# If the distance between two charges is doubled, what will happen the force between the charges?

Feb 26, 2018

the force between 2 charges, by coulomb's law is given by,

$F = \frac{{q}_{1} {q}_{2}}{4 \pi {\epsilon}_{0} {r}^{2}}$ where, ${q}_{1}$ and ${q}_{2}$ are the magnitude of the charges. and $r$ is the distance between them.

So, according to the question, if the distance between two charges is doubled, that is $\textcolor{red}{r '} = 2 r$ what happens to the force between them.?

the new force, $F ' = \frac{{q}_{1} {q}_{2}}{4 \pi {\epsilon}_{0} {\textcolor{red}{r '}}^{2}}$

$F ' = \frac{{q}_{1} {q}_{2}}{4 \pi {\epsilon}_{0} {\textcolor{red}{2 r}}^{2}}$

F' = (q_1q_2)/(4piepsilon_0 (4r^2)

F' = (q_1q_2)/(4(4piepsilon_0 r^2)

$F ' = \frac{F}{4}$

therefore the new force if 4 times the old one :)