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

Mar 28, 2016

${F}_{\text{net"=28.10^9" N}}$

#### Explanation: $\text{The force between two charges is given by formula:}$

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

$r : \text{distance between two charges}$
${q}_{1} : \text{the first charge}$
${q}_{2} : \text{the second charge}$
${F}_{\text{AB":" represents the force between charges A and B}}$
${F}_{\text{CB":"represents the force between charges C and B}}$
${F}_{\text{AB" and F_"CB" " are opposite forces}}$
${F}_{\text{AB"=k*((-4)*(-4))/2^2" "F_"AB}} = \frac{16 \cdot k}{4} = 4 \cdot k$

${F}_{\text{CB"=k*((-2)*(-4))/3^2=" "F_"AB}} = \frac{8 k}{9}$

${F}_{\text{net"=F_"AB"-F_"CB}}$

${F}_{\text{net}} = 4 \cdot k - \frac{8 \cdot k}{9}$

${F}_{\text{net}} = \frac{36 \cdot k - 8 \cdot k}{9}$

${F}_{\text{net"=(28*k)/9" }} k = 9 \cdot {10}^{9}$

${F}_{\text{net}} = \frac{28 \cdot \cancel{9} \cdot {10}^{9}}{\cancel{9}}$

${F}_{\text{net"=28.10^9" N}}$