Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109

Part A

What would be the force between two 1.9-mm-diameter copper spheres 1.4cm apart? Assume that each copper atom has an equal number of electrons and protons.

Part B

Would this amount of force be detectable?

a. Yes

b. No

A) F = 1.09 10 5 N, b) Yes

Explanation:

Part A

For this exercise we need the number of free electrons in copper, as the valence of copper + 1 there is a free electron for each atom. Let's use the concept of density to find the mass of copper in the sphere

ρ = m / V

. m = ρ V = ρ 4/3 π r³

The radius is half the diameter

r = 1.9 10⁻² / 2 = 0.95 10⁻² m

ρ = 8960 kg / m3

m = 8960 4/3 π (0.95 10⁻²) ³

m = 3.2179 10⁻² kg

The molecular weight of copper is 63,546 g / mol which has 6,022 10²³ atoms

With this we can use a rule of proportions to enter the number of atom is this mass

#_atom = 6.022 10²³ 3.2179 10⁻² / 63.546 10⁻³

#_atom = 3,049 10²³

Therefore there is the same number of electrons, as they indicate that the charge of the protone and the electon differs by 1/10⁹ the total charge for each spherical is

q = e / 10⁹ #_atom

q = e / 10⁹ 3,049 1023

q = 3,049 10⁴ (-1.6 10⁻¹⁹)

q = - 4,878 10-5 C

Electric force is

F = k q₁q₂ / r²

F = k q² / r²

Let's calculate

F = 8.99 10⁹ (4.878 10⁻⁵) ²2 / (1.4 10⁻²) ²

F = 1.09 10 5 N

This is a force of repulsion.

Part B

The magnitude of this force is in very easy to detect