# Two charges of  -2 C  and  4 C  are at points  (4, -2, -1)  and  ( -1 , 3, -8 ), respectively. Assuming that both coordinates are in meters, what is the force between the two points?

$F = 726 , 464 , 646.5 \text{ ""Newtons}$

#### Explanation:

Compute the distance $r$ between the two charges

Let ${P}_{2} \left({x}_{2} , {y}_{2} , {z}_{2}\right) = \left(4 , - 2 , - 1\right)$
Let ${P}_{1} \left({x}_{1} , {y}_{1} , {z}_{1}\right) = \left(- 1 , 3 , - 8\right)$

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

$r = \sqrt{{\left(4 - - 1\right)}^{2} + {\left(- 2 - 3\right)}^{2} + {\left(- 1 - - 8\right)}^{2}}$

$r = \sqrt{25 + 25 + 49}$

$r = \sqrt{99} \text{ }$meters

$F = \frac{{k}_{e} \cdot \left\mid {q}_{1} \cdot {q}_{2} \right\mid}{r} ^ 2$

$F = \frac{\left(8.99 x {10}^{9}\right) \cdot \left\mid - 2 \cdot 4 \right\mid}{\sqrt{99}} ^ 2$

$F = 726 , 464 , 646.5 \text{ ""Newtons}$

God bless....I hope the explanation is useful.