Question

What is the repulsive force between two pith balls that are 13.0 cm apart and have equal charges of −24.0 nC?

Answer 1

Answer:

The repulsive force is $3.067\times10^{-4}N$.

Step-by-step explanation:

Consider the provided information.

The coulomb's law to calculate the repulsive force: $F=\frac{kQ_1Q_2}{r^2}$

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

Substitute the respective values in the above formula.

$F=\frac{9\times10^9\frac{N\cdot m^2}{c^2} \times(-24\times10^{-9}C)^2}{[(13 cm)(\frac{1m}{100cm} )]^2}$

$F=\frac{9\times10^9\frac{N\cdot m^2}{c^2} \times(-24\times10^{-9}C)^2}{(0.13 m)^2}$

$F\approx0.0003067N$

$F=3.067\times10^{-4}N$

Hence, the repulsive force is $3.067\times10^{-4}N$.