# A charge of 1 C is at (-2,7,1) and a charge of 2 C is at (8,-1,4) . If both coordinates are in meters, what is the force between the charges?

Jan 31, 2018

$F = 1.04 \times {10}^{8} N$

#### Explanation:

Charge $1 C$ placed at $\left(- 2 , 7 , 1\right)$
Charge $2 C$ placed at $\left(8 , - 1 , 4\right)$

Distance between the charges
r=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2
r=sqrt((8-(-2))^2+((-1)-7)^2+(4-1)^2
$r = \sqrt{{\left(8 + 2\right)}^{2} + {\left(- 1 - 7\right)}^{2} + {\left(4 - 1\right)}^{2}}$
$r = \sqrt{{10}^{2} + {\left(- 8\right)}^{2} + {\left(3\right)}^{2}}$
$r = \sqrt{100 + 64 + 9} = \sqrt{173}$

Force between charges
$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$
where $k = 9 \times {10}^{9} N - {m}^{2} / {C}^{2}$

$F = \frac{9 \times {10}^{9} \times 1 \times 2}{\sqrt{173}} ^ 2$
$F = \frac{18 \times {10}^{9}}{173} = 0.104 \times {10}^{9}$
$F = 1.04 \times {10}^{8} N$