# Two charges of  -1 C  and  -2 C are positioned on a line at points  -6  and  4 , respectively. What is the net force on a charge of  3 C at  1 ?

May 14, 2016

${\vec{F}}_{R} = - 5.45 \cdot {10}^{9} \text{ "N" attractive force}$

#### Explanation: $\text{data given:}$

$k = {9.10}^{9} N \cdot {m}^{2} \cdot {C}^{-} 2$

$F = k \cdot \frac{{Q}_{1} \cdot {Q}_{2}}{r} ^ 2 \text{ Coulomb's law}$

${Q}_{1} = - 1 C$

${Q}_{2} = - 2 C$

${Q}_{3} = 3 C$

$\text{solution:}$

${\vec{F}}_{1} = k \frac{{Q}_{1} \cdot {Q}_{3}}{r} ^ 2 \text{ ; "vec F_1=k((-1)*3)/7^2" ; } {\vec{F}}_{1} = - \frac{3 k}{49}$

${\vec{F}}_{2} = k \frac{{Q}_{2} \cdot {Q}_{3}}{R} ^ 2 \text{ ; "vec F_2=k((-2)*3)/3^2" ; } {\vec{F}}_{2} = \frac{- 6 k}{9}$

${\vec{F}}_{2} > {\vec{F}}_{1}$

${\vec{F}}_{R} = {\vec{F}}_{2} - {\vec{F}}_{1}$

${\vec{F}}_{R} = - \frac{6 k}{9} + \frac{3 k}{49}$

${\vec{F}}_{R} = \frac{- 294 k + 27 k}{441}$

${\vec{F}}_{R} = \frac{- 267 k}{441} \text{ ; } {\vec{F}}_{R} = - \frac{267 \cdot 9 \cdot {10}^{9}}{441}$

${\vec{F}}_{R} = - 5.45 \cdot {10}^{9} \text{ "N" attractive force}$