Question

Two charges of `2` `C` and `4` `C` are positioned on a line at points `-5` and `-2`, respectively. What is the net force on a charge of `3` `C` at `1`?

Answer 1

The forces on the `3` `C` charge due to each of the other charges can simply be added together.

The net force is `F = (kxx2xx3)/6^2+(kxx4xx3)/3^2 = 1.35 xx 10^10` `C`

Explanation:

The charges in question are all positive, so the forces will be repulsive. The `3` `C` charge will be experiencing a force to the right, if we imagine the number line.

The distance from the `2` `C` charge to the `3` `C` charge is `6` `m` (we are not told the units of distance being used, but metres are the appropriate SI unit). The distance from the `4` `C` charge to the `3` `C` charge is `3` `m`.

The net force, then, is:

`F = (kxx2xx3)/6^2+(kxx4xx3)/3^2`

where `k = 9xx10^9` `Nm^2C^-2`

`F = (kxx2xx3)/6^2+(kxx4xx3)/3^2 = 1.35 xx 10^10` `C`

This is a very large force, but these are very large charges, quite close together, and the constant is large.