# Two charges of  -6 C  and  4 C are positioned on a line at points  -2  and  9 , respectively. What is the net force on a charge of  -1 C at  1 ?

Dec 30, 2015

${F}_{3} = 6.5625 \cdot {10}^{9} N$

#### Explanation:

Consider the figure. Let the charges $- 6 C , 4 C$ and $- 1 C$ be denoted by ${q}_{1} , {q}_{2}$ and ${q}_{3}$ respectively.
Let the positions at which charges are placed be in the units of meters.

Let ${r}_{13}$be the distance between the charges ${q}_{1} \mathmr{and} {q}_{3}$.
From figure
${r}_{13} = 1 - \left(- 2\right) = 1 + 2 = 3 m$

Let ${r}_{23}$be the distance between the charges ${q}_{2} \mathmr{and} {q}_{3}$.
From figure
${r}_{23} = 9 - 1 = 8 m$

Let ${F}_{13}$ be the force due to charge ${q}_{1}$ on the charge ${q}_{3}$
${F}_{13} = \frac{k {q}_{1} {q}_{3}}{r} _ {13}^{2} = \frac{9 \cdot {10}^{9} \cdot \left(6\right) \left(1\right)}{3} ^ 2 = 6 \cdot {10}^{9} N$
This force is repulsive and is towards charge ${q}_{2}$.

Let ${F}_{23}$ be the force due to charge ${q}_{2}$ on the charge ${q}_{3}$
${F}_{23} = \frac{k {q}_{2} {q}_{3}}{r} _ {23}^{2} = \frac{9 \cdot {10}^{9} \cdot \left(4\right) \left(1\right)}{8} ^ 2 = 0.5625 \cdot {10}^{9} N$
This force is attractive and is towards charge ${q}_{2}$.

The total force or net force on charge ${q}_{3}$ is the sum of above two forces.
Since the above two forces ${F}_{13}$ and ${F}_{23}$ are in same direction therefore they can be added directly.
Let ${F}_{3}$ be the total force on the charge ${q}_{3}$.
$\implies {F}_{3} = {F}_{13} + {F}_{23} = 6 \cdot {10}^{9} + 0.5625 \cdot {10}^{9} = 6.5625 \cdot {10}^{9} N$
$\implies {F}_{3} = 6.5625 \cdot {10}^{9} N$
Since ${F}_{13} \mathmr{and} {F}_{23}$ are toward the charge ${q}_{2}$ therefore the force ${F}_{3}$ is also toward the charge ${q}_{2}$.