# Two charges of  -1 C  and  -3 C  are at points  (-2 ,4,-3 )  and  ( -3, 5,-81 ), respectively. Assuming that both coordinates are in meters, what is the force between the two points?

Jun 14, 2017

$4.41 \times {10}^{6}$ $\text{N}$

#### Explanation:

The electric force $F$ between two point charges ${q}_{1}$ and ${q}_{2}$ separated by a distance $r$ is given by the equation

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

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

To find the distance between the two objects, we simply find the distance between the two coordinate points by

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

$= \sqrt{{\left(- 3 \text{m"--2"m")^2 + (5"m"-4"m")^2 + (-81"m"--3"m}\right)}^{2}}$

= color(red)(78.2 color(red)("m"

Since we have all our known variables, let's plug them in to the equation

F = (8.988 xx 10^9 ("N"·cancel("m"^2))/(cancel("C"^2)))((|(-1cancel("C"))(-3cancel("C"))|)/(color(red)(78.2)cancel(color(red)("m")))^2)

= color(blue)(4.41 xx 10^6 color(blue)("N"