# Two charges of  -3 C  and  4 C are positioned on a line at points  -3  and  4 , respectively. What is the net force on a charge of  7 C at  6 ?

Mar 12, 2016

${F}_{\text{net}} = 60 , 66 \cdot {10}^{9} N$

#### Explanation: $\text{The force between two charges is given by Coulomb's formula: }$
$F = k \cdot \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2 \text{ } k = 9 \cdot {19}^{9}$
$\text{The force between A and C is :"F_(AC)" (the blue vector)}$
$\text{The Force between B and C is :"F_(BC)" (the red vector)}$
${F}_{A C} = k \frac{\left(- 3\right) \cdot 7}{9} ^ 2 = - \frac{21 k}{81}$
${F}_{B C} = k \cdot \frac{4 \cdot 7}{2} ^ 2 = \frac{28 k}{4}$
$\text{Net Force on a charge of 7C:"F_"net} = {F}_{A C} + {F}_{B C}$
${F}_{\text{net}} = - \frac{21 k}{81} + \frac{28 k}{4}$
${F}_{\text{net}} = k \left(\frac{28}{4} - \frac{21}{81}\right)$
${F}_{\text{net}} = \frac{2184}{324} \cdot k$
${F}_{\text{net}} = 6 , 74 \cdot k$
${F}_{\text{net}} = 6 , 74 \cdot {9.10}^{9}$
${F}_{\text{net}} = 60 , 66 \cdot {10}^{9} N$