# A charge of -2 C is at (-6, 4 ) and a charge of 3 C is at ( -1, 5 ) . If both coordinates are in meters, what is the force between the charges?

Feb 3, 2018

The force is $= 2.08 \cdot {10}^{9} N$

#### Explanation:

The force is

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

The charge ${q}_{1} = - 2 C$

The charge ${q}_{2} = 3 C$

The Coulomb's constant is $k = 9 \cdot {10}^{9} N {m}^{2} {C}^{-} 2$

The distance between the $2$ charges is

$r = \sqrt{{\left(- 1 + 6\right)}^{2} + {\left(5 - 4\right)}^{2}} = \sqrt{26} m$

Therefore,

The force is

$| | \vec{F} | | = 9 \cdot {10}^{9} \cdot \left(2\right) \cdot \frac{3}{26} = 2.08 \cdot {10}^{9} N$

Since the chages are charged positive anf negative, the charges are attracted to one another with a force of $= {2.0810}^{9} N$ and the direction is the line joining the $2$ charges.