# Two charges of  -7 C  and  4 C  are at points  (4, 7 ,-8)  and  ( 2 ,-3, -8 ), respectively. Assuming that both coordinates are in meters, what is the force between the two points?

Jul 10, 2017

${F}_{e} = 2.42 \times {10}^{9}$ $\text{N}$

#### Explanation:

We're asked to find he magnitude of the electric force ${F}_{e}$ between two point charges, which we'll call ${q}_{1}$ and ${q}_{2}$.

To do this, we can use the equation

${F}_{e} = k \frac{| {q}_{1} {q}_{2} |}{{r}^{2}}$

where

• $k$ is Coulomb's constant, equal to $8.988 \times {10}^{9} \left({\text{N"•"m"^2)/("C}}^{2}\right)$

• $r$ is the distance between the two point charges, in $\text{m}$

This distance can be found via the distance formula:

r = sqrt((4-2)^2 + (7-(-3))^2 + (-8-(-8))^2) = color(red)(10.2 color(red)("m"

Plugging in known values, we have

F_e = (8.988 xx 10^9("N"•cancel("m"^2))/(cancel("C"^2)))((|(-7cancel("C"))(4cancel("C"))|)/((color(red)(10.2)cancel(color(red)("m")))^2))

= color(blue)(2.42xx10^9 color(blue)("N"