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

Sep 21, 2017

$9.3 \cdot {10}^{8} N$

#### Explanation:

Two charges of −2C and $3 C$ are positioned on a line at points
3 and 4, respectively.

To find the net force on a charge of -3C at 0.

Force on -3C at 0 due to -2C placed at 3.
${F}_{1} = \frac{k {q}_{1} {q}_{2}}{r} ^ 2 = \frac{k \left(- 2\right) \left(- 3\right)}{3} ^ 2 = \frac{2 k}{3}$

Force on -3C at 0 due to 3C placed at 4.
${F}_{2} = \frac{k {q}_{1} {q}_{2}}{r} ^ 2 = \frac{k \left(3\right) \left(- 3\right)}{4} ^ 2 = \frac{- 9 k}{16}$
Total Force
$F = {F}_{1} + {F}_{2} = \frac{2 k}{3} - \frac{9 k}{16} = \frac{32 - 27}{48} k = \frac{5}{48} k$
$F = \frac{5}{48} \cdot 9 \cdot {10}^{9} = \frac{45}{48} \cdot {10}^{9} N = 9.3 \cdot {10}^{8} N$