Answer:
[tex]8.2\cdot 10^{-6} N[/tex], attractive
Explanation:
The magnitude of 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
And the force is:
- Repulsive if the two charges have same sign
- Attractive if the two charges have opposite signs
In this problem:
[tex]q_1=1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the electron
[tex]q_2=1.6\cdot 10^{-19}C[/tex]is the magnitude of the charge of the proton
[tex]r=5.3\cdot 10^{-12} m[/tex] is the separation between the two particles
So the magnitude of the force is
[tex]F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(5.3\cdot 10^{-12})^2}=8.2\cdot 10^{-6} N[/tex]
And since the two charges have opposite signs, the force is attractive.
