# Electric Force/ Electrical Fields

1. Feb 6, 2009

### xswtxoj

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

An electron is released a short distance above earth's surface. A 2ndelectron directly below it excerts an electrostatic force on the 1st electron just great enough to cancel the gravitational force on it How far below the 1st electron is the 2nd?

2. Relevant equations

F= Ke (q1)(q2)/r^2

3. The attempt at a solution

2. Feb 6, 2009

### DukeLuke

Since the two forces are acting in opposite directions and you want them to add to zero you can set your gravity force equation equal to the electric force equation. Then you should be able to solve for the distance "r".

3. Feb 6, 2009

### xswtxoj

Can you write it out for me? I don't quite understand what you mean!

4. Feb 6, 2009

### DukeLuke

Well the force of gravity on the 1st electron is can be found using $F=ma$ where "a" = g = acceleration due to gravity 9.8 $\frac{m}{s^2}$. You already have the equation for the force between the two electrons due to their charge. Since you want the forces to be equal(becuase they are opposing and cancel each other out) you can set both of these equations equal to each other. Then the only variable you won't know in this combined equation should be r which is the distance between the paritcles' centers. Your equation should look something like

$$mg = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2}$$

5. Feb 7, 2009

### xswtxoj

how would u solve this problem if no numbers are given? i really dont get this?

6. Feb 7, 2009

### xswtxoj

what u said so i set up the equation and to cancel is like Keqq= Frsq then qq=mg4 pi Eo rsq, then cancel all like terms, to have Ke=F=mg4 pi Eo?

7. Feb 7, 2009

8. Feb 7, 2009

### xswtxoj

considering that its an electron as 1.6 x 10^-19, and ur finding the distances, how would u apply that to the 2 equations above?

9. Feb 7, 2009

### LowlyPion

DukeLuke gave you the equation. So look up the mass.

Solve for R.

10. Feb 7, 2009

### xswtxoj

can the equation be used as q1q2/ 4 pi (8.85x 10 ^-12 ) r sq,,,, the 8.85 be used for E0?

11. Feb 7, 2009

### LowlyPion

Yes. Of course.

12. Feb 7, 2009

### xswtxoj

is i use the $$mg = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2}$$ then would it be: 9.8= 1.6 E -19 sq/ 4 (3.14)(8.85 E -12) r2, then cross multiply to get 9.8 x 4 x 3.14 x 8.85xE-12 (r sq) = 1.6x10-19 sq then get 1.089E -9 (r sq) = 2.56x -38 then divide to isolate r sq, then solve for r to get 4.84E -15