Question

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of -30.0 nC?

Answer 1

`|F| ~~1.26xx10^-3N`

Explanation:

The force between two charges is given by Coulomb's Law:

`|F|=k(|q_1||q_2|)/r^2`

Where `k` is a constant equal to `8.99*10^9(N*m^2)/C^2`

We are given that `r=8.00cm=0.08m` and `q_1=q_2=-30.0nC=-30.0*10^-9C`

`=> |F|=(8.99*10^9(N*m^2)/C^2)((30.0*10^-9C)(30.0*10^-9C))/(0.08m)^2`

`~~1.26xx10^-3N`