Physics Circle home Student Login

Home›Chapters›Electrostatics PYQs›Electrostatic Potential Energy›Question 4

JEE Main 2026 (28 Jan, Shift 1) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2026 (28 Jan, Shift 1)

Q4JEE Main 2026 · 28 JAN · SHIFT 1Easy

Electrostatic Potential Energy

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 ‾ μJ\underline{\qquad}\ \mu\text{J}.

14πϵ0=9×109 N⋅m2/C2\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ N}\cdot\text{m}^2/\text{C}^2

Show answer and solution

Answer: (c)

  1. Vcorner=k(q1+q2)r=9×109×3×10−90.03=900V_{corner} = \dfrac{k(q_1 + q_2)}{r} = \dfrac{9\times10^9 \times 3\times10^{-9}}{0.03} = 900 V
  2. W=qV=3×10−9×900=2.7×10−6W = qV = 3\times10^{-9} \times 900 = 2.7\times10^{-6} J
  3. W=2.7 μW = 2.7\ \muJ. So the answer is (c).
Report a mistake

More questions on Electrostatic Potential Energy