Question

A small, 29 gram pith balls carrying 2 nC of charge is dangling from an insulating string. Another charged pith balls carrying an unknown amount of charge is brought close to the first ball, causing it to shift away. When everything is in equilibrium, the two balls are horizontally separated by distance of 5 cm, and the string is tilted about 11 degrees away from the vertical direction. How much charge is on the second ball

Answer 1

`q_2=7.67* 10^(-6)\ C`

Step-by-step explanation:

Given that,

Mass of a pitch ball, m = 29 g = 0.029 kg

Charge on one pitch ball, q₁ = 2 nC

Another charged pith balls carrying an unknown amount of charge is brought close to the first ball, causing it to shift away. When everything is in equilibrium, the two balls are horizontally separated by distance of 5 cm or 0.05 m, and the string is tilted about 11 degrees away from the vertical direction.

At equilibrium,

`T\cos\theta=mg\ .....(1)\\\\\text{and}\\\\T\sin\theta=(kq_1q_2)/(d^2)\ ....(2)`

Where q₂ is the change on the second ball

Dividing equation (2) by (1) we get :

`\tan\theta=(kq_1q_2)/(d^2mg)`

Solving for q₂, we get :

`q_2=(\tan\theta d^2 mg)/(kq_1)\\\\q_2=(\tan(11)* 0.05^2* 0.029* 9.8)/(9* 10^9* 2* 10^(-9))\\\\q_2=7.67* 10^(-6)\ C`

So, the charge on the second ball is
`7.67* 10^(-6)\ C`.