# Two charges of  -2 C  and  -7 C are positioned on a line at points  3  and  4 , respectively. What is the net force on a charge of  -5 C at  0 ?

Oct 18, 2016

$\text{Net force on charge of -5C is "2.969" } N$

#### Explanation: ${\vec{F}}_{1} : \text{Force between "q_A" and } {q}_{C}$

${\vec{F}}_{2} : \text{Force between "q_B" and } {q}_{C}$

${\vec{F}}_{C} : \text{ vectorial sum of "F_1" and "F_2" (net force on charge of -5C)}$

${\vec{F}}_{1} = \cancel{9} {.10}^{9} \frac{{q}_{A} \cdot {q}_{C}}{\cancel{3}} ^ 2 = {10}^{9} \cdot {q}_{A} \cdot {q}_{C} = \left(- 2\right) \cdot \left(- 5\right) \cdot {10}^{9}$

${\vec{F}}_{1} = 10 \cdot {10}^{9} \text{ } N$

${\vec{F}}_{2} = 9 \cdot {10}^{9} \frac{{q}_{B} \cdot {q}_{C}}{4} ^ 2 = 9 \cdot {10}^{9} \frac{\left(- 7\right) \cdot \left(- 5\right)}{16}$

${\vec{F}}_{2} = \frac{35 \cdot 9 \cdot {10}^{9}}{16} = 19.69 \cdot {10}^{9} \text{ } N$

${\vec{F}}_{C} = {\vec{F}}_{1} + {\vec{F}}_{2}$

${\vec{F}}_{C} = 10 \cdot {10}^{9} + 19.69 \cdot {10}^{9}$

${\vec{F}}_{C} = 29.69 \cdot {10}^{9}$

${\vec{F}}_{C} = 2.969 \cdot {10}^{10} \text{ } N$