# A charge of -2 C is at (-2 , 1 ) and a charge of -1 C is at (1 ,-3) . If both coordinates are in meters, what is the force between the charges?

Jan 14, 2016

The force between the charges is given by Coulomb's Law, $F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$ , and when the distance is correctly calculated is found to be $1.08 \cdot {10}^{9} N$.

#### Explanation:

The first step is to find the distance between the two charges:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
$d = \sqrt{{\left(1 - \left(- 2\right)\right)}^{2} + {\left(\left(- 3\right) - 1\right)}^{2}}$
$d = \sqrt{{3}^{2} + {\left(- 4\right)}^{2}} = \sqrt{25} = 5$

So the points are $5 m$ apart.

Now use Coloumb's Law. The constant k has the value $9 \cdot {10}^{9} \left(N {m}^{2}\right) {C}^{-} 2$.

$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$ = $\frac{9 \cdot {10}^{9} \cdot - 2 \cdot - 1}{5} ^ 2$ = $\frac{2.7 \cdot {10}^{10}}{25}$

$F = 1.08 \cdot {10}^{9} N$

This is quite a large force, but it's unsurprising because a coulomb is quite a large amount of electric charge, and because the constant in Coulumb's Law is so large (electrostatic forces are much stronger than gravitational forces, for example).