Question

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

Answer 1

The net force will be `4.50 xx10^9N` acting to the right (toward the `-3C` charge at -1.

Explanation:

We solve this problem by separately calculating the force acting on the `-3C` charge due to each of the other two charges, then add them together. In this, I will assume the line referred to is measured in metres.

First, the force due to the `-4C` charge. I will call this `F_4`

`F_4=(1/(4pi epsilon_o)) ((q_1q_2)/r^2) = ((9xx10^9)(3)(4))/4^2=6.75xx10^9 N`

(Notice I have deliberately left out the signs on the charges. This is because I prefer to let the formulas determine the magnitude of the force. I already know the direction. Since these are like charges, they repel. So, `F_4` points away from the `-4C` charge - the to right.)

Next `F_3`, the force due to the `-3C` charge

`F_3=(1/(4pi epsilon_o)) ((q_1q_2)/r^2) = ((9xx10^9)(3)(3))/6^2=2.25xx10^9 N`

This force also repels the -3C charge at -1, meaning it acts to the left.

Since the two forces act in opposite directions, we subtract the two values to get the net force. Since `F_4` is the greater force, the net force acts to the right.

`6.75xx10^9 N-2.25xx10^9N = 4.50xx10^9N`