A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

Solution

Physics sir · 4 steps

Each plate supplies half the field between them, so take one plate away and half the force goes with it.

1. Field of one plate. A large charged plate with surface charge density σ makes a uniform field σ/2ε₀ on either side of it. This does not depend on how far away you are.
2. Both plates together. Between the plates of a capacitor, the +σ plate and the -σ plate push the field in the same direction, so their fields add.

   E_(between) = σ/2ε₀ + σ/2ε₀ = σ/ε₀

3. Remove one plate. Now only one plate is left, so the field at the particle drops to σ/2ε₀. That is half of before. Since F = qE, the force also halves.
4. So. The new force is 10/2 = 5 N, which is option (a).

The idea to keep. Between capacitor plates the field σ/ε₀ is the sum of two equal halves, σ/2ε₀ from each plate. Remove a plate and you remove its half.

Answer from the NTA final answer key. Solution by PrepWiser.