Answer:
0.00899 N
Explanation:
The magnitude of the electrostatic force between two charges is given by the equation:
[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 charges
r is the distance between the two charges
And the force is:
- Repulsive if the two charges have same sign
- Attractive if the two charges have opposite sign
In this problem we have:
[tex]q_1=1.0\mu C = 1.0\cdot 10^{6}C[/tex] (charge of object 1)
[tex]q_2=1.0\mu C = 1.0\cdot 10^{6}C[/tex] (charge of object 2)
r = 1 m (separation between the objects)
So, the electric force is
[tex]F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N[/tex]
