# What is the force, in terms of Coulomb's constant, between two electrical charges of 18 C and 25 C that are 15 m  apart?

Jan 8, 2018

The force is $= + \left(0.72 k\right) N$

#### Explanation:

The force between $2$ charges ${q}_{1}$ and ${q}_{2}$ is

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

Where $r$ is the distance between the charges.

And the constant $k = 9 \cdot {10}^{9} N {m}^{2} {C}^{-} 2$

The charge ${q}_{1} = 18 C$

The charge ${q}_{2} = 25 C$

The distance is

$r = 15 m$

Therefore,

The force is $= k \cdot \frac{\left(18\right) \cdot \left(25\right)}{{15}^{2}} = + \left(0.72 k\right) N$

The positive sign indicates that the force is repulsive