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

Answer:

Force, $F=8.71\times 10^{-4}\ N$

Explanation:

Given that,

Charges on pith balls, $q_1=q_2=-28\ nC=-28\times 10^{-9}\ C$

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 :

$F=k\dfrac{q_1^2}{d^2}$

$F=9\times 10^9\times \dfrac{(-28\times 10^{-9})^2}{(0.09)^2}$

F = 0.000871 N

or

$F=8.71\times 10^{-4}\ N$

So, the repulsive force between the pith balls is $8.71\times 10^{-4}\ N$. Hence, this is the required solution.