Question

If two equal charges, each of 1 C are separated in the air by a distance of 1 km, what would be the magnitude and kind of force between them?

Answer 1

The foce between an equal charges are repulsive. According to the Coulomb's law:

$$F = k\dfrac{q_1q_2}{r^2}$$

where $q_1 = q_2 = 1C$ are the charges, $k = 9\times10^9N\cdot m^2/C^2$ is the constant, and $r = 1000m$ is the distance between the charges. Thus, obtain:

$$F = 9\times 10^9\cdot \dfrac{1\cdot 1}{1000^2}=9000N$$

Answer. 9000N.