Question

What is the repulsive force between two pith balls that are 9.00 cm apart and have equal charges of -28.0 nC?

Answer 1

Force,
`F=8.71* 10^(-4)\ N`

Step-by-step explanation:

Given that,

Charges on pith balls,
`q_1=q_2=-28\ nC=-28* 10^(-9)\ C`

Distance between balls, d = 9 cm = 0.09 m

Let F is the repulsive force between two pith balls. We know that the repulsive force between two charges is given by :

`F=k(q_1^2)/(d^2)`

`F=9* 10^9* ((-28* 10^(-9))^2)/((0.09)^2)`

F = 0.000871 N

or

`F=8.71* 10^(-4)\ N`

So, the repulsive force between the pith balls is
`8.71* 10^(-4)\ N`. Hence, this is the required solution.