# What is the force, in terms of Coulomb's constant, between two electrical charges of 24 C and -6 C that are 12 m  apart?

Dec 30, 2015

$F = 9 \cdot {10}^{9} N$

#### Explanation:

The electrostatic force between two charges is given by
$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$
Where $F$ is the force between the charges, $k$ is the constant and its value is $9 \cdot {10}^{9}$,${q}_{1}$ and ${q}_{2}$ are the magnitudes of the charges and $r$ is the distance between the two charges.

Here F=??, $k = 9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$, ${q}_{1} = 24 C$, ${q}_{2} = - 6 C$ and $r = 12 m$.

$\implies F = \frac{9 \cdot {10}^{9} \cdot 24 \cdot 6}{12} ^ 2 = \frac{1296 \cdot {10}^{9}}{144} = 9 \cdot {10}^{9} N$
$\implies F = 9 \cdot {10}^{9} N$
Note:- The negative sign of the charges is not taken in the problems.