Answer:
0.0078 N
Explanation:
The electrostatic force between two charges is given by Coulomb's law:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the two charges
r is the separation between the two charges
The force is attractive if the two charges have opposite sign, and repulsive if the two charges have same sign.
In this problem, we have:
[tex]q_1 = -1 \mu C = -1 \cdot 10^{-6}C[/tex] located at [tex]x_1=-3 m[/tex]
[tex]q_2=+9 \mu C = +9\cdot 10^{-6}C[/tex] located at [tex]x_2=0 m[/tex]
[tex]q_3 = -5 \mu C = -5\cdot 10^{-6} C[/tex] located at [tex]x_3=+3 m[/tex]
The force between charge 1 and charge 2 is:
[tex]F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N[/tex]
And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.
The force between charge 1 and charge 3 is:
[tex]F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N[/tex]
And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.
Therefore, the net force on charge 1 is:
[tex]F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N[/tex]
towards the right.
