# What is the force, in terms of Coulomb's constant, between two electrical charges of 24 C and 32 C that are 12 m  apart?

Jan 10, 2016

If ${q}_{1}$ and ${q}_{2}$ are charges and $r$ is the distance between them then the force $F$ between them is given by

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

Where $k$ is a constant and is known as Coulomb's force constant

Here Let ${q}_{1} = 24 C$, ${q}_{2} = 32 C$ and $r = 12 m$,

$F = \frac{k \cdot 24 \cdot 32}{12} ^ 2$

$\implies F = \frac{768 k}{144} =$

$\implies F = 5.33 k N$

Since the force is positive so the force between the charges is repullsive