# Two charges of  6 C  and  -7 C are positioned on a line at points  -7  and  8 , respectively. What is the net force on a charge of  -3 C at  2 ?

Nov 25, 2017

$0.361$ Relative net repulsive force from right to left.

#### Explanation:

$6 {C}_{- 7} \ldots \ldots \ldots - 3 {C}_{2} \ldots \ldots . - 7 {C}_{8}$

Coulomb's Law states:

$F = {k}_{e} \frac{{q}_{1} {q}_{2}}{r} ^ 2$

Coulomb's Constant k_e = 8.99×10^9 N m^2 C^(−2)

Without specifying the distance units we can take the relative value without the constant. Distance_1 = 2 - (-7) = 9, Distance_2 = 8 - 2 = 6

${F}_{1 - 2} = \frac{{q}_{1} {q}_{2}}{r} ^ 2 = \left(6\right) \frac{- 3}{9} ^ 2 = - 0.222$ (attractive)

${F}_{2 - 3} = \frac{{q}_{2} {q}_{3}}{r} ^ 2 = \left(- 3\right) \frac{- 7}{6} ^ 2 = 0.583$ (repulsive)

${F}_{\ne t} = - 0.222 + 0.583 = 0.361$ Net repulsive from right to left.