# What is the force, in terms of Coulomb's constant, between two electrical charges of 18 C and -14 C that are 15 m  apart?

Jan 13, 2016

This is solved using Coulomb's Law, ${F}_{e} = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

#### Explanation:

The force, in this case attraction (since the charges are opposite), is given by

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

where $k$ is the constant $8.99 \cdot {10}^{9} \frac{N {m}^{2}}{C} ^ 2$
${q}_{1}$ and ${q}_{2}$ are the charges (18C and -14C)
and r is the distance between them, in this case 15 m.

I think you will find it to be an absurdly large number.