# A charge of 1 C is at (3,-2) and a charge of -3 C is at ( -4,-1) . If both coordinates are in meters, what is the force between the charges?

Mar 5, 2017

$5.4 \times {10}^{8} N$ attractive force.

#### Explanation:

The distance between them can be found by subtracting $x$'s and $y$'s and doing Pythagoras':

$r = \sqrt{{\left(3 - - 4\right)}^{2} + {\left(- 2 - - 1\right)}^{2}} = \sqrt{50} = 5 \sqrt{2}$

The electrical force is given by

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

where $k \approx 9 \times {10}^{9}$ is Coulomb's constant, $Q$ is the charges, and $r$ is the distance which we just worked out.

Using the values we know,

$F = 9 \times {10}^{9} \times \frac{1 \times - 3}{5 \sqrt{2}} ^ 2 = \frac{- 2.7 \times {10}^{10}}{50}$

$= - 5.4 \times {10}^{8} N$

Since the charges are opposite, they are attractive, so even though the force has a negative symbol, that means they are pulling together, not pushing apart.