Question

How far apart (in mm) must two point charges of 70.0 nC (typical of static electricity) be to have a force of 1.30 N between them?

Answer 1

Answer:r=5.824 mm

Step-by-step explanation:

Given

Charge
`q_1=70 nC`

`q_2=70 nC`

Force between them
`F=1.30 N`

Electrostatic Force is given by

`F=(kq_1q_2)/(r^2)`

where
`q_1` and
`q_2` are the charge on the Particles

r=distance between them

`k=Coulomb\ constant =9* 10^9 N-m^2/C^2`

`1.3=(9* 10^9* 70* 70* 10^(-18))/(r^2)`

`r^2=(9* 10^9* 70* 70* 10^(-18))/(1.3)`

`r=\sqrt{33.923* 10^(-6)}`

`r=5.824* 10^(-3)`

`r=5.824 mm`