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

Force between two equal charges will be 608.4 N

Step-by-step 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=(1)/(4\pi \varepsilon _0)(q_1q_2)/(r^2)=(Kq_1q_2)/(r^2)`, here K is constant which value is
`9* 10^9Nm^2/C^2`

So force
`F=(9* 10^9* 0.65* 0.65)/(2500^2)=608.4N`