# What must be the distance between point charge 5

What must be the distance between point charge $q_1=26.0$ μC and point charge $q_2=47.0$ μC for the electrostatic force between them to have a magnitude of 5.70 N?

#### Solution:

We can use Coulomb’s law to figure out the distance between the two point charges. Coulomb’s law states:

$F=k\dfrac{(q_1)(q_2)}{r^2}$

(Where $F$ is the electrostatic force, $k$ is Coulomb’s constant, $q_1$ is the charge of the first particle, $q_2$ is the charge of the second particle, and $r$ is the distance between the two charges)

Substitute the values we know into our equation.

$F=k\dfrac{(q_1)(q_2)}{r^2}$

$5.7=(8.99\times 10^9)\dfrac{(26\times 10^{-6})(47\times 10^{-6})}{r^2}$

(isolate for r)

$5.7r^2=(8.99\times 10^9)(26\times 10^{-6})(47\times 10^{-6})$

$r^2=1.9273298$

$r=\sqrt{1.9273298}=1.39$ m

$r=1.39$ m