# Two charges of  -2 C  and  6 C are positioned on a line at points  3  and  4 , respectively. What is the net force on a charge of  -5 C at  0 ?

Mar 5, 2017

$6.9 \times {10}^{9} N$ attractive force.

#### Explanation:

The electrical force is given by

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

where $k \approx 9 \times {10}^{9}$ is Coulomb's constant, $Q$ is the charge and $r$ is the distance between them.

The force between the $- 2 C$ and $- 5 C$ charges is

$F = 9 \times {10}^{9} \times \frac{- 2 \times - 5}{3} ^ 2 = {10}^{10} N$

and, between the $6 C$ and $- 5 C$ charges, is

$F = 9 \times {10}^{9} \times \frac{6 \times - 5}{4} ^ 2 = - 1.69 \times {10}^{10}$

The net force is the sum of the forces, so

${F}_{\text{net}} = {10}^{10} + \left(- 1.69 \times {10}^{10}\right) = - 6.9 \times {10}^{9} N$

This is in the negative direction, which is the same direction as the force between the $- 5 C$ and $6 C$ charges, which must attract because the are opposites. Therefore, the net force is attractive, too.