Question

A negative charge of - 0.0005 C exerts an attractive force of 7.0 N on a second charge that is 10 m away. What is the magnitude of the second charge?

Answer 1

the magnitude of the second charge is 0.000156 C.

Step-by-step explanation:

Given;

mangitude of the first charge, q₁ = 0.0005 C

attractive force between the two charges, F = 7.0 N

distance between the two charges, r = 10 m

let the magnitude of the second charge = q₂

The magnitude of the second charge is calculated by applying Coulomb's law;

`F = (kq_1q_2)/(r^2)`

Where;

K is Coulomb's constant = 9 x 10⁹ Nm²/C²

`q_2 = (Fr^2)/(kq_1) \\\\q_2 = (7 * \ 10^2)/((9* 10^9)(0.0005)) \\\\q_2 = 0.000156 \ C`

Therefore, the magnitude of the second charge is 0.000156 C.