# 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 ?

Jan 28, 2017

The net force will be $4.50 \times {10}^{9} N$ acting to the right (toward the $- 3 C$ charge at -1.

#### Explanation:

We solve this problem by separately calculating the force acting on the $- 3 C$ 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 $- 4 C$ charge. I will call this ${F}_{4}$

${F}_{4} = \left(\frac{1}{4 \pi {\epsilon}_{o}}\right) \left(\frac{{q}_{1} {q}_{2}}{r} ^ 2\right) = \frac{\left(9 \times {10}^{9}\right) \left(3\right) \left(4\right)}{4} ^ 2 = 6.75 \times {10}^{9} N$

(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 $- 4 C$ charge - the to right.)

Next ${F}_{3}$, the force due to the $- 3 C$ charge

${F}_{3} = \left(\frac{1}{4 \pi {\epsilon}_{o}}\right) \left(\frac{{q}_{1} {q}_{2}}{r} ^ 2\right) = \frac{\left(9 \times {10}^{9}\right) \left(3\right) \left(3\right)}{6} ^ 2 = 2.25 \times {10}^{9} N$

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.

$6.75 \times {10}^{9} N - 2.25 \times {10}^{9} N = 4.50 \times {10}^{9} N$