JEE Main 2026 · 21 January · morning · Single correct · +4 / −1
A point charge of 10⁻⁸C is placed at origin. The work done in moving a point charge 2μC from point A(4,4,2)m to B(2,2,1)m is ____ J. (1/4πεₒ=9× 10⁹ in SI units)
- (a) 30× 10⁻⁶
- (b) 0
- (c) 15× 10⁻⁶
- (d) 45× 10⁻⁶
Answer and solution
Answer (a) 30× 10⁻⁶
Solution
Physics sir · 3 steps
Work depends only on the start and end distances from the charge, so find r_A and r_B first.
- Distances from origin. r_A = √(4² + 4² + 2²) = √36 = 6 m and r_B = √(2² + 2² + 1²) = √9 = 3 m.
- Potential at each. With V = kQ/r and kQ = 9 × 10⁹ × 10⁻⁸ = 90 V m: V_A = 90/6 = 15 V and V_B = 90/3 = 30 V.
W = q(V_B - V_A) = 2 × 10⁻⁶ × (30 - 15) = 30 × 10⁻⁶ J
- So. The work done in moving the charge from A to B is 30 × 10⁻⁶ J, option (a). It is positive because a positive charge is pushed closer to a positive charge.
The idea to keep. The electrostatic field is conservative, so work between two points is q(V_(final) - Vᵢₙᵢₜᵢₐₗ). For a point charge, you only need each point's distance from it.
Answer from the NTA final answer key. Solution by PrepWiser.