# The force of repulsion that two like charges exert on each other is 3.4 N. What will the force be if the distance between the charges is increased to 4 times its original value?

Jan 7, 2016

Let ${q}_{1}$ and ${q}_{2}$ be the charges, $K$ be the Coulomb's constant and $r$ be the distance between the charges when the force is $3.4 N$ and let $F$ denote this force.

$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

$\implies 3.4 = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

Now, every other condition remains same except the distance between the charges. The distance between the charges has been increased $4$ times its original value. Let $r '$ be then new distance between the charges, then
$r ' = 4 r$
Let the new force between the charges be denoted by $F '$.
$F ' = \frac{k {q}_{1} {q}_{2}}{r} _ {'}^{2}$

$\implies F ' = \frac{k {q}_{1} {q}_{2}}{4 r} ^ 2$

$\implies F ' = \frac{k {q}_{1} {q}_{2}}{16 {r}^{2}}$

$\implies F ' = \left(\frac{1}{16}\right) \left(\frac{k {q}_{1} {q}_{2}}{r} ^ 2\right)$

Since $3.4 = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

$\implies F ' = \left(\frac{1}{16}\right) \left(3.4\right)$

$\implies F ' = 0.2125 N$

Hence the force between the two charges becomes $0.2125 N$