Question

A simple pendulum consists of a small sphere of mass m suspended by a thread of length ℓ. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is

(A) T = 2π[ℓ/{g – (qE/m)}](1/2) 

(B) T = 2π(ℓ/g)(1/2) 

(C) T = 2π[ℓ/{g + (qE/m)}](1/2) 

(D) T = 2π[(mℓ/qE)](1/2)

Answer 1

The correct option (A) T = 2π[ℓ/{g – (qE/m)}]^(1/2)

Explanation:

The sphere of mass m has charge + q.

The electric field E is directed vertically up.

hence sphere will experience force in upward direction.

This forces will be F = qE

Due to this force, acceleration produced in the upward direction in sphere is a = (F/m) = (qE/m)

The net acceleration on sphere is g' = g – a = g – (qE/m)

As = Time period = T = 2π√(ℓ/g')

∴ T = 2π√[ℓ/{g – (qE/m)}]