Question

What is the repulsive force between two pith balls that are 10.0 cm apart and have equal charges of -40.0 NC?

Answer 1

So the repulsive force between the pith ball will be
`1.44* 10^(-3)N`

Step-by-step explanation:

We have given that the pith ball have the equal charge q = -40 nC =
`-40* 10^(-9)C`

Distance between the charges = 10 cm =0.1 m

According to coulombs law
`F=(KQ_1Q_2)/(R*2)`

`F=(9* 10^9* -40* 10^(-9)* -40* 10^(-9))/(0.1^2)=1440000* 10^(-9)=1.44* 10^(-3)N`

So the repulsive force between the pith ball will be
`1.44* 10^(-3)N`