Question

We know that the magnitudes of 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 0.00060%. With what force would two copper coins, placed 0.76 m apart, repel each other? Assume that each coin contains 3.2 × 1022 copper atoms. (Hint: A neutral copper atom contains 29 protons and 29 electrons.)

Answer 1

q = difference of the charge between proton and electron = 0.000006 x 1.6 x 10^-19 C

n = number of electrons or proton = 29

N = Number of atoms = 3.2 x 10²²

Q = Total difference of charge = N n q = (3.2 x 10²²) (29) (0.000006 x 1.6 x 10^-19) = 0.89 C

r = distance between two coins = 0.76 m

Force between the coins is given as

F = k Q²/r²

F = (9 x 10⁹) (0.89)² /(0.76)²

F = 1.23 x 10¹⁰ N