Physics · Electrostatics

JEE Main 2025 — 24 January, Evening Shift — Question 53

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 :

Option A:
$\frac{K q_{0}^{2}}{a}(4 \sqrt{2}-2)$
Option B:
$\frac{\mathrm{Kq}_{0}^{2}}{\mathrm{a}}(3-\sqrt{2})$
Option C:
$\frac{\mathrm{Kq}_{0}^{2}}{\mathrm{a}}(4-2 \sqrt{2})$
Option D:
$\frac{\mathrm{Kq}_{0}^{2}}{\mathrm{a}}(3 \sqrt{2}-2)$
Correct

Answer: D

Step-by-step solution

$\quad \mathrm{U}_{1}=\frac{4 \mathrm{Kq}_{0}^{2}}{\mathrm{a}}+\frac{2 \mathrm{Kq}_{0}^{2}}{\sqrt{2} \mathrm{a}}=\frac{\mathrm{Kq}_{0}^{2}}{\mathrm{a}}(4+\sqrt{2})$

$\mathrm{U}_{2}=\frac{\mathrm{Kq}_{0}^{2}}{\left(\frac{\mathrm{a}}{\sqrt{2}}\right)}(4+\sqrt{2})=\frac{\mathrm{Kq}_{0}^{2}}{\mathrm{a}}(4 \sqrt{2}+2)$

$\mathrm{U}_{2}-\mathrm{U}_{1}=\frac{\mathrm{Kq}_{0}^{2}}{\mathrm{a}}(3 \sqrt{2}-2)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4 × 10^-8 C charge and then removed to a…

Answer: D

Step-by-step solution

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