# Forces between charged particles

1. Sep 27, 2014

### PyMaster

1. The problem statement, all variables and given/known data
Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 109.

What would be the force between two 1.7-mm-diameter copper spheres 1.7cm apart? Assume that each copper atom has an equal number of electrons and protons.
Express your answer using two significant figures.

2. Relevant equations
M = ρ V
ρcopper = 8920 kg/m^3
Vsphere = 4*pi / 3 * r^3
Molar mass of Copper = 63.5 g/mol
1 mol = 6.02*10^23

e=1.6*10^-19 C
k=9*10^9 N*m^2 / C^2
total negative charge = -e * #electrons
total positive charge = e * #protons

3. The attempt at a solution
Msphere = (8920 kg/m^3)(4pi/3)(1.7/2 * 10^-3)^3 *1000g/1kg= 0.0229 g

0.0229 g Copper * 1/63.5 mol/g * 6.02*10^23 atoms/mol = 2.17 *10^20 atoms of Copper

total negative charge = -e*(29)*(2.17*10^20)
total positive charge = (29)*(-e+10^-9)
net charge = the sum of the two above

The charge is the same for both copper spheres, so just plug and chug after this point. I am getting the wrong answer. I am certain it has to due with the calculation of the total charge on each of the copper spheres.

I am not sure how to tie the "Suppose the magnitude of the proton charge differs from the magnitude of the electron charge by a mere 1 part in 10^9." part into the actual charge of the copper atoms.

Last edited by a moderator: Sep 27, 2014
2. Sep 27, 2014

### Staff: Mentor

Saying that something differs by one part in x, where x is some number, means that one of them is larger (or smaller) by an amount that is 1/x of the other.

So in this case you might decide to let the proton charge be smaller than the electron charge by one part in 109, so then
$$q_p = \left(1 - \frac{1}{10^9} \right) q_e$$
Of course you could make the proton charge the larger one, too, if you wish.

You could also just work with the differences in the charges, since $\Delta Q = (q_e - q_p) = \frac{1}{10^9}q_e$