Question

A small glass ball rubbed with silk gains a charge of +2.0 uc. the glass ball is placed 12 cm from a small charged rubber ball that carries a charge of -3.5 uc.
a. what is the magnitude of the electric force between the two balls?

Answer 1

The magnitude of the electric force between two obejcts with charge $q_1$ and $q_2$ is given by Coulomb's law:
$F= k_e \frac{q_1 q_2}{r^2}$
where 
$k_e = 8.99 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb's constant
and r is the distance between the two objects.

In our problem, the distance is $r=12 cm=0.12 m$, while the magnitudes of  the two charges are
$q_1 = 2.0 \mu C=2.0 \cdot 10^{-6}C$
$q_2 = 3.5 \mu C = 3.5 \cdot 10^{-6} C$
(we can neglect the sign of the second charge, since we are interested only in the magnitude of the force).

So, using the formula and the data of the problem, we find
$F=(8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{(2.0 \cdot 10^{-6} C)(3.5 \cdot 10^{-6} C)}{(0.12 m)^2}=4.37 N$