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Where is the electric field the stronger, slight annulus

  1. Dec 9, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider a slight annulus of radius b which lies in the x-y plane (its center is at the origin). Find the point on the positive z-axis in which the magnitude of the electric field is the greatest. The total charge of the annulus is Q.


    2. Relevant equations None given.



    3. The attempt at a solution
    I've sketched the situation and I realize that if z=0 the electric field is null. Furthermore the electric field only has its component on the z-axis. So its magnitude is its projection onto the z-axis.
    I tried to find the electric field for all points on the z-axis but without success.
    I consider a differential part of the annulus of length dl, so its charge is [tex]\lambda dl[/tex]. I have that [tex]2\pi b \lambda=Q[/tex].
    [tex]d\vec E = \frac{dQ}{r^2}\sin \theta[/tex] according to my draft (or is it cos?).
    [tex]\vec E = 2\pi b \lambda \int \frac{\sin \theta}{r^2} dr[/tex]. But I'm stuck here, and I think I already made an error. [tex]\theta[/tex] represent the angle 0dlP where P is any point on the z-axis.
    I don't know why I integrated this part, does [tex]E=\frac{2\pi b \lambda}{b^2}=\frac{2\pi \lambda}{b}[/tex] instead? (I could eventually replace lambda by what it's worth with respect to Q.) I'm sure not, E must depends on [tex]\theta[/tex], and if [tex]\thetha=0[/tex], E=0.
    Can someone help me? Thanks!
     
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  3. Dec 10, 2009 #2

    cepheid

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    I'm not totally sure what r is in your notation. It looks like it is meant to be the separation between the element of charge and the point on the z axis.

    You're dealing with a circle in the xy-plane, so I think that it makes the most sense to work in what is essentially cylindrical coordinates with r being the distance from the origin (in the xy-plane, or in any z = const. plane), theta being the angle of the position vector of a point lying on the circle, and z being z.

    So you have a circle defined by r = b, z = 0. To cover every point on the circle, you only have to integrate over theta. Furthermore, every electric field vector makes a right triangle with base b and height z so that the magnitude of the separation between the charge point and the field point is always given by (z2 + b2)1/2. Its direction has both an r component and a z component, but it's pretty clear that the r components of all the dE's will cancel by symmetry, so that all you have left to calculate is Ez by integrating dEz over all theta.
     
  4. Dec 10, 2009 #3

    fluidistic

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    Ok I follow you entirely. My r is then (z2 + b2)1/2, but ok, let's stick with your notation.
    I have problems to find dEz. In fact I have difficulties to find dEz in function of dE.
    I see that dEz=dEsin(alpha) where alpha is the angle between 0-b and b-z. Where b is any point on the circle and z is any point on the positive z-axis.
    It complicates much more the problem.

    [tex]dE=k\frac{dQ_z}{b^2+z^2}[/tex].
    [tex]E=k\frac{2 \pi b }{b^2+z^2} \int _0^{2\pi}cos (\alpha)d\alpha[/tex] but I don't trust the integral. I just put it by intuition but I don't understand why it should be so.
    I don't know how to get a cosine, instead of a sine, for the mentioned above (3rd line of this reply).
    Can you help me a bit more please? I feel incapable.
     
  5. Dec 10, 2009 #4

    cepheid

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    Hi fluidistic,

    It looks like you're confusing your angles. Specifically, alpha is not the angle over which you are supposed to integrate. Alpha is a constant[edit: what I should have said was, "alpha depends only on z"]. I've included an image that illustrates the geometry of the situation. This will also make sure that we're both on the same page.



    You want to eventually integrate over theta, which is the azimuthal angle. So, how to tackle this problem? Basically, the magnitude of the electric field at point z due to the differential charge dQ at a point on the annulus is given by:

    [tex] dE(z) = k\frac{dQ}{z^2 + b^2} = k\frac{\lambda d\ell}{z^2 + b^2} = k\frac{\lambda bd\theta}{z^2 + b^2} [/tex] ​

    From the right triangle in the figure, we can see that this vector can be decomposed into two components, one of which is radially inward, and the other one of which is in the positive z direction:

    [tex] d\textbf{E}(z) = dE_z \hat{\textbf{z}} + dE_r \hat{\textbf{r}} = dE\cos\alpha \hat{\textbf{z}} - dE\sin\alpha \hat{\textbf{r}} [/tex] ​

    Now, it's clear by symmetry that the r components will cancel each other out as you integrate around the circle, so you can ignore them completely. But if you're not convinced, then bear in mind that:

    [tex] \hat{\textbf{r}} = \cos \theta \hat{\textbf{x}} + \sin \theta \hat{\textbf{y}} [/tex] ​

    and the integrals of both of these functions over all theta are zero. So, we know that:

    [tex] E_z(z) = \int_0^{2\pi} k\cos\alpha\frac{\lambda bd\theta}{z^2 + b^2} [/tex]​

    none of which depends on theta at all (making the integral trivial). The other nice thing is that you can easily express alpha in terms of z. The trignometric ratios are obvious from looking at the triangle:

    [tex] \cos \alpha = \frac{z}{(z^2 + b^2)^{\frac{1}{2}}} [/tex]

    [tex] \sin \alpha = \frac{b}{(z^2 + b^2)^{\frac{1}{2}}} [/tex] ​

    Can you take it from here?
     
    Last edited by a moderator: May 4, 2017
  6. Dec 11, 2009 #5

    fluidistic

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    Thank you very much for your reply. :smile:
    Your alpha is not the one I used (I used the other angle in the right triangle as alpha), but let's stick with your notation.
    I could follow you, fortunately. However... I don't think I reached the answer.
    [tex]E_z (z)=\frac{2\pi k \lambda b}{(z^2+b^2)^{3/2}}[/tex].
    But [tex]E_z (0)=\frac{kQ}{2b^3}\neq 0[/tex], so either I didn't understand you somewhere, either I made an error!
     
  7. Dec 11, 2009 #6

    cepheid

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    I think you might just be missing a factor of z. Starting from here:

    We obtain

    [tex] E_z(z) = k \left[\frac{z}{(z^2 + b^2)^{\frac{3}{2}}}\right]\lambda b \int_0^{2\pi}\, d\theta [/tex]​

    where the quantity in square brackets is (z2 + b2)-1 * cos(α) from the original integral expression. This becomes:

    [tex] E_z(z) = k \left[\frac{z}{(z^2 + b^2)^{\frac{3}{2}}}\right]\lambda b (2\pi) [/tex]

    [tex] = kQ \left[\frac{z}{(z^2 + b^2)^{\frac{3}{2}}}\right] [/tex]​
     
  8. Dec 11, 2009 #7

    fluidistic

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    Woops, I feel stupid about this arithmetic error!
    I reached, as answer, [tex]z=\frac{1}{\sqrt 5}[/tex]. I don't know if you could confirm (it's just math now). You've entirely solved my problem, although I tried to do some part of it. I'll study this problem over and over, to try to get it alone. I thank you infinitely for all.
     
  9. Dec 11, 2009 #8

    cepheid

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    I think that the answer would have to be in terms of b. I'm assuming you took the derivative?
     
  10. Dec 11, 2009 #9

    fluidistic

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    Woops, I made 2 errors I think. Yes I took the derivative, I messed up with the rest. Now I find [tex]z=\frac{3+\sqrt{9-4b^2}}{2}[/tex], which also seems unreal, ahahah. I'll try to do it correctly.
     
  11. Dec 11, 2009 #10

    cepheid

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    I got something similar to what you got, until I realized that I forgot to use the chain rule when differentiating the (z^2 + b^2)^-3/2 part of it. In the end, if you do it properly, you end up with something simpler, not a full quadratic equation. You can also plot E(z) to check your answer graphically.
     
  12. Dec 11, 2009 #11

    fluidistic

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    Ok thank you very much. Yes, forgetting to differentiate this term was one of my error. The other one was to treat b as 1 if I remember well.
     
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