JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Capacitance

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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 :

(A)5 N✓ Correct
(B)10 N
(C)20 N
(D)Zero

Answer: (A)

Solution

Between two parallel plates of a capacitor, the electric field is

E = \frac{Q}{A ϵ _{0}}

, giving force

F = q E = 10

N. When one plate is removed, each plate creates a field

E^{'} = \frac{Q}{2 A ϵ _{0}}

(half the original). Thus the force becomes

F^{'} = q E^{'} = \frac{q Q}{2 A ϵ _{0}} = 5

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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