# Question bc9cb

Jan 21, 2017

Force becomes four times.

#### Explanation:

If $r$ is the distance between two charges ${q}_{1} \mathmr{and} {q}_{2}$ the force of interaction is given by Coulombs Law

$| \vec{F} | = {k}_{e} \setminus \frac{{q}_{1} {q}_{2}}{{r}^{2}}$ .....(1)
where ${k}_{e}$ is Coulomb's constant = 8.99×10^9 N m^2 C^-2#

Given that both charges are doubled. The new force is
$| {\vec{F}}_{\text{New}} | = {k}_{e} \setminus \frac{\left(2 {q}_{1}\right) \left(2 {q}_{2}\right)}{{r}^{2}}$

$\implies | {\vec{F}}_{\text{New}} | = {k}_{e} \setminus \frac{4 {q}_{1} {q}_{2}}{{r}^{2}}$ ....(2)
Using equation (1) we have

$| {\vec{F}}_{\text{New}} | = 4 | \vec{F} |$