# A charge of 5 C is at (-6, 1 ) and a charge of -3 C is at (-2, 1) . If both coordinates are in meters, what is the force between the charges?

Mar 20, 2016

The force between the charges is $8 \setminus \times {10}^{9}$ N.

#### Explanation:

Use Coulomb's law:
$F = \setminus \frac{k \setminus \left\mid {q}_{1} {q}_{2} \right\mid}{{r}^{2}}$

Calculate $r$, the distance between the charges, using the Pythagorean theorem
${r}^{2} = \setminus \Delta {x}^{2} + \setminus \Delta {y}^{2}$
${r}^{2} = {\left(- 6 - \left(- 2\right)\right)}^{2} + {\left(1 - 1\right)}^{2}$
${r}^{2} = {\left(- 6 + 2\right)}^{2} + {\left(1 - 1\right)}^{2}$
${r}^{2} = {4}^{2} + {0}^{2}$
${r}^{2} = 16$
$r = 4$

The distance between the charges is $4$m. Substitute this into Coulomb's law. Substitute in the charge strengths as well.

$F = \setminus \frac{k \setminus \left\mid {q}_{1} {q}_{2} \right\mid}{{r}^{2}}$
$F = k \setminus \frac{\setminus \left\mid \left(5\right) \left(- 3\right) \right\mid}{{4}^{2}}$
$F = k \setminus \frac{15}{16}$
F = 8.99×10^9(\frac{15}{16}) (Substitute in the value of Coulomb's constant)
$F = 8.4281 \setminus \times {10}^{9}$ N
$F = 8 \setminus \times {10}^{9}$ N (As you're working with one significant figure)