Physics · Electrostatics

JEE Main 2026 — 28 January, Morning Shift — Question 39

Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm . The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is $\_\_\_\_$ $\mu \mathrm{J}$ . $\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} . \mathrm{m}^{2} / \mathrm{C}^{2}$

Option A:
2.7
Correct
Option B:
5.4
Option C:
3.3
Option D:
27

Answer: A

Step-by-step solution

$\left(\mathrm{q}_{2}\right)$ $\mathrm{W}=\left(\frac{\mathrm{kq}_{1}}{\ell}+\frac{\mathrm{kq}_{2}}{\ell}\right) \mathrm{q}_{3}$ $=\frac{9 \times 10^{9}}{3 \times 10^{-2}}\left(3 \times 10^{-9}\right) \times 3 \times 10^{-9}$ $=27 \times 10^{-7} \mathrm{~J}=2.7 \mu \mathrm{~J}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electrostatic Potential Energy and Electric Potential