Question

2 charged particles reacted with a force of 0.40 N with a distance of 1.6 cm, the 2nd charge contains 60 nC. Find the value of the first charge.

Answer 1

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

$q_2=\frac{Fr^2}{kq_1}$

Put value

$$q_2=\frac{0.40\times0.016^2}{9\times10^9\times60\times10^{-9}}=18.97\times10^{-6}C$$