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

Mar 25, 2017

$\text{Net Force is } - 3.97 \cdot {10}^{9} u n i t s$
$\text{The negative sign shows that charges at the points A and B }$$\text{attract the charge at the point of C.}$

#### Explanation:

$\text{Both of charge at the points A and B apply a force on}$
$\text{the charge at the point of C.}$

$\text{let "F_A" be a force that A applies C.}$

$\text{The charge at the point of A attracts the charge at C}$

$\textcolor{g r e e n}{{F}_{A}} = \frac{k \cdot {q}_{A} \cdot {q}_{C}}{{d}_{\text{AC}}^{2}} = - \frac{K \cdot 5 \cdot 2}{3 + 2} ^ 2 = - \frac{K \cdot 2.5}{25} = - \frac{2 K}{5}$

$\text{let "F_B" be a force that B applies C.}$

$\text{The charge at the point of B attracts the charge at C}$

$\textcolor{b l u e}{{F}_{B}} = \frac{k \cdot {q}_{B} \cdot {q}_{C}}{{d}_{\text{BC}}^{2}} = - \frac{K \cdot 1 \cdot 2}{5 + 2} ^ 2 = - \frac{K \cdot 2.1}{49} = - \frac{2 K}{49}$

$\textcolor{red}{{F}_{C}} = \textcolor{g r e e n}{{F}_{A}} + \textcolor{b l u e}{{F}_{B}}$

$\textcolor{red}{{F}_{C}} = - \frac{2 K}{5} - \frac{2 K}{49}$

$\textcolor{red}{{F}_{C}} = - 2 K \left(\frac{1}{5} + \frac{1}{49}\right)$

$\textcolor{red}{{F}_{C}} = - 2 K \left(\frac{49 + 5}{49 \cdot 5}\right)$

$\textcolor{red}{{F}_{C}} = - 2 K \frac{54}{245}$

$K = 9 \cdot {10}^{9}$

$\textcolor{red}{{F}_{C}} = - 2 \cdot 9 \cdot {10}^{9} \left(\frac{54}{245}\right)$

color(red)(F_C)=-3.97.10^9 "