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

Feb 10, 2016

$F \cong 0 , 3699 \cdot {10}^{9} N$

#### Explanation:

$g i v e n \mathrm{da} t a :$
$\text{r :distance between two charges}$
${q}_{1} \text{ :2C charge}$
${q}_{2} \text{ :3C charge}$
$\text{k :"9*10^9" Coulomb's constant}$
$\text{formula for solving this problem:}$
$F = \frac{k \cdot {q}_{1} \cdot {q}_{2}}{r} ^ 2 \text{ (Coulomb's law)}$
$\text{find distance between two points.}$
$r = \sqrt{{\left(2 - \left(- 3\right)\right)}^{2} + {\left(- 5 - 7\right)}^{2}}$
r=sqrt(5^2+(-11)^2
$r = \sqrt{25 + 121}$
$r = \sqrt{146}$
${r}^{2} = 146 m$
$\text{fill data into formula}$
$F = \frac{9 \cdot {10}^{9} \cdot 2 \cdot 3}{146}$
$F = \frac{54 \cdot {10}^{9}}{146}$
$F \cong 0 , 3699 \cdot {10}^{9} N \text{ Force between two point}$