Question

What happens to the force between two charges if the distance between them is cut in half?

Answer 1

Step 1: Assumption

As given the distance between them which is cut in half is $\frac{\mathrm{r}}{2}$

Let $F\text{'}$ be the force between the charges when the distance is cut in half

Step 2: Formula used

The force between the two charged particles can be written as:
$F=\frac{k{q}_1{q}_2}{r^2}$
where, $q_1$ and $q_2$ are the charges, $r$ is the distance, $k$ is constant of proportionality(Coulomb's constant), $F$ is the force between the charges.

Step 3: Calculation of the force when the distance between the charges is cut in half

When the distance is halved the force becomes

$$\begin{gathered} F=\frac{k{q}_1{q}_2}{r^2} \\ F\text{'}=\frac{k{q}_1{q}_2}{{\left(\frac{r}{2}\right)}^2} \\ F\text{'}=\frac{4k{q}_1{q}_2}{{\displaystyle r^2}} \\ F\text{'}=4F \end{gathered}$$

So, when the distance between the two charges is halved, the force is increased four times.