Question

Two charges are found to be 0.20 m apart. The charge of the first is twice the charge of the second. If the magnitude of the second charge is 1 nC, find the force experienced by the first charge due to the second charge.

Answer 1

Given:

$r=\rm 0.20\: m$

$q_1=2q_2=2*10^{-9}\:\rm C$

$q_2=1*10^{-9}\:\rm C$

The Coulomb's law says

$$F=k\frac{q_1q_2}{r^2}\\ =9*10^9*\frac{2*10^{-9}*1*10^{-9}}{(0.20)^2}=4.5*10^{-7}\: \rm N$$