Question

Two Nerf balls have the same charge. Each ball has an excess of n = 108 protons. The balls are initially separated by a distance, d = 2.8 m. The Coulomb constant is k = 8.988 × 10^9 N m²/C². An expression for the charge on one of the balls, q, in terms of the charge on a proton e and the number of excess protons n.

a. q = enk/d
b. q = (enkd)^(1/2)
c. q = enk/d^2
d. q = enk/(2d)

Answer 1

The expression for the charge on one of the balls, q, in terms of the charge on a proton e and the number of excess protons n is q = enk/(2d).

Step-by-step explanation:

The expression for the charge on one of the balls, q, in terms of the charge on a proton e and the number of excess protons n is given by q = enk/(2d).

Using the values given, q = (108 × 1.6 × 10^-19 C × 8.988 × 10^9 N m²/C²) / (2 × 2.8 m) = 8.659 × 10^-17 C. Since each Nerf ball has an excess of n = 10³⁸ protons and the charge of a single proton is e = 1.6 × 10⁻¹⁹ C, you simply multiply the charge of one proton by the number of excess protons to find the total charge q on one ball. Therefore, the correct option is q = enk/(2d).