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

Answer:

the magnitude of the second charge is 0.000156 C.

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;

$F = \frac{kq_1q_2}{r^2}$

Where;

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

$q_2 = \frac{Fr^2}{kq_1} \\\\q_2 = \frac{7 \times \ 10^2}{(9\times 10^9)(0.0005)} \\\\q_2 = 0.000156 \ C$

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