Question

A negative charge of - 0.0005 C exerts an attractive force of 9.0 N on a second charge that is 10 m away. What is the magnitude of the second charge?

Answer 1

Answer:

q₂ = 0.0002 C

Explanation:

Given that,

Charge, q₁ = -0.0005 C

The force between two charges, F = 9 N

The distance between charges, d = 10 m

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

$F=\dfrac{kq_1q_2}{r^2}\\\\q_2=\dfrac{Fr^2}{kq_1}\\\\q_2=\dfrac{9\times (10)^2}{9\times 10^9\times 0.0005}\\\\q_2=0.0002\ C$

So, the magnitude of the second charge is 0.0002 C.