# Two charges of  -2 C  and  7 C are positioned on a line at points  5  and  -2 , respectively. What is the net force on a charge of  3 C at  1 ?

Feb 9, 2016

${F}_{N e t} = {F}_{A} + {F}_{B} = 50.5687 \setminus \times {10}^{9} N$

#### Explanation:

Let us label the charges as: ${Q}_{A} = - 2 C$; ${Q}_{B} = + 7 C$ and $Q = + 3 C$

Note: You have not mentioned the units of the distance. I assume it to be meters.

Let us label the positions of these charges as : ${x}_{A} = + 5$m, ${x}_{B} = - 1$m; and $x = + 1$m.

The charge $Q$ is placed between ${Q}_{A}$ and ${Q}_{B}$. Because $Q$ and ${Q}_{A}$ are charges of opposite signs they interact attractively and $Q$ is pulled towards ${Q}_{A}$. Since $Q$ and ${Q}_{B}$ are of the same sign they interact repulsively and $Q$ is pushed away from ${Q}_{B}$. So $Q$ feels a net force toward ${Q}_{A}$.

Now apply Coulomb's law to get their magnitudes,
${F}_{A} = k \setminus \frac{| {Q}_{A} Q |}{{\left({x}_{A} - x\right)}^{2}}$
$\setminus q \quad$ $= \left(8.99 \setminus \times {10}^{9} \frac{N . {m}^{2}}{C} ^ 2\right) \left(\setminus \frac{| \left(- 2 C\right) \left(+ 3 C\right) |}{{\left(5 m - 1 m\right)}^{2}}\right)$
$\setminus q \quad$$= 3.37125 \setminus \times {10}^{9} N$

${F}_{B} = k \setminus \frac{| {Q}_{B} Q |}{{\left(x - {x}_{B}\right)}^{2}}$
$\setminus q \quad$ $= \left(8.99 \setminus \times {10}^{9} \frac{N . {m}^{2}}{C} ^ 2\right) \left(\setminus \frac{| \left(+ 7 C\right) \left(+ 3 C\right) |}{{\left(1 m - \left(- 1 m\right)\right)}^{2}}\right)$
$\setminus q \quad$$= 47.1975 \setminus \times {10}^{9} N$

${F}_{N e t} = {F}_{A} + {F}_{B} = 50.5687 \setminus \times {10}^{9} N$