# 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 ?

Jul 15, 2017

${F}_{\text{net}} = 4.725 \times {10}^{10}$ $N$ 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.

$F = \frac{k {q}_{1} {q}_{2}}{r} ^ 2$

For the $5$ $C$ charge:

$F = \frac{9 \times {10}^{9} \times 5 \times 1}{1} ^ 2 = 4.5 \times {10}^{10}$ $N$ 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:

$F = \frac{9 \times {10}^{9} \times - 1 \times 1}{2} ^ 2 = 2.25 \times {10}^{9}$ $N$, also in the positive direction

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

${F}_{\text{net}} = 4.5 \times {10}^{10} + 2.25 \times {10}^{9} = 4.725 \times {10}^{10}$ $N$ in the positive direction.