Question

Two charges of `-2 C` and `6 C` are positioned on a line at points `3` and `4`, respectively. What is the net force on a charge of `-5 C` at `0`?

Answer 1

`6.9xx10^9N` attractive force.

Explanation:

The electrical force is given by

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

where `k ~~ 9xx10^9` is Coulomb's constant, `Q` is the charge and `r` is the distance between them.

The force between the `-2C` and `-5C` charges is

`F = 9xx10^9 xx (-2 xx -5)/3^2 = 10^10N`

and, between the `6C` and `-5C` charges, is

`F = 9xx10^9 xx (6 xx -5)/4^2 = -1.69 xx 10^10`

The net force is the sum of the forces, so

`F_("net") = 10^10 + (-1.69xx10^10) = -6.9xx10^9N`

This is in the negative direction, which is the same direction as the force between the `-5C` and `6C` charges, which must attract because the are opposites. Therefore, the net force is attractive, too.