# Two charges of  2 C  and  3 C are positioned on a line at points  -5  and  6 , respectively. What is the net force on a charge of  -8 C at  -2 ?

Dec 25, 2015

Let us use Coulomb's Law for electrostatic interaction.

#### Explanation:

Charges distribution is given by this picture:

where blue charges will be our fonts, and the red charge will be our object charge.

According to Coulomb's Law, the force between two charges 1 and 2 is given by:

${F}_{12} = \frac{1}{4 \pi {\varepsilon}_{0}} \frac{{q}_{1} \cdot {q}_{2}}{{d}^{2}} = K \frac{{q}_{1} \cdot {q}_{2}}{{d}^{2}}$

where $K = 9 \cdot {10}^{9} {\text{N" cdot "m"^2 / "C}}^{2}$ is electrostatic constant, ${q}_{1}$ and ${q}_{2}$ are the values of charges in coulombs, and $d$ is the distance between both charges.
Electrostatic force is repulsive if both charges have the same sign, and attractive if they have opposite signs.

Let us calculate both forces:

• Force from 1 to 3:

${F}_{13} = 9 \cdot {10}^{9} \text{N" cdot "m"^2 / "C"^2 cdot {2 "C" cdot (-8) "C"}/(3 "m")^2 = -16 cdot 10^9 "N}$

• Force from 2 to 3:

${F}_{23} = 9 \cdot {10}^{9} \text{N" cdot "m"^2 / "C"^2 cdot {3 "C" cdot (-8) "C"}/(8 "m")^2 = -3,375 cdot 10^9 "N}$

Both forces are attractive (that's why both signs are negative). We must substract them, and the resultant force has the same orientation than the bigger one:

${F}_{\text{Total" = (16 cdot 10^9 "N") - (3,375 cdot 10^9 "N") = 12,625 cdot 10^9 "N}}$

${\vec{F}}_{\text{Total}}$ goes from charge 3 to charge 1 (because ${F}_{13}$ is bigger than ${F}_{23}$)