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

Jan 13, 2017

The net force is $1.07 \times {10}^{11} N$, acting to the right.

Explanation:

We solve this problem by separately calculating the force acting on the $4 C$ 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 $- 1 C$ charge. I will call this ${F}_{1}$

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

(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 unlike charges, they attract. So, ${F}_{1}$ points toward the $- 1 C$ charge - the to left.)

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

${F}_{1} = \left(\frac{1}{4 \pi {\epsilon}_{o}}\right) \left(\frac{{q}_{1} {q}_{2}}{r} ^ 2\right) = \frac{\left(9 \times {10}^{9}\right) \left(3\right) \left(4\right)}{1} ^ 2 = 1.08 \times {10}^{11} N$

This force acts toward the $- 3 C$ charge, meaning to the right.

Since the two forces act in opposite directions, we subtract the two values to get the net force

$1.08 \times {10}^{11} N - 1 \times {10}^{9} N = 1.07 \times {10}^{11} N$, acting to the right.