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

Jan 27, 2016

$F = 9.0 \times {10}^{9} \frac{3}{25} = 7.2 \times {10}^{8} N$

#### Explanation:

The force between 2 charged particles is

$F = K \frac{{q}_{1} {q}_{2}}{r} ^ 2$ where
${q}_{i} = \text{Charges of the particles} , K = 9.0 \times {10}^{9} N \cdot {m}^{2} / {C}^{2}$
Inserting the value of q_1 = -3; q_2 = -1; r = (4- (-1)) = 5
$F = 9.0 \times {10}^{9} \frac{3}{25} = 7.2 \times {10}^{8} N$
Note: The force between 1 coulomb charges separated by 5 meters is huge - roughly about a billion Newtons. What does this tell you?
Well the unit of Coulomb is pretty large, and for that size 5 meters separation generated and amazing (unrealistic amount of force)