# Two charges of  -5 C  and  -3 C are positioned on a line at points  7  and  4 , respectively. What is the net force on a charge of  -1 C at  0 ?

Jan 27, 2018

The net force is $= 2.61 \cdot {10}^{9} N$

#### Explanation:

The net force between the charges is

$F = k \frac{{Q}_{1} {Q}_{3}}{r} _ {1}^{2} + k \frac{{Q}_{2} {Q}_{3}}{r} _ {2}^{2}$

Where $r$ is the distance between the charges

The Coulomb Constant is $k = 9 \cdot {10}^{9} N {m}^{2} / {C}^{2}$

The charge ${Q}_{1} = - 5 C$ at $\left(7 , 0\right)$

The charge ${Q}_{2} = - 3 C$ at $\left(4 , 0\right)$

The charge ${Q}_{3} = - 1 C$ at $\left(0 , 0\right)$

The net force is

$F = {910}^{9} \cdot \left(- 5\right) \cdot \frac{- 1}{7} ^ 2 + {910}^{9} \left(- 3\right) \cdot \frac{- 1}{4} ^ 2 = 2.61 \cdot {10}^{9} N$

The positive sign indicates that the force is repulsive