We know that the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differ from each other by 0.00011%. With what force would two copper pennies, placed 1.4 m apart, repel each other? Assume that each coin contains 3 1022 copper atoms. (Hint: A neutral copper atom contains 29 protons and 29 electrons.)(From your answer, it should be obvious that the magnitude of the charge on a proton and the charge on an electron cannot possibly differ by as much as the given percentage.)

Question

I will be happy to critique your thinking or work on this. The charge on the electron and proton are not equal, they are opposites.

Answer 1

The total positive charge in a neutral copper penny is
Q+ = (3 × 1022 atoms)(29 protons/atom)((1.60 × 10−19 C/proton) = 1.39 × 105 C
and the total negative charge is Q− = −1.39 × 105C. If the positive charge were greater b y
a factor of 5.0 × 10−7 and the negative charge were less by the same factor, the net charge
on the penny would be (1.39 × 105)(1.0 × 10−6) = 0.139 C. Use Coulomb’s law to find the
force of one penny on another, separated by 1.0m.