# Two charges of  -6 C  and  -3 C are positioned on a line at points  -5  and  8 , respectively. What is the net force on a charge of  4 C at  2 ?

Aug 13, 2017

The net force is $1.40 \times {10}^{9} \text{ N}$ to the left.

#### Explanation:

The force between two charges is given by Coulomb's law:

$| \vec{F} | = k \frac{| {q}_{1} | | {q}_{2} |}{r} ^ 2$

where ${q}_{1}$ and ${q}_{2}$ are the magnitudes of the charges, $r$ is the distance between them, and $k$ is a constant equal to $8.99 \times {10}^{9} {\text{ Nm"^2//"C}}^{2}$, sometimes referred to as Coulomb's constant.

• Since unspecified, I'll assume radii values to be in meters so that the units work out.

Here is a diagram of the situation:

To find the net force on the $4 C$ charge, we consider the forces exerted on it by the $- 6 C$ and $- 3 C$ charges.

We have the following information:

• $\mapsto {Q}_{1} = - 6 \text{C at } - 5$
• $\mapsto {Q}_{2} = 4 \text{C at } 2$
• $\mapsto {Q}_{3} = - 3 \text{C at } 8$
• $\mapsto k = 8.99 \times {10}^{9} {\text{ Nm"^2//"C}}^{2}$

$\textcolor{b l u e}{{\vec{F}}_{\text{net on 3}} = \sum \vec{F} = {\vec{F}}_{1 o n 2} + {\vec{F}}_{3 o n 2}}$

Recall that like charges repel, while opposite charges attract. Each of the charges acting on ${Q}_{2}$ are negative. Therefore, the force vectors can be drawn in as follows:

where ${\vec{F}}_{1 \text{ on } 2}$ is the force of ${Q}_{1}$ on ${Q}_{2}$ and ${\vec{F}}_{3 \text{ on } 2}$ is the force of ${Q}_{3}$ on ${Q}_{2}$

So, we can calculate ${\vec{F}}_{1 \text{ on } 2}$ and subtract from the value we obtain from ${\vec{F}}_{3 \text{ on } 2}$ to find the net force on the ${Q}_{2}$ charge.

$\left\mid {\vec{F}}_{1 \text{ on } 2} \right\mid = k \cdot \frac{| {q}_{1} | | {q}_{2} |}{{r}_{12}^{2}}$

$= {\left(8.99 \times {10}^{9} \text{ Nm"^2//"C"^2)*(6C*4C)/(7"m}\right)}^{2}$

$\textcolor{b l u e}{= 4.40 \times {10}^{9} \text{ N}}$

$\left\mid {\vec{F}}_{3 \text{ on } 2} \right\mid = k \cdot \frac{| {q}_{3} | | {q}_{2} |}{{r}_{32}^{2}}$

$= \left(8.99 \times {10}^{9} {\text{ Nm"^2//"C}}^{2}\right) \cdot \frac{3 C \cdot 4 C}{6} ^ 2$

$\textcolor{b l u e}{= 3.00 \times {10}^{9} \text{ N}}$

Therefore, we have:

$\textcolor{b l u e}{{\vec{F}}_{n e t} = - 1.40 \times {10}^{9} \text{ N}}$

That is, $1.40 \times {10}^{9} \text{ N}$ to the left.

$- - - - - - - -$

Here is what the forces exerted on charges by other charges of the same or different sign look like: