Question

a 5.2nC charge exerts a repulsive force of 30mN on a second charge which is located at a distance of 60m away from it. what is the charge (magnitude and sign) of the second charge?

Answer 1

Charge,
`q_2=2.3\ C`

Step-by-step explanation:

It is given that,

Charge 1,
`q_1=5.2\ nC=5.2* 10^(-9)\ C`

Repulsive force,
`F=30\ mN=30* 10^(-3)\ N`

Distance between two charges, d = 60 m

We need to find the magnitude of second charge. The electric force between two charge is given by :

`F=k(q_1q_2)/(d^2)`

`q_2` is second charge

`q_2=(Fd^2)/(kq_1)`

k is electrostatic constant

`q_2=(30* 10^(-3)* (60)^2)/(9* 10^9* 5.2* 10^(-9)) \\\\q_2=2.3\ C`

So, the magnitude of second charge is 2.3 C. Hence, this is the required solution.