# Electric Forces

## Homework Statement

An electron is released a short distance above the surface of the Earth. A second electron directly below it exerts an electrostatic force on the first electron just great enough to cancel the gravitational force on it. How far below the first electron is the second? M

## Homework Equations

F = G(MeMp)/r^2
F = Ke|q1||q2|/r^2

## The Attempt at a Solution

I am not sure how to attempt this problem but this is what i thought.

F = (6.67 x 10^-11) (9.11 x 10^-31)(1.67 x 10^-27)/r^2

my problem there is i dont know how to find F to solve for r. its the same way if i use the other formula I still dont know F to solve for r.

I don't understand this at all and what numbers am i suppose to find to do this.

Kurdt
Staff Emeritus
Gold Member
All you have to do for this is make the gravitational force equal to the electrostatic force and solve for r. That is:

$$G\frac{M_Em_e}{r_e^2}=K\frac{e^2}{r^2}$$

Remember re is the radius of the Earth and r is the distance between the electrons so they are not the same thing if thats what was confusing you originally.

Okay, I understand the formula a bit but not sure of what the varibles are.

M_E = mass of electron?
m_e = mass of earth?
e = ?????

Kurdt
Staff Emeritus
Gold Member
e is the charge on an electron.

Thanks for the help on solving this.

I have one question, How did you get the equation K(e^2/r^2)?

ranger
Gold Member
Thanks for the help on solving this.

I have one question, How did you get the equation K(e^2/r^2)?

Since e is the charge of the electron and your system has two electrons, it is shorter to write K(e^2/r^2) rather than K(e*e/r^2).

Okay, I get it now thanks for the explanation.