# Two charges of  1 C  and  -2 C  are at points  (0,-1,2)  and  ( -2,5,7), respectively. Assuming that both coordinates are in meters, what is the force between the two points?

Jul 18, 2017

${F}_{e} = 2.77 \times {10}^{8}$ $\text{N}$

#### Explanation:

We're asked to find the magnitude of the electric force ${F}_{e}$ between two point charges.

To do this, we can use the equation

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

where

• ${q}_{1}$ and ${q}_{2}$ are the point charges ($1$ $\text{C}$ and $- 2$ $\text{C}$) , in no particular order

• $k$ 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 here can be found by the distance formula:

$r = \sqrt{{\left(0 - \left(- 2\right)\right)}^{2} + {\left(- 1 - 5\right)}^{2} + {\left(2 - 7\right)}^{2}}$

$= 8.06$ $\text{m}$

Plugging in known values, we have

F_e = (8.988xx10^9("N"·"m"^2)/("C"^2))(|(1cancel("C"))(-2cancel("C"))|)/((8.06cancel("m")^2))

= color(red)(2.77xx10^8 color(red)("N"