# A charge of 3 C is at (3, -1 ) and a charge of -4 C is at ( -3, 7 ). If both coordinates are in meters, what is the force between the charges?

Jul 6, 2018

$F = - 1.08 \cdot {10}^{9} N$

#### Explanation:

Charge ${q}_{1} = 3 C$ is placed at $\left(3 , - 1\right)$
Charge ${q}_{2} = - 4 C$ is placed at $\left(- 3 , 7\right)$

Distance between both charges is
r=sqrt((x_2-x_1)^2+(y_2-y_1)^2
r=sqrt((-3-(3))^2+(7-(-1))^2
r=sqrt((-6)^2+(8)^2
$r = \sqrt{36 + 64}$
$r = \sqrt{100} m = 10 m$ and ${r}^{2} = 100$

Force between two charges ${q}_{1}$ and ${q}_{2}$ is
$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$ where $k = 9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$
$\implies F = \frac{\left(9 \cdot {10}^{9}\right) \left(3\right) \left(- 4\right)}{100}$
$\implies F = \frac{\left(9 \cdot {10}^{9}\right) \left(- 12\right)}{100}$
$\implies F = \frac{\left(- 108 \cdot {10}^{9}\right)}{100} N$
$\implies F = - 1.08 \cdot {10}^{9} N$