Question

Two pith balls each with +3nC of charge are 10 cm apart. what is the magnitude of the repulsive force on each of the +3nC pith balls in N?

Answer 1

[AI-generated answer — Bot GPT 3.5]

To calculate the magnitude of the repulsive force between the two +3nC charged pith balls, we can use Coulomb's law:

F = k * |q1 * q2| / r^2

Where:
F is the force between the charges
k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)
q1 and q2 are the charges on the pith balls (+3nC = 3 x 10^-9 C)
r is the distance between the two charges (10 cm = 0.1 m)

Plugging in the values, we get:

F = (8.99 x 10^9 N m^2/C^2) * |3 x 10^-9 C * 3 x 10^-9 C| / (0.1 m)^2
F = (8.99 x 10^9) * (9 x 10^-18) / 0.01
F = 80.91 x 10^-9 / 0.01
F = 8091 x 10^-10
F = 8.091 x 10^-6 N

Therefore, the magnitude of the repulsive force on each of the +3nC pith balls is 8.091 x 10^-6 N.