Question

A 0.56-C charge is placed 15.0 m from a 1.56-C charge. What is the force between them?

Answer 1

The force between them:

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

$$F=\frac{(8.99 \cdot 10^9)(0.56)(1.56)}{15^2}=3.49 \cdot 10^7 N$$