# Electric Field between two charges

1. Jan 27, 2007

### brandon26

1. The problem statement, all variables and given/known data
Two charges of value +4C and -16C are seperated by a distance of 3m.

At what spot along the line is the electric field is zero?

2. Relevant equations

I tried using the equation E = Q / 4pieE(o)r^2

3. The attempt at a solution

2. Jan 27, 2007

### cristo

Staff Emeritus
Well, what specifically did you do? If you show your work, it will be easier to help!

3. Jan 27, 2007

### Hootenanny

Staff Emeritus
I assure you that it can be solved, perhaps if you showed you working...

Edit: Damn, your quick cristo. It seems roles have been reversed...

4. Jan 27, 2007

### cristo

Staff Emeritus
Haha, I beat you to it for once, about time too! Although I wouldn't say roles were reversed just yet

5. Jan 27, 2007

### robphy

Before writing equations, you might ask yourself:
- will it be between the charges? if not, on which side will it be?
- will it be equidistant from each charge? if not, which charge is it closer to?

(Hint: think physically: what would a third charge feel at this point you seek?)

(Hint: Although you can write down one [fancy] equation that you can solve, you probably won't. The above suggest cases to consider and some intuition to almost guess the numerical solution.)

6. Jan 27, 2007

### brandon26

GOD! This website is soo useless!!!

7. Jan 27, 2007

### cristo

Staff Emeritus
Brandon, please see here: FAQ: Why hasn't anybody answered my question?, with specific reference to point 1)
You must show your work before we can help you; this means the actual equations you used, not simply a brief "I used this method" type expression.

Last edited: Jan 27, 2007
8. Jan 28, 2007

### thephy

Robphy is helping you to think and solve the problem by yourself.

Details:
1) since 16C >4C, the third spot should be closer to +4C
2) the spots between the two charges could not be 0, since in this case, the field of the two charge is in the same direction.

Thus, the spot should in the left of +4C.

The equation should be 4/x^2=16/(3+x)^2.

It does have a solution.

9. Jan 28, 2007

### brandon26

this is exactly what I got, except for one small difference. The charge of the second particle is negative, why have u ignored it??

10. Jan 28, 2007

### brandon26

actually...i js noticed your second point... both charges act in the same direction so the null point cant be inbetween the charges.

But why have you ignored the negative sign ??

11. Jan 28, 2007

### robphy

Draw a free-body diagram.

12. Jan 28, 2007

### brandon26

I have put it still doesnt make sense. You cannot ignore the negative sign, can you?