Answer:
[tex]8.76\cdot 10^{-15}m[/tex]
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=9\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:
- Attractive if the charges have opposite sign
- Repulsive if the charges have same sign
In this problem, we have:
F = 3 N is the force between the proton and the electron
[tex]q_1=q_2=1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the proton and the electron
Therefore, solving for r, we find the separation between the two particles:
[tex]r=\sqrt{\frac{kq_1 q_2}{F}}=\sqrt{\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})^2}{3}}=8.76\cdot 10^{-15}m[/tex]
