Question

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?

Answer 1

`F_e = 2.42 xx 10^9` `"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(|q_1q_2|)/(r^2)`

where

• `k` is Coulomb's constant, equal to `8.988 xx 10^9("N"•"m"^2)/("C"^2)`

• `r` is the distance between the two point charges, in `"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"`