#### Question

One end of a 10 cm long silk thread is fixed to a large vertical surface of a charged non-conducting plate and the other end is fastened to a small ball of mass 10 g and a charge of 4.0× 10^{-6} C. In equilibrium, the thread makes an angle of 60° with the vertical (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.

#### Solution

(a)

In equilibrium state, the thread makes an angle of 60^{o} with the vertical.

The tension in the thread is resolved into horizontal and vertical components.

Then, tension in the string in equilibrium,

T cos 60° = mg

`"T" xx 1/2 = (10 xx 10^-3) xx 10`

T =(10 × 10^{-3} )× 10 × 2 = 0.20 N

(b) As it is displaced from equilibrium, net force on the ball,

`"F" = sqrt (("mg")^2 + ( ("q" sigma)/(2∈_0))^2)`

As F = ma

`"a" = sqrt (("g"))^2 + (("q" sigma)/("m"2∈_0))^2`

The surface charge density of the plate (as calculated in the previous question), *σ* = 7.5×10^{-7} C/m^{2}

Charge on the ball, *q* = 4×10^{-6 }C

Mass of the ball, m =

The time period of oscillation of the given simple pendulum,

`"T" = 2 pi sqrt ("l"/"g")`

`= 2 pi sqrt ((10 xx 10^-2)/9.8)`

= 0.45 sec