Electric Field finding zero

1. Jan 18, 2010

Keithkent09

1. The problem statement, all variables and given/known data
In the figure below (q1 = -2.3 µC, q2 = 6.80 µC), determine the point (other than infinity) at which the electric field is zero. (Figure is attached)

2. Relevant equations
E=k*Q/r^2
F-k*q_1*q_2/r^2
F-Q*E

3. The attempt at a solution
I thought about finding where the force is zero and then the electric field would be zero because of F=Q*E. But I did not think that that was correct.

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2. Jan 18, 2010

tiny-tim

Hi Keith!

(try using the X2 tag just above the Reply box )
No, that's fine … F = qE, so they're zero together.

(btw, i think it's usually little q for charge, except big Q for the charge of a black hole )

3. Jan 18, 2010

Keithkent09

Alright I get that part.
So then I set the two forces equal to each other and k*q_1*q_2/x^2=k*q_2*q_1/(d-x)^2
Is that the correct set up?

4. Jan 18, 2010

tiny-tim

erm … what about q (or Q) ?

5. Jan 18, 2010

Keithkent09

Um you mean the F=qE part?
If that is the case then the q is just the overall charge right and I am given two qs so I am not sure what that translates as...

6. Jan 18, 2010

tiny-tim

No, q isn't "the overall charge", it's a hypothetical "test" charge.

… cancels out to give 1/x2 = 1/(d-x)2.

You need the forces on a hypothetical charge q to be equal and opposite.

7. Jan 18, 2010

Keithkent09

Yeah I did that earlier and I got x to be .5 using d=1 meter.
However this answer is not correct. Was my algebra just bad or is there something more to the problem. Sorry I am just having a tough time.

8. Jan 18, 2010

tiny-tim

That's because your k*q_1*q_2/x^2=k*q_2*q_1/(d-x)^2 is wrong.

You need the force on q from q1 to be equal to the force on q from q2

9. Jan 18, 2010

Keithkent09

So that leaves me with q_1/x^2=q_2/(d-x)^2?

10. Jan 18, 2010

tiny-tim

Yes.

11. Jan 18, 2010

Keithkent09

Alright good. I got x=.57. This does not seem to be the right answer and I think this leaves me to be in between the two charges instead of to the left of them. What is the next step or did I do something incorrectly again?

12. Jan 18, 2010

tiny-tim

ah, they have opposite charge, so the equilibrium position will be outside, so you need to use d+x instead of d-x.

13. Jan 18, 2010

Keithkent09

Do I include the negative signs for the charges or is it the absolute value of those charges I cannot remember?

14. Jan 19, 2010

tiny-tim

Good morning Keith!

(just got up :zzz: …)

To find the direction of a force, you must include the sign of the charge.