Question

Suppose two equal charges of 0.65 C each are separated by a distance of 2.5 km in air. What is the magnitude of the force acting between them, in newtons?

Answer 1

Answer:

Force between two equal charges will be 608.4 N

Explanation:

We have given charges $q_1=0.65C\ and\ q_2=0.65C$

Distance between the charges = 2.5 km = 2500 m

According to coulombs law force between two charges is given by

$F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}$, here K is constant which value is $9\times 10^9Nm^2/C^2$

So force $F=\frac{9\times 10^9\times 0.65\times 0.65}{2500^2}=608.4N$