Answer:
[tex]1\cdot 10^{-8} N[/tex] to the left
Explanation:
The magnitude of the electrostatic force between two charges is given by the following equation:
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where:
[tex]k=9\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant
[tex]q_1, q_2[/tex] are the magnitude of the two charges
r is the distance between the two charges
Moreover, the force is:
- Attractive if the charges have opposite sign
- Repulsive if the charges have same sign
In this problem, we have:
[tex]q_1=2nC=2\cdot 10^{-9}C[/tex] is the magnitude of charge 1
[tex]q_2=5nC =5\cdot 10^{-9}C[/tex] is the magnitude of charge 2
r = 3 m is the distance between the two charges
Substituting, we find the force on both charges:
[tex]F=\frac{(9\cdot 10^9)(5\cdot 10^{-9})(2\cdot 10^{-9}}{3^2}=1\cdot 10^{-8} N[/tex]
Here, the two charges are both positive, so the force is repulsive; since the 2 nC charge is on the left, this means that the force on this charge is to the left (away from the 5 nC charge).
