# Two charges of  -2 C  and  -3 C are positioned on a line at points  4  and  -1 , respectively. What is the net force on a charge of  5 C at  0 ?

Nov 7, 2017

$\frac{115}{8} {k}_{e} {C}^{2}$ to the negative direction.

#### Explanation:

Electrostatic force between two point charges can be calculated using Coulomb's law.
$F = {k}_{e} \frac{{q}_{1} \cdot {q}_{2}}{r} ^ 2$, where ${k}_{e}$ is the Coulomb's constant($k = 8.99 \cdot {10}^{9} N {m}^{2} {C}^{-} 2$), ${q}_{1}$ and ${q}_{2}$ are the charges and $r$ is the distance.

Force between $- 2 C$ (at the point 4) and $5 C$ (at the point 0) is
${F}_{1} = {k}_{e} \frac{2 C \cdot 5 C}{4} ^ 2 = \frac{5}{8} {k}_{e} {C}^{2}$. The direction of ${F}_{1}$ is positive, for the two point charges have opposite signs.

Force between $- 3 C$ (at the point -1) and $5 C$ (at the point 0) is
${F}_{2} = {k}_{e} \frac{3 C \cdot 5 C}{1} ^ 2 = 15 {k}_{e} {C}^{2}$. This is also an attractive force, so the direction is negative.

Therefore, the net force on a charge of $5 C$ at $0$ is $15 {k}_{e} {C}^{2} - \frac{5}{8} {k}_{e} {C}^{2} = \frac{115}{8} {k}_{e} {C}^{2}$, and its direction is negative.