Q 12-02-046JEE MainJEE Main 2026 (28 Jan, Shift 1)Easy
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 ______ $\mu$J.
$\dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{N}\cdot\text{m}^2/\text{C}^2$
Answer: (C) 2.7
Potential at the third corner (both charges 3 cm away):
$$V = \frac{9\times10^9\times(1+2)\times10^{-9}}{0.03} = 900\ \text{V}$$
$$W = qV = 3\times10^{-9}\times900 = 2.7\times10^{-6}\ \text{J} = 2.7\ \mu\text{J}$$
Solution by Sreeraj P, M.Sc Physics