JEE Main 2026 — Electrostatics Question with Solution

Two point charges of 1 nC and 2 nC are placed at two corners of an equilateral triangle with side 3 cm. The work done in bringing a 3 nC charge from infinity to the third corner equals the charge times the potential at that point.
Both source charges are at distance 0.03 m from the third corner.
The potential due to the 1 nC charge is $V_1=\frac{9\times 10^9\times 1\times 10^{-9}}{0.03}=300$ V
and due to the 2 nC charge is $V_2=\frac{9\times 10^9\times 2\times 10^{-9}}{0.03}=600$ V.
Total potential: $V_{total}=300+600=900$ V
Work done: $W=3\times 10^{-9}\times 900=2.7\times 10^{-6}$ J = 2.7 μJ