# Two charges of  -4 C  and  -3 C are positioned on a line at points  -8  and  16 , respectively. What is the net force on a charge of  2 C at  -1 ?

Dec 24, 2015

No units are given for distance so I have assumed they are in metres for which:

$F = - 1.28 \times {10}^{9} \text{N}$

#### Explanation: The force between two charges ${q}_{1}$ and ${q}_{2}$ separated by a distance $r$ is given by Coulombs' Law:

$F = \frac{1}{4 \pi {\epsilon}_{0}} . \frac{{q}_{1} {q}_{2}}{r} ^ 2$

This simplifies down to:

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

$k = 9 \times {10}^{9} \text{m""/""F}$ and is the constant of proportionality.

The net force on ${q}_{2}$ will be the vector sum of the other two forces:

${F}_{r e s} = - {F}_{12} + {F}_{32}$

$= - k \frac{4 \times 2}{7} ^ 2 + k \frac{3 \times 2}{17} ^ 2$

$= k \left(\frac{6}{289} - \frac{8}{49}\right)$

$= - k \times 0.1423$

${F}_{r e s} = - 9 \times {10}^{9} \times 0.1423 = - 1.28 \times {10}^{9} \text{N}$

The size of the force reflects how large a unit the coulomb is.