# Two charges of  -2 C  and  5 C are positioned on a line at points  4  and  -1 , respectively. What is the net force on a charge of  -9 C at  0 ?

#### Explanation:

The electrostatic force $F$ between two point charges ${Q}_{1}$ & ${Q}_{2}$ separated by a distance $r$ is given as

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

Where, $K$ is constant & $K = 9 \setminus \times {10}^{9} \setminus \frac{N {m}^{2}}{C} ^ 2$

Now, the electrostatic force ${F}_{1}$ of repulsion between point charges ${Q}_{1} = - 2 C$ & ${Q}_{2} = - 9 \setminus C$ separated by a distance $r = 4$ is given as

${F}_{1} = \setminus \frac{K \left(2\right) \left(9\right)}{{4}^{2}} = \frac{9}{8} K$

Similarly, the electrostatic force ${F}_{2}$ of attraction between point charges ${Q}_{1} = 5 C$ & ${Q}_{2} = - 9 \setminus C$ separated by a distance $r = 1$ is given as

${F}_{2} = \setminus \frac{K \left(5\right) \left(9\right)}{{1}^{2}} = 45 K$

Since ${F}_{1}$ & ${F}_{2}$ on $- 9 C$ are in the same direction hence the net force on $- 9 C$ considering direction of both the forces

${F}_{\setminus \textrm{\ne t}} = {F}_{1} + {F}_{2}$

$= \frac{9}{8} K + 45 K$

$= \frac{369}{8} K$