# 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?

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.