# Two charges of  3 C  and  -7 C are positioned on a line at points  6  and  5 , respectively. What is the net force on a charge of  2 C at  -2 ?

May 13, 2017

${\vec{F}}_{n e t} = 1.12 \times {10}^{10}$ $N$

#### 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 $2 C$ charge, we consider the force exerted on it by the $- 7 C$ and $3 C$ charges. Let's call the $2 C$ charge ${q}_{1}$, the $- 7 C$ charge ${q}_{2}$ and the $3 C$ 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. Therefore, the force vectors can be drawn in as follows: So we can calculate ${\vec{F}}_{21}$ and subtract from that ${\vec{F}}_{31}$ to find the net force on the ${q}_{1}$ charge.

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

$= \left(8.99 \times {10}^{9} N {m}^{2} / {C}^{2}\right) \cdot \frac{7 C \cdot 2 C}{7} ^ 2$

$= 2.57 \cdot {10}^{9}$ $N$

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

$= \left(8.99 \times {10}^{9} N {m}^{2} / {C}^{2}\right) \cdot \frac{3 C \cdot 2 C}{8} ^ 2$

$= 8.43 \cdot {10}^{8}$ $N$

Therefore, we have:

${\vec{F}}_{n e t} = 1.73 \cdot {10}^{9}$ $N$