Question

Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?

Answer 1

The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:

`F= k_e ( q_1 q_2)/(r^2)`
where
`k_e=8.99\cdot10^9~N m^2 C^(-2)` is the Coulomb's constant,
`q_1=q_2=29~nC=29\cdot 10^(-9)~C` is the intensity of the two charges, and
`r=5.5~cm=0.055~m` is the distance between them.

Substituting these numbers into the equation, we get

`F=2.5~10^(-3)~N`

The force is repulsive, because the charges have same sign and so they repel each other.