Question

1. Electron (-) has a charge of 1.6 x 10-19 C. Proton (+) has a charge of 1.6 x 10-19

C. In a hydrogen atom the distance between a proton and an electron is
5.3 x 10-12 meter.
a) Calculate the force acting between the proton and the electron. Show your
work. Remember to include a correct force unit in your final answer.

Answer 1

`8.2\cdot 10^(-6) N`, attractive

Step-by-step explanation:

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

`F=k(q_1 q_2)/(r^2)`

where:

`k=8.99\cdot 10^9 Nm^(-2)C^(-2)` is the Coulomb's constant

`q_1, q_2` 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:

`q_1=1.6\cdot 10^(-19)C` is the magnitude of the charge of the electron

`q_2=1.6\cdot 10^(-19)C`is the magnitude of the charge of the proton

`r=5.3\cdot 10^(-12) m` is the separation between the two particles

So the magnitude of the force is

`F=(8.99\cdot 10^9)((1.6\cdot 10^(-19))^2)/((5.3\cdot 10^(-12))^2)=8.2\cdot 10^(-6) N`

And since the two charges have opposite signs, the force is attractive.