Question

A charge of `1 C` is at `(3,-2)` and a charge of `-3 C` is at `( -4,-1)`. If both coordinates are in meters, what is the force between the charges?

Answer 1

`5.4xx10^8N` attractive force.

Explanation:

The distance between them can be found by subtracting `x`'s and `y`'s and doing Pythagoras':

`r = sqrt((3--4)^2+(-2--1)^2) = sqrt50 = 5sqrt2`

The electrical force is given by

`F = k(Q_1Q_2)/r^2`

where `k~~9xx10^9` is Coulomb's constant, `Q` is the charges, and `r` is the distance which we just worked out.

Using the values we know,

`F = 9xx10^9 xx (1 xx -3)/(5sqrt2)^2 = (-2.7xx10^10)/50`

`=-5.4xx10^8N`

Since the charges are opposite, they are attractive, so even though the force has a negative symbol, that means they are pulling together, not pushing apart.