If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
[tex]F= k \frac{q_1 q_2}{r^2} [/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
[tex]r'= \frac{r}{2} [/tex]
the magnitude of the force changes as follows:
[tex]F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k \frac{q_1 q_2}{r^2}=4 F [/tex]
so, the force increases by a factor 4.
