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.

Step-by-step solution

Electric field due to each plate, E 1 = E 2 = σ 2 ε 0 = Q 2 A ε 0 Net electric field between the plates, E net = E 1 + E 2 = Q A ε 0 Force on charged particle between the plates, F 1 = q E n e t = q Q A ε 0 = 10   N Now, in second case, the net electric field, E = σ 2 ε 0   = Q 2 A ε 0 = 5   N Force on charged particle, F 2 = q E = q Q 2 A ε 0 = 5   N