How far apart are two electrons if they exert a force of repulsion of 1.0 N on each other?

Question

3.8 x 10-5
1.5 x 10-16
3.8 x 10-16
1.5 x 10-14

do u get this one if you do can you help

Answer 1

The force between two charged particles is

Ef = Kc * q1*q2/r^2 where

Kc is electrostatic constant = 8.988x10^9N.m^2/C^2
q1 is charge of particle 1
q2 is charge of particle 2
r is distance between charges

so
r = sqrt(Kc.q^2/Ef)

in your case
Ef = 1.0N and
q1 = q2 = q = -1.6 x 10^-19 C

r = 1.5x10^-14m

Answer 2

The calculations presented represent the theoretical results of classical electrostatics (Coulomb's law). In reality, classical mechanics breaks down at such distances, and no two electron's would ever actually repel each other with such force, no matter how fast you slammed them together. As soon as they approached to the distance where the electric potential energy divided by the speed of light squared (E/c^2) exceeded the combined rest mass of the electron's, one or more electron-positron pairs would be created, and the original electron pair's approach would be halted, and they'd scatter off in different directions.
.

Answer 3

If i'm no longer flawed you employ the equation: F = G (m1m2)/r^2 so which you may have: 2N = 6.67x10^-11 (9.11x10^-31*9.11x10^-31) / r^2 so sparkling up for r. I have been given r= a million.05x10^-35 meters

Answer 4

F=k*e*e/sqr(r)
r=e*sqrt(k/F)
e - charge of electron
k -constant - you have to kow it.

F=1.0 N

Evaluate this result youself, pleas.