Homework Help: Uniformly Charge on a Wire - Electric Field

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1. Feb 15, 2015

Aristotle

1. The problem statement, all variables and given/known data
(Just for number 1 only - finding electric field)

2. Relevant equations
dE = k dq/R^2
sin theta = y/R = y / sqrt (a^2 + y^2)
dq= lamda*dy

3. The attempt at a solution
I'm confused at the point of calculating the integral from -L/2 to L/2. I got the final integral solution from the "Special integrals table". Are my steps correct for the problem?...for some reason I keep getting zero for my answer and it just doesn't seem correct. Thank you!

2. Feb 15, 2015

Staff: Mentor

Where in your calculation do you account for the difference in sign of the charge elements above and below the x-axis?

3. Feb 15, 2015

Aristotle

Well I know that the point charge has a small dE towards the negative side of the wire while the positive side repels from the charge. Breaking it into component forms of both dE experienced by the point charge, I realized that the dEx would simply cancel each other out. From there I just took the integral of dEy. Was this wrong?

4. Feb 15, 2015

Staff: Mentor

How would your result for the y-component integration differ from that for a line of charge that was all positive? What distinguishes your integral from that one?

5. Feb 15, 2015

Aristotle

If the wire WAS all positive, then in that case the y component would cancel out resulting in both x components of the dE in the positive x direction. From that only one integration was done to find a small section of dy. Are you implying that I have to add the integral of both dEs of the y-components?

6. Feb 15, 2015

Aristotle

So ∫(from 0 to L/2) dE1 + ∫(from -L/2 to 0) dE2 ?

7. Feb 15, 2015

Staff: Mentor

Yes. You have to split the integral because of the change in sign of the charge elements. This applies to the summation of the y components. Otherwise you're treating them as all due to positive charges only, and by symmetry the y-components from the top half would cancel the y-components from the bottom half.

8. Feb 15, 2015

Aristotle

Oh wow, I really appreciate your help. Your questions got me thinking! Makes much more sense now when you think about a positive wire compared to one with half positive and negative.

9. Feb 15, 2015

Staff: Mentor

10. Feb 16, 2015

Aristotle

Sorry to bother you again...so I did the integration this time with the different limits but still got an answer of zero. Could this possibly be due to my R- afterall they are the same for both Ey1 and Ey2...

11. Feb 16, 2015

Staff: Mentor

Usually these sorts of issues are related to sign changes that are tied to the direction of integration: the order of the integration limits versus the sign of the differential element (dy in this case) and that of the variable being integrated. It can be tricky to untangle at times.

It may be expedient to do the integral for half the wire (say the positive half), check the sign of the result versus what you expect from your diagram, then invoke symmetry to declare that second integral returns the same result.

12. Feb 16, 2015

Aristotle

So you're saying in this case it's best to "assume" that both dE are the same and by symmetric have E= 2 ∫(from 0 to L/2) for dE1 ? Hate me for this, but I'm a little confused on what you're inferring :/ if that's the case, the result from it is -2k*lamda[-1/ sqrt((a)^2 + (L^2)/4) + 1/a ].

13. Feb 16, 2015

BvU

You want to check the integrals table. I miss a factor y in the numerator.
 Sorry, wrong component.

You can check your result -2k*lamda[-1/ sqrt((a)^2 + (L^2)/4) + 1/a ] here .

Your q is $\ {\lambda L\over 2}\$ , your d is $\ {L\over 2}\$ so your $\ p = {\lambda L^2\over 4}$

With $\ {1\over\sqrt{1+({L\over 2a})^2} }\approx 1 - {L^2\over 8 a^2}\$ you should get the $\ E = {kp\over a^3} \$ (and I think you do).

Last edited: Feb 16, 2015
14. Feb 16, 2015

Aristotle

Our professor hasn't covered electric dipole potential yet so I don't think that will be useful to me. My main concern is to why we have to 2*∫(from 0 to L/2) for the dE of the positive side when they aren't really symmetric in terms of the dE of the negative.

15. Feb 16, 2015

Staff: Mentor

You can demonstrate the symmetry geometrically on your diagram by selecting suitable representative charge elements from both ends of the charge distribution and showing that their contributions to E in the y-direction are equal and in the same direction.

This fact alone allows you to reduce your work to a single integral over a portion of the diagram where the signs of things are straightforward to deal with: from 0 the L/2.

Note how the positive charge elements are making contributions in the negative y direction, so you need to be careful about handling the signs of your sin() expressions if you choose to perform both integrations separately.

16. Feb 16, 2015

BvU

You'd be surprised ! Part 2 of your original post is asking precisely what the field far away is. So by the time he does mention a dipole field you have already calculated one !

And the other thingy is made clear with the good doctor's nice drawing, I should hope.

17. Feb 16, 2015

Aristotle

Ah I see, makes a little more sense now ...so since both the dEs are going in the negative y direction with the same magnitudes, its safe to say that the positive side & negative side of the wire can be found through integration by symmetry. Thank you for the visual! Just a curious thought though...when would it be the case to "integrate seperately"? I feel like with a linear wire, at least one of the component cancels each other out and symmetry seems to be always applicable.

18. Feb 16, 2015

Aristotle

Will keep in mind haha

19. Feb 16, 2015

Staff: Mentor

You can't apply symmetry if you cannot find a symmetry to exploit. So for example if the charge distributions above and below the x-axis were described by dissimilar functions you'd have to handle them separately.

20. Feb 16, 2015

Aristotle

Forgive my
Hm, by dissimilar functions would you be referring to if the wire itself was non-uniform charge distributions--such as some positive & negative charges above x-axis and below? Another thing, it's been a while but...when taking symmetry in account: 2* ∫(from 0 to L/2), why does it not equal to when ∫(from -L/2 to L/2)? --Shouldnt they equal similar answers regardless?

21. Feb 16, 2015

Staff: Mentor

Well they too could exhibit symmetry if they were symmetrically arranged and sized. But if, for example, one followed a cubic density distribution while the other a square, that would throw a wrench into the symmetry. Or if one end was longer than the other, or,.....
Well, they were similar. It just happened to end up equal and opposite for the two halves resulting in a sum of zero.

It's due to all the sneaky sign changes that you have to uncover. When you set up your integral you have to be aware of how the geometry will affect the signs of the expression you write. Your expression for the sine of the angle is a case in point. It automatically changes sign when y goes from - to +. So you need to incorporate the knowledge of the expected direction of the resulting dE contribution into how you write that expression for each portion of the integration.

22. Feb 16, 2015

Aristotle

Hah I see -- explanation on point! Thank you again for the help!