# Electric Forces

1. Feb 18, 2007

### BunDa4Th

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Feb 18, 2007

### Kurdt

Staff Emeritus
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.

3. Feb 18, 2007

### BunDa4Th

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 = ?????

4. Feb 18, 2007

### Kurdt

Staff Emeritus
e is the charge on an electron.

5. Feb 18, 2007

### BunDa4Th

Thanks for the help on solving this.

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

6. Feb 18, 2007

### ranger

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).

7. Feb 18, 2007

### BunDa4Th

Okay, I get it now thanks for the explanation.