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

Answer:

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$

Explanation:

We have given that the pith ball have the equal charge q = -40 nC =$-40\times 10^{-9}C$

Distance between the charges = 10 cm =0.1 m

According to coulombs law $F=\frac{KQ_1Q_2}{R*2}$

$F=\frac{9\times 10^9\times -40\times 10^{-9}\times -40\times 10^{-9}}{0.1^2}=1440000\times 10^{-9}=1.44\times 10^{-3}N$

So the repulsive force between the pith ball will be $1.44\times 10^{-3}N$