Question

What feels a stronger repulsive force ... two pith balls each with +3nC of excess charge that are 10 cm apart, or two pith balls each with +2nC of charge that are 6 cm apart?

Answer 1

[AI-generated answer — Bot GPT 3.5]

To calculate the repulsive force between two charges, we can use Coulomb's Law:

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

where F is the force, k is the electrostatic constant (9.0 x 10^9 N m^2 / C^2), q1 and q2 are the charges, and r is the distance between the charges.

For the first scenario, with two pith balls each with +3nC of excess charge and 10 cm apart:

F1 = (9.0 x 10^9) * ((3nC) * (3nC)) / (0.1m)^2

F1 = 9.0 x 10^9 * 9nC^2 / 0.01m^2

F1 = 9.0 x 10^9 * 9 * 10^-9 C^2 / 10^-2 m^2

F1 = 9.0 x 10^9 * 9 * 10^-7 N

For the second scenario, with two pith balls each with +2nC of charge and 6 cm apart:

F2 = (9.0 x 10^9) * ((2nC) * (2nC)) / (0.06m)^2

F2 = 9.0 x 10^9 * 4nC^2 / 0.0036m^2

F2 = 9.0 x 10^9 * 4 * 10^-9 C^2 / 3.6 x 10^-3 m^2

F2 = 9.0 x 10^9 * 4 * 10^-6 N

Comparing the two forces F1 and F2, we can see that F1 is larger than F2. Therefore, the stronger repulsive force exists between two pith balls each with +3nC of excess charge that are 10 cm apart.