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

Jan 10, 2016

The distance formula for Cartesian coordinates is

d=sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2
Where ${x}_{1} , {y}_{1} , {z}_{1}$, and${x}_{2} , {y}_{2} , {z}_{2}$ are the Cartesian coordinates of two points respectively.
Let $\left({x}_{1} , {y}_{1} , {z}_{1}\right)$ represent $\left(- 2 , 4 , 8\right)$ and $\left({x}_{2} , {y}_{2} , {z}_{2}\right)$ represent $\left(2 , 4 , 8\right)$.

d=sqrt((2-(-2))^2+(4-4)^2+(8-8)^2

d=sqrt((2+2)^2+(0)^2+(0)^2

$d = \sqrt{{\left(4\right)}^{2} + 0 + 0} = \sqrt{16} = 4$

Hence the distance between the two charges is $4$ meters.

The electrostatic force between two charges is given by
$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

Where $F$ is the force between the charges, $k$ is the constant

and its value is $9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$,${q}_{1}$ and ${q}_{2}$ are the

magnitudes of the charges and $r$ is the distance between the two charges.

Here F=??, $k = 9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$, ${q}_{1} = - 2 C$, ${q}_{2} = 4 C$ and $r = d = 4$.

$\implies F = \frac{9 \cdot {10}^{9} \cdot \left(- 2\right) \cdot 4}{4} ^ 2 = \frac{9 \cdot {10}^{9} \cdot \left(- 8\right)}{16} = \frac{- 72 \cdot {10}^{9}}{16} = - 4.5 \cdot {10}^{9} N$

$\implies F = - 4.5 \cdot {10}^{9} N$

Negative sign shows that the force between the two charges is attractive.