JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
A two point charges 4q and -q are fixed on the x-axis at x = - d/2 and x = d/2 respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will :
- (a) increase by 3q²/(4π ε ₀d)
- (b) increase by 2q²/(3π ε ₀d)
- (c) decrease by q²/(4π ε ₀d)
- (d) decrease by 4q²/(3π ε ₀d)
Answer and solution
Answer (d) decrease by 4q²/(3π ε ₀d)
Solution
Physics sir · 2 steps
The path is a distraction: potential energy depends only on where the charge starts and ends.
- Path does not matter. The electrostatic force is conservative, so the change in energy is q(V_f - Vᵢ), whatever the path. Take k = 1/4πε₀. The semicircle is centred on -q, so the distance to -q stays d/2 the whole way.
Point Distance to 4q Distance to -q Potential V Origin (start) d/2 d/2 k(8q/d - 2q/d) = 6kq/d x = d (end) 3d/2 d/2 k(8q/3d - 2q/d) = 2kq/3d Δ U = q(V_f - Vᵢ) = kq²/d(2/3 - 6) = -16/3·q²/(4πε₀ d)
- So. Δ U = -4q²/(3πε₀ d). The energy decreases by 4q²/(3πε₀ d), which is option (d).
The idea to keep. For work or energy change in a static field, ignore the path and compute q(V_f - Vᵢ) at the two end points. Check whether any distance stays fixed; it saves half the work.
Answer from the NTA final answer key. Solution by PrepWiser.