Electric Charge problem

1. Feb 9, 2013

kidi3

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

http://snag.gy/Zg3SO.jpg

At the moment is it question A

3. The attempt at a solution
A)
I've realized that Force applied by F_31 + F_41 = F_21

SO i've writte this equation up but Mathmathica, says there is an errror, and gives me an incorrect result.

http://snag.gy/rQ2kz.jpg

Mathmatica says it should {{D -> -0.225008}, {D -> 0.185008}}

The correct answer should be 2,27cm.

2. Feb 9, 2013

TSny

Did you take into account the directions of the forces F_31 and F_41? They have both x and y components.

3. Feb 9, 2013

kidi3

No..
y -component must be zero since it is zero for q2
and the sum of q3 and q4 would negate the y component.

4. Feb 9, 2013

TSny

Correct. So, you only need to work with the x components of the forces.

5. Feb 9, 2013

kidi3

But aren't I doing it already?

6. Feb 9, 2013

TSny

I don't think so. It appears to me that on the left hand side of your equation you are using the full magnitude of the forces F_31 and F_41, rather than their x-components.

7. Feb 9, 2013

kidi3

Ahh.. so you want me to skip
d/cos($\theta$)
and just use d.. ?

Last edited: Feb 9, 2013
8. Feb 9, 2013

TSny

No, that part is correct, r = d/cosθ is the correct distance between charges 4 and 1. Draw a sketch showing the direction of the force F41. How would you express the x-component of F41 in terms of F41 and θ?

9. Feb 9, 2013

kidi3

well the distance it moves is d.. so I have to write an expression for d..

So it would be rcosθ = d??

10. Feb 9, 2013

TSny

Sorry, but I don't understand. What is moving a distance d? I thought you were trying to find the distance D so that the net force on q1 is zero.

Your first equation which you derived looks good except it doesn't take into account that you want the x-components of the forces to add to zero. (You already know the y-components add to zero by symmetry.)

11. Feb 9, 2013

kidi3

well q1 lies on a distance d from q3 and q4...

and isn't that distance given by r cos(θ)= d

where r is the Hypotenuse of the triangle`..

12. Feb 9, 2013

TSny

Yes. All of that is correct. So, your expression (d/cosθ)2 in the denominator of your equation is correct.

Force is a vector quantity. The equation F = q1q2/(4∏εor2) gives the magnitude of the force. You need to determine the x-component of the force vector. Again, a sketch of the forces will help.

13. Feb 9, 2013

kidi3

... hmm would it then be
F_x = q1q2/(4∏εo(d))

Since both lies in a distance d?... or how..

14. Feb 9, 2013

TSny

No. Consider the x-component of F31. Can you use trigonometry or similar triangles to find Fx in the attached picture?

Attached Files:

• Force component.jpg
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15. Feb 9, 2013

kidi3

hmmm.. is it then cos(\theta)=r/x <=> x = cos(theta)/r

16. Feb 9, 2013

TSny

Not sure what you are writing here. You need to find Fx, rather than x.

17. Feb 9, 2013

kidi3

Ahh.. I see what you mean now..
My drawings weren't correct..
But by using youre drawing I see that Cos(theta) = F_x/F <=> F_x = Fcos(theta)

So the expression for F_X = (q1q2/(4∏εo(d^2))) cos(theta)

18. Feb 9, 2013

TSny

Yes. Good.

[EDIT: oops, you did mean r instead of d in the denominator, right? And the charges are q1 and q3 ]

Last edited: Feb 9, 2013
19. Feb 9, 2013

kidi3

Hmm.. but i still get an incorrect answer.. :(

http://snag.gy/LDwAe.jpg

When i Tries to calculate D i get complex number..

20. Feb 9, 2013

TSny

It seems to work out ok. Note that you don't want to plug in a negative value for q3 since you have set up your equation to state that the magnitudes of the forces balance out.

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