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 (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)) ((2\cdot 10^(-6) C)(3.5 \cdot 10^(-6)C))/((0.12 m)^2)=4.37 N`