Answer :
Answer:
magnitude: 100 N direction: east
Explanation:
The electrostatic force between two charges is given by
[tex]F=k\frac{q_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the charges
In this problem, we have
[tex]q_1 = 5 \mu C = 5\cdot 10^{-6} C[/tex]
[tex]q_2 = 2 \mu C = 2\cdot 10^{-6} C[/tex]
[tex]r=3\cdot 10^{-2} m[/tex]
Substituting into the equation, we find the magnitude of the force
[tex]F=(9\cdot 10^9 N m^2 C^{-2})\frac{(5\cdot 10^{-6}C)(2\cdot 10^{-6}C)}{(3\cdot 10^{-2}m)^2}=100 N[/tex]
Concerning the direction, let's notice that:
- Both charges are positive, so the force between them is repulsive
- The charge q1 is west of the charge q2
- So, the force applied by q1 on q2 must be to the east (away from charge q1)
Answer:
Magnitude 100 N, direction East
Explanation:
Edge test answer confirmed