# Calculate the magnitude of the electrostatic force 1

A particle of charge $+3.00\times 10^{-6}$ C is 12.0 cm distant from a second particle of charge $-1.50\times 10^{-6}$ C. Calculate the magnitude of the electrostatic force between the particles.

#### Solution:

We will use Coulomb’s law to figure out the electrostatic force between the particles. 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:

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

$F=(8.99\times 10^9)\dfrac{(3\times 10^{-6})(1.5\times 10^{-6})}{0.12^2}$

(Note that we only use the magnitudes of the charges since there is an attraction between them. Also note that 12 cm = 0.12 m)

$F=2.81$ N

$F=2.81$ N