A simple pendulum of mass m and charge +q is suspended vertically by a massless thread of length l. At the point of suspension, a point charge +q is also fixed. If the pendulum is displaced slightly from equilibrium position, its time period will be

Options

A. `T=2pisqrt(l/(g+(kq^(2))/(ml^(2))))`
B. `T=2pi`
C. `T=2pisqrt(l/g)`
D. will be greater then `2pisqrt(l/g)`

Answer

C

Explanation

Torque of electrostatic force is zero.