JEE Main 2025 · 2 April · morning · Single correct · +4 / −1
Consider two infinitely large plane parallel conducting plates as shown below. The plates are uniformly charged with a surface charge density +σ and -2 σ. The force experienced by a point charge +q placed at the mid point between two plates will be:
- (a) (3 σ q)/(4 ε₀)
- (b) (3 σ q)/(2 ε₀)
- (c) (σ q)/(4 ε₀)
- (d) (σ q)/(2 ε₀)
Answer and solution
Answer (b) (3 σ q)/(2 ε₀)
Solution
Physics sir · 3 steps
Treat each plate as its own sheet of charge; between them, both fields push +q the same way, so they add.
- Field of one sheet. An infinite sheet with charge density σ makes a field of σ/2ε₀ on both sides. It does not depend on distance, so the exact midpoint does not matter.
- Directions at the middle. The +σ plate pushes away from itself, towards the -2σ plate, with σ/2ε₀. The -2σ plate pulls towards itself, which is the same direction, with 2σ/2ε₀ = σ/ε₀.
E = σ/2ε₀ + 2σ/2ε₀ = 3σ/2ε₀
- So. The force on +q is F = qE = (3σ q)/2ε₀, directed towards the -2σ plate. That is option (b).
The idea to keep. Each infinite sheet gives σ/2ε₀, whatever the distance. Fix the direction of each field at the point first, then add or subtract.
Answer from the NTA final answer key. Solution by PrepWiser.