Question

The Force between two identical charge's 1.90 x10^-2 C is separated by 1 cm. What is the electric froce between the two of charges?

Answer 1

Force

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

$$F=\frac{9\times10^9\times1.90\times10^{-2}\times1.90\times10^{-2}}{0.01^2}=3.25\times10^{10}N$$