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

Correct answer: D. Kq_0^2 a (3 2 -2) aligned & u_ =(2 (K q_0)/(a)+ K q_0 2 a ) q_0 × 2 & u_0=(2 K q_0 2 a +(K q_0)/(a)) q_0 × 2 aligned aligned & So, Δ u=u_2-u_1=2 q_0 (k q_0)/(a)[2 2 +1-2- 1 2 ] & Δ u=(2 q_0^2)/(4 π _0 a)[ 4- 2 -1 2 ]=(2 q_0^2)/(4 π _0 a) (3- 2 ) 2 & Δ u=(2 k q_0^2)/(a)[ 3- 2 2 ]=(k q_0^2)/(a)(3 2 -2) aligned

JEE Main 202524 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Evening Shift
Subject: Physics
Chapter: Electrostatics

JEE Main 2025 · 24 January · Evening Shift · Physics · Electrostatics

In the first configuration (1) as shown in the figure, four...

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 :

Options

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 $\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)$

Correct answer

D. $\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)$

Step-by-step solution

\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}

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