Question

Two charges of `5` `C` and `-1` `C` are positioned on a line at points `-2` and `1`, respectively. What is the net force on a charge of `1` `C` at `-1`?

Answer 1

`F_"net" = 4.725xx10^10` `N` in the positive direction.

Explanation:

If we imagine the situation (or, better, draw a large, clear, labeled diagram!) we know that the `5` `C` charge is `1` `m` from the `1` `C` charge and the `-1` `C` charge is `2` `m` from it in the opposite direction.

(We were not told the units of distance on the line but we need to assume it was `m` to be able to solve the problem)

It's also worth realising that the force on the `1` `C` charge due to the `5` `C` charge will be large and repulsive (i.e. in the positive direction along the line), and the force on it due to the `-1` `C` charge will be smaller but attractive and therefore also in the positive direction.

`F=(kq_1q_2)/r^2`

For the `5` `C` charge:

`F=(9xx10^9xx5xx1)/1^2=4.5xx10^10` `N` in the positive direction.

(if we drew our number line in the usual left-to-right direction, this force is acting to the right)

For the `-1` `C` charge:

`F=(9xx10^9xx-1xx1)/2^2=2.25xx10^9` `N`, also in the positive direction

The net force is simply the sum of these two forces:

`F_"net" = 4.5xx10^10 + 2.25xx10^9 = 4.725xx10^10` `N` in the positive direction.