Question

Two negative charges that are both - 3.8 C push each other apart with a force of 19.0 N. How far apart are the two charges?

Answer 1

Answer:

The separation distance between the two charges must be 82704.2925 m

Explanation:

Given:

Two negative charges that are both q = -3.8 C

Force of 19 N

Question: How far apart are the two charges, s = ?

First, you need to get the electrostatic force of this two negative charges:

$F=\frac{kq}{s^{2} } \\s=\sqrt{\frac{kq}{F} }$

Here

k = electric constant of the medium = 9x10⁹N m²/C²

Substituting values:

$s=\sqrt{\frac{9x10^{9}*(-3.8)^{2} }{19} } =82704.2925m$