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 electric force between the two balls is given by:
$F=k_e \frac{q_1 q_2}{r^2}$
where
$k_e = 8.99 \cdot 10^9 Nm^2 C^{-2}$ is the Coulomb's constant
q1 and q2 are the charges of the two balls
r is the separation between the two balls

In our problem, we have 
$q_1 = 2 \mu C=2 \cdot 10^{-6} C$
$q_2 = 3.5 \mu C = 3.5 \cdot 10^{-6} C$ (we can neglect the sign of the charge, since we are only interested in the magnitude of the force)
$r=12 cm=0.12 m$
Therefore if we use the previous equation, we can find the electrostatic force between the two balls:
$F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2\cdot 10^{-6} C)(3.5 \cdot 10^{-6}C)}{(0.12 m)^2}=4.37 N$