Question

What is the electric force between a glass ball that has +2.5 x 10^-6 C of charge and a rubber ball that has -5.0 x 10^-6 C of charge when they are separated by a distance of 0.0050 m?

Answer 1

As per Coulomb's law we know that force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$

here we know that

$q_1 = 2.5 \times 10^{-6} C$

$q_2 = -5.0 \times 10^{-6} C$

$r = 0.0050 m$

now from above formula we will have

$F = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(5 \times 10^{-6})}{(0.0050)^2}$

$F = 4500 N$

so they will attract towards each other as they are opposite in nature with force F = 4500 N