Q 12-02-053JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
Two charges $7\ \mu\text{C}$ and $-4\ \mu\text{C}$ are placed at $(-7\ \text{cm}, 0, 0)$ and $(7\ \text{cm}, 0, 0)$ respectively. Given $\epsilon_0 = 8.85\times10^{-12}\ \text{C}^2\text{N}^{-1}\text{m}^{-2}$, the electrostatic potential energy of the charge configuration is
Answer: (A) $-1.8\ \text{J}$
Separation $r = 14\ \text{cm} = 0.14\ \text{m}$, and $\dfrac{1}{4\pi\epsilon_0} \approx 9\times10^9$:
$$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} = \frac{9\times10^9\times7\times10^{-6}\times(-4\times10^{-6})}{0.14} = \frac{-0.252}{0.14} = -1.8\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics