02(42−2)
- D
aKq02(32−2)
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