# Two charges of  -2 C  and  3 C are positioned on a line at points  5  and  -6 , respectively. What is the net force on a charge of  -1 C at  0 ?

Feb 12, 2016

${F}_{n} = 3 \cdot {10}^{7}$

#### Explanation:

$F : \text{ force between two charges}$
$F = k \cdot \frac{{q}_{1} \cdot {q}_{2}}{r} ^ 2 \text{ Coulomb's law}$
$x : \text{ distance between the charge of 3C and -1C}$
$x = 6 - 0 = 6$
$y : \text{ distance between the charge of -1C and -2C}$
$y : 5 - 0 = 5$
${F}_{1} : \text{ Force between the charge of 3C and -1C}$
${F}_{1} = k \cdot \frac{3 \cdot \left(- 1\right)}{6} ^ 2$
${F}_{1} = \frac{- 3 \cdot k}{36}$
${F}_{2} : \text{ Force between the charge of -1C and -2C}$
${F}_{2} = \frac{k \cdot \left(- 1\right) \cdot \left(- 2\right)}{5} ^ 2$
${F}_{2} = \frac{2 \cdot k}{25}$
${F}_{n} = \frac{- 3 \cdot k}{36} + \frac{2 \cdot k}{25}$
${F}_{n} = \frac{- 75 \cdot k + 72 \cdot k}{36 \cdot 25}$
${F}_{n} = \frac{- \cancel{3} \cdot k}{\cancel{36} \cdot 25}$
${F}_{n} = \frac{k}{12 \cdot 25} \text{ , } k = 9 \cdot {10}^{9}$
${F}_{n} = \frac{\cancel{9} \cdot {10}^{9}}{\cancel{12} \cdot 25} \text{ ; } {F}_{n} = \frac{3 \cdot {10}^{9}}{4 \cdot 25}$
${F}_{n} = 3 \cdot {10}^{7}$