# Two charges of  -2 C  and  3 C are positioned on a line at points  3  and  4 , respectively. What is the net force on a charge of  -9 C at  0 ?

Jul 22, 2017

$2.81 \cdot {10}^{9} N$

#### Explanation:

Coulomb's law: $F = \frac{{k}_{e} {Q}_{1} {Q}_{2}}{r} ^ 2$, where:
$F$ = the force ($N$)
${k}_{e}$ = the coulomb constant($\approx 8.99 \cdot {10}^{9} N$ ${m}^{2}$ ${C}^{- 2}$)
Q_1 &Q_2 = the charge of each particle ($C$)
$r$ = the distance ($m$)

Assuming $1$ $\text{unit} = 1 m$

Also, assuming a positive resultant force = repulsion, and a negative resultant force = attraction.

For $- 9 C$ and $- 2 C$:
$F = \frac{\left(8.99 \cdot {10}^{9}\right) \left(- 9\right) \left(- 2\right)}{{3}^{2}} = 1.798 \cdot {10}^{10} N$

For $- 9 C$ and $3 C$:
$F = \frac{\left(8.99 \cdot {10}^{9}\right) \left(- 9\right) \left(3\right)}{{4}^{2}} = = - 1.5170625 \cdot {10}^{10} N$

Net force = $\left(1.798 \cdot {10}^{10}\right) - \left(1.5170625 \cdot {10}^{10}\right) = 2.809375 \cdot {10}^{9} \approx 2.81 \cdot {10}^{9} N$