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

Aug 4, 2017

${\vec{F}}_{n e t} = 6.76 \cdot {10}^{9}$ $N$ (to the right)

#### 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 $8.99 \times {10}^{9} N {m}^{2} / {C}^{2}$, sometimes referred to as Coulomb's constant.

Here is a quick diagram of the situation:

To find the net force on the $3 C$ charge, we consider the force exerted on it by the $9 C$ and $2 C$ charges. Let's call the $3 C$ charge ${q}_{1}$, the $9 C$ charge ${q}_{2}$ and the $2 C$ charge ${q}_{3}$. Then:

${\vec{F}}_{n e t} = \sum \vec{F} = {\vec{F}}_{2 o n 1} + {\vec{F}}_{3 o n 1}$

Recall that like charges repel, while opposite charges attract. Each of our charges are positive. Therefore, the force vectors can be drawn in as follows:

23 should say 21 in the above diagram

where $21 \left(23\right)$ is the force of ${q}_{1}$ on ${q}_{2}$ and $31$ is the force of ${q}_{3}$ on ${q}_{1}$

So we can calculate ${\vec{F}}_{31}$ and subtract from that ${\vec{F}}_{21}$ to find the net force on the ${q}_{1}$ charge.

$| {\vec{F}}_{31} | = k \cdot \frac{| {q}_{1} | | {q}_{3} |}{{r}_{31}^{2}}$

$= \left(8.99 \times {10}^{9} N {m}^{2} / {C}^{2}\right) \cdot \frac{3 C \cdot 2 C}{2} ^ 2$

$= 1.35 \cdot {10}^{10}$ $N$

$| {\vec{F}}_{21} | = k \cdot \frac{| {q}_{1} | | {q}_{2} |}{{r}_{21}^{2}}$

$= \left(8.99 \times {10}^{9} N {m}^{2} / {C}^{2}\right) \cdot \frac{3 C \cdot 9 C}{6} ^ 2$

$= 6.74 \cdot {10}^{9}$ $N$

Therefore, we have:

${\vec{F}}_{n e t} = 6.76 \cdot {10}^{9}$ $N$ (to the right)