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?

As per Coulomb's law we know that force between two charges is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]q_1 = 2.5 \times 10^{-6} C[/tex]
[tex]q_2 = -5.0 \times 10^{-6} C[/tex]
[tex]r = 0.0050 m[/tex]
now from above formula we will have
[tex]F = \frac{(9 \times 10^9)(2.5 \times 10^{-6})(5 \times 10^{-6})}{(0.0050)^2}[/tex]
[tex]F = 4500 N[/tex]
so they will attract towards each other as they are opposite in nature with force F = 4500 N