Question

Two-point charges are attracted from each other with a force of 6kN. If the distance between them is reduced to one fourth of its original value, determine the new force of attraction between them.

Answer 1

According to the Coulomb's law, the force between two point charges is proportional to the squared distance between them:

$$F_1\sim\dfrac{1}{r_1^2}$$

where $F_1 = 6kN$ is the force and $r_1$ is the distance in first case. Similarly:

$$F_2\sim\dfrac{1}{(r_1/4)^2}$$

where $F_2$ is the new force and $r_1/4$ is the reduced distance. Taking the ration of these two expressions, obtain:

$$\dfrac{F_2}{F_1} = \dfrac{r_1^2}{(r_1/4)^2}\\ F_2 = F_1\dfrac{r_1^2}{(r_1/4)^2} = \dfrac{F_1r_1^2\cdot 4^2}{r_1^2} = 16F_1\\ F_2 = 6kN\cdot 16 = 96kN$$

Answer. 96kN.