Question

What is the repulsive force between two pith balls that are 12.0 cm apart and have equal charges of −27.0 nC?

Answer 1

The repulsive force is
`4.556*10^(-4)N`.

Explanation:

Consider the provided information.

Use the coulomb's law to calculate the repulsive force:
`F=(kQ_1Q_2)/(r^2)`

Where the value of k is 9.00×10⁹ Nm²/C²

Substitute the respective values in the above formula.

`F=(9*10^9(N\cdot m^2)/(c^2) *(-27*10^(-9)C)^2)/([(12 cm)((1m)/(100cm) )]^2)`

`F=(9*10^9(N\cdot m^2)/(c^2) *(-27*10^(-9)C)^2)/((0.12 m)^2)`

`F=0.000455625N`

`F=4.556*10^(-4)N`

Hence, the repulsive force is
`4.556*10^(-4)N`.