# Two charges of  -7 C  and  -2 C are positioned on a line at points  -3  and  6 , respectively. What is the net force on a charge of  -8 C at  -2 ?

Dec 24, 2017

The force is $= + {506.2510}^{9} N$

#### Explanation:

The net force is

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

The charge ${q}_{1} = - 7 C$

The charge ${q}_{2} = - 2 C$

The charge ${q}_{3} = - 8 C$

The distance ${r}_{1} = - 2 - \left(- 3\right) = 1 m$

The distance ${r}_{2} = 6 - \left(- 2\right) = 8 m$

The constant $k = 9 \cdot {10}^{9} N {m}^{2} {C}^{-} 2$

Therefore,

$F = 9 \cdot {10}^{9} \left(\frac{\left(- 7\right) \cdot \left(- 8\right)}{1} ^ 2 + \frac{\left(- 8\right) \cdot \left(- 2\right)}{8} ^ 2\right)$

$= + {910}^{9} \left(56 + \frac{1}{4}\right) N$

$= + 506.25 \cdot {10}^{9} N$

The $+$ sign indicates that the force is repulsive.