Question

At what distance would the repulsive force between two electrons have a magnitude of 4.0 N?

Answer 1

The distance between two electrons for the repulsive force to have a magnitude of 4.0 N is approximately 3.02 x 10^-5 meters.

Step-by-step explanation:

The repulsive force between two electrons can be calculated using Coulomb's law. According to Coulomb's law, the force between two charges is given by the equation:

F = k x q1 x q2 / r^2

Where F is the force, k is the electrostatic constant (8.99 x 10^9 N m^2 / C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.

In this case, we are given that the magnitude of the repulsive force between two electrons is 4.0 N. We can rearrange the equation to solve for r as follows:

r = sqrt(k x q1 x q2 / F)

Plugging in the given values, we get:

r = sqrt((8.99 x 10^9 N m^2 / C^2) x (1.6 x 10^-19 C)^2 / 4.0 N)

r = sqrt(9.14 x 10^-9 m^2)

r = 3.02 x 10^-5 m

Therefore, the distance between two electrons for the repulsive force to have a magnitude of 4.0 N is approximately 3.02 x 10^-5 meters.

Answer 2

`7.6\cdot 10^(-15) m`

Step-by-step explanation:

The electrostatic force between two electrons is given by:

`F=k(e^2)/(r^2)`

where

`k=8.99\cdot 10^9 Nm^2C^(-2)` is the Coulomb's constant

`e=-1.6\cdot 10^(-19)C` is the electron charge

r is the distance between the two electrons

In this problem, we know F=4.0 N, so we can re-arrange the equation to calculate r:

`r=\sqrt{(ke^2)/(F)}=\sqrt{((8.99\cdot 10^9)(-1.6\cdot 10^(-19))^2)/(4.0 N)}=7.6\cdot 10^(-15) m`