We are told that the magnitude of the negative charge on the electron is exactly equal to the magnitude of the postive charge on the proton. But is it? Suppose that these magnitudes differ from each other by 0.000040 %. Each copper coin contains 3 X 1022 copper atoms.

(a)With that slight difference, with what force would two copper coins, separated by 1.4 m, repel each other?

To find the force with which two copper coins, separated by 1.4 m, repel each other due to the slight difference in the magnitudes of electron and proton charges, we can use Coulomb's Law.

Coulomb's Law states that the force (F) between two charged objects is proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, it can be expressed as:

F = k * (q1 * q2) / r^2

Where:
F is the force between the two objects,
k is the electrostatic constant (9 × 10^9 N m^2/C^2),
q1 and q2 are the magnitudes of the charges, and
r is the distance between the charges.

In this case, the difference in the magnitudes of electron and proton charges is given as 0.000040%. Let's assume the magnitude of the electron charge (q1) is slightly larger than the magnitude of the proton charge (q2).

Now, we need to calculate the actual magnitudes of the electron and proton charges based on this difference. Given that the magnitudes differ by 0.000040%, we can calculate the magnitude of the electron charge as:

q1 = (1 + 0.000040/100) * q2

The magnitude of the proton charge (q2) would be the same as the elementary charge, which is approximately 1.602 × 10^(-19) C.

Substituting the magnitudes of the charges and the distance into Coulomb's Law, we get:

F = (9 × 10^9 N m^2/C^2) * [(1 + 0.000040/100) * (1.602 × 10^(-19) C)^2] / (1.4 m)^2

Evaluating this equation will give us the force with which the two copper coins repel each other.