An electron at Earth's surface experiences a gravitational force of magnitude F = (9.11*10-31 kg) ⋅ (9.8 m/s2). Part A How far away can a proton and an electron be from each other, such that they still exert an electric force on each other of magnitude F? (Your answer should show why gravity is unimportant on the molecular scale!)

r = 5,085 m

Explanation:

The force exerted by on the surface of the Earth on an electron is its weight

W = F = 9.11 10⁻³¹ 9.8

W = 8.9 10⁻³⁰ N

The electric force between an electron and a proton is given by Coulomb's Law

Fe = k q₁ q₂ / r²

Fe = - k q² / r²

They ask us that W = Fe

W = k q² / r²

r = √ k q² / W

Let's calculate

r = √ 8.99 10⁹ (1.6 10⁻¹⁹) ² / 8.9 10⁻³⁰

r = √ 25.86

r = 5,085 m

Let's look for the relationship of this distance with the harmonic distance

R / R_atomic = 5,085 / 10⁻¹⁰

R / R_Atomic = 5 10¹⁰

We see that this distance is 10¹⁰ times the interatomic distance, so the gravitational attraction force is very small at atomic scale