# A charge of 5 C is at (-6, 3 ) and a charge of -2 C is at (-5 , 7 ) . If both coordinates are in meters, what is the force between the charges?

##### 1 Answer
Dec 23, 2016

Recall that Coulomb's law tells you the electric force of attraction or repulsion between two point charges:

$\boldsymbol{{F}_{E} = \frac{k {q}_{1} {q}_{2}}{{r}^{2}} = \frac{{q}_{1} {q}_{2}}{4 \pi {\epsilon}_{0} {r}^{2}}}$

where $k = \frac{1}{4 \pi {\epsilon}_{0}}$ is a constant, ${\epsilon}_{0} = 8.854 \times {10}^{- 12} {\text{C"^2"/N"cdot"m}}^{2}$ is the vacuum permittivity, and $r$ is the distance between the two point charges.

Since you were given coordinates, recall the distance formula to calculate $r$:

$r = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$= \sqrt{{\left(- 5 - - 6\right)}^{2} + {\left(7 - 3\right)}^{2}}$

$= \sqrt{{\left(1\right)}^{2} + {\left(4\right)}^{2}} = \sqrt{17}$,

or, $\textcolor{g r e e n}{{r}^{2} = 17}$.

The constant $k$ is:

$\textcolor{g r e e n}{k} = \frac{1}{4 \pi {\epsilon}_{0}} = \frac{1}{\left(4\right) \left(3.1415926535 \cdots\right) \cdot \left(8.854 \times {10}^{- 12} {\text{C"^2"/N"cdot"m}}^{2}\right)}$

$= \textcolor{g r e e n}{8.987 \times {10}^{9}}$ $\textcolor{g r e e n}{{\text{N"cdot"m"^2"/C}}^{2}}$

So the electric force ${F}_{E}$ is:

$\textcolor{b l u e}{{F}_{E}} = \frac{k {q}_{1} {q}_{2}}{{r}^{2}}$

$= \left(\left(8.987 \times {10}^{9} {\text{N"cdotcancel"m"^2"/"cancel("C"^2))(5 cancel"C")(-2 cancel"C"))/(17 cancel("m}}^{2}\right)\right)$

$= \textcolor{b l u e}{- 5.287 \times {10}^{9}}$ $\textcolor{b l u e}{\text{N}}$

This negative value means that the two point charges are highly attracted to each other. This makes sense because they're oppositely-signed charges.