# Two charges of  2 C  and  8 C are positioned on a line at points  -3  and  6 , respectively. What is the net force on a charge of  -3 C at  -2 ?

Feb 3, 2016

$\Delta F = 50 , 625 \cdot {10}^{9} \cdot {C}^{2}$

#### Explanation: ${q}_{a} = 2 C$ charge on the point of A
${q}_{b} = - 3 C$ charge on the point of B
${q}_{c} = 8 C$ charge on the point of C
$k = 9 \cdot {10}^{9} \frac{N \cdot {m}^{2}}{C} ^ 2$
$\text{formula needed to solve this problem is Coulomb's law}$
$F = k \cdot \frac{{q}_{1} \cdot {q}_{2}}{d} ^ 2$
$F : \text{Force between two charges acting each other}$
${q}_{1} , {q}_{2} : \text{charges}$
$d : \text{distance between two charges}$

$s t e p : 1$
color(red)(F_(AB))=k*(q_A*q_B)/(d_(AB)^2
$\textcolor{red}{{F}_{A B}} = 9 \cdot {10}^{9} \frac{2 C \cdot \left(- 3 C\right)}{1} ^ 2$
$\textcolor{red}{{F}_{A B}} = - 54 \cdot {C}^{2} \cdot {10}^{9}$

$s t e p : 2$
color(blue)(F_(CB))=k*(q_C*q_B)/(d_(CB)^2
$\textcolor{b l u e}{{F}_{C B}} = 9 \cdot {10}^{9} \frac{\cancel{8} C \cdot \left(- 3 C\right)}{\cancel{8}} ^ 2$
$\textcolor{b l u e}{{F}_{C B}} = - 27 \cdot {10}^{9} \cdot {C}^{2} / 8$
$\textcolor{b l u e}{{F}_{C B}} = - 3 , 375 \cdot {10}^{9} \cdot {C}^{2}$

$s t e p : 3$
$\Delta F = \textcolor{b l u e}{{F}_{C B}} - \textcolor{red}{{F}_{A B}}$
Delta F=-3,375*10^9*C^2-(-54*10^9*C²)
$\Delta F = - 3 , 375 \cdot {10}^{9} \cdot {C}^{2} + 54 \cdot {10}^{9} \cdot {C}^{2}$
$\Delta F = 50 , 625 \cdot {10}^{9} \cdot {C}^{2}$