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Electric field of a semi circle ring

  1. Feb 24, 2015 #1
    1. The problem statement, all variables and given/known data

    5jXScWt.jpg

    Suppose the charge Q on the ring of Fig. 28 was all distributed uniformly on only the upper half of the ring, and no charge was on the lower half. Determine the electric field [itex]\vec {E}[/itex] at P.

    2. Relevant equations

    [tex] E = k\frac {q} {r^2} [/tex]

    3. The attempt at a solution

    Starting with the charge density,

    [tex]\lambda = \frac{Q}{\pi a} = \frac{dq}{dl}[/tex]
    electric field for dq
    [tex] dE = k\frac{dq}{x^2+a^2}[/tex]
    Let's find its x component
    [tex]dE_x = dEcos\theta = k\frac{dq}{x^2+a^2}cos\theta = kx\frac{dq}{(x^2+a^2)^{3/2}}[/tex]
    [tex]dE_x = \frac{k\lambda x}{(x^2+a^2)^{3/2}}dl[/tex]

    Wıth the limits being [itex]0[/itex] and [itex]\pi a[/itex]

    [tex]E_x = \frac{k\lambda x \pi a}{(x^2+a^2)^{3/2}} = \frac{kqx}{(x^2+a^2)^{3/2}}[/tex]
    This is correct according to my textbook.
    However, I end up with a wrong y component when I apply the same logic for the vertical.
    [tex]E_y = -\frac{k\lambda a \pi a}{(x^2+a^2)^{3/2}} = -\frac{kqa}{(x^2+a^2)^{3/2}}[/tex]

    But the correct answer for the vertical is
    [tex]E_y = -\frac{kq2a/\pi}{(x^2+a^2)^{3/2}}[/tex]

    Having searched the internet for this question, I found that the solution for the y axis includes an arc length element, which is something like this.
    [tex]\lambda = \frac {dq} {a d\theta}[/tex]
    Then I suppose, it integrates from [itex]0[/itex] to [itex]\pi[/itex] for a semi circle.

    So my question is, how does that differ if one uses [itex]d\theta[/itex] or [itex]dl[/itex] for integration? Also, why did I get the x component correct, even though the fact that I used [itex]dl[/itex]?
     
  2. jcsd
  3. Feb 24, 2015 #2

    haruspex

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    For the x component, the contribution of each element was independent of theta. Not so for the y component.
     
  4. Feb 24, 2015 #3

    TSny

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    You didn't show the details of your work for the y component.

    Note that using sinθ instead of cosθ will give you the component of ##d\vec{E}## that is perpendicular to the x axis. But this perpendicular component is not parallel to the y axis (in general).
     
  5. Feb 24, 2015 #4
    It is much like the work I've done for the x component, but let me write it any way.

    [tex]dE_y = dEsin\theta = k\frac{dq}{x^2+a^2}sin\theta = ka\frac{dq}{(x^2+a^2)^{3/2}}[/tex]
    [tex]dE_y = \frac{k\lambda a}{(x^2+a^2)^{3/2}}dl[/tex]
    [itex]0[/itex] to [itex]\pi[/itex]
    [tex]E_y = \frac{k\lambda a \pi a}{(x^2+a^2)^{3/2}} = \frac{kqa}{(x^2+a^2)^{3/2}}[/tex]

    As you say, I'm probably missing something geometrical, but I can't see how dE*sin(theta) is not parallel to the y axis in this particular situation. If its cosine component is parallel to the x axis, shouldn't its sine component be parallel to the y axis?
     
  6. Feb 24, 2015 #5

    TSny

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    Consider an infinitesimal charge element located at one end of the semicircular ring. In what direction does the perpendicular component dE*sin(theta) point? (I'm imagining the lower half of the ring chopped off and removed so the semicircle that is left has two ends.)
     
  7. Feb 24, 2015 #6

    haruspex

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    There are three dimensions here. If the cosine gives the component parallel to the x axis then the sine gives the component orthogonal to the x axis, but that does not make it always parallel to the y axis.
     
  8. Feb 24, 2015 #7
    I think it doesn't have a y component because the ends of the circle is at the same height with point P, so I believe there is no y since there is no elevation.
     
  9. Feb 24, 2015 #8
    Yes, now I got it. I've been playing with my hands for the last 10 minutes, making a circle with one and trying to understand what is going on. I see that all sine components cannot be added together because they do not point the same direction. Am I right?
     
  10. Feb 24, 2015 #9

    TSny

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    Yes. Good.
     
  11. Feb 25, 2015 #10
    I understand that we are in a 3d coordinate system, and sine component is perpendicular to x axis. Also, that perpendicular component is not always parallel to the y axis. So how can I proceed to the y components knowing these?
    Thank you so much for your help by the way.
     
  12. Feb 25, 2015 #11

    TSny

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    Pick an element of charge and imagine a line drawn from the origin to the element. This line will make some angle φ to the y axis. Imagine a plane that contains this line and the x axis. Can you see that the electric field produced by the element of charge will lie in this plane? So, the component of the electric field that is perpendicular to the x axis will also lie in this plane. See if that helps in finding the y component of the field.
     
  13. Feb 25, 2015 #12
    cLJeY0T.jpg

    If my setup looks correct, then I would like to ask how to find a relationship between beta and theta.
     
  14. Feb 25, 2015 #13

    TSny

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    You won't need a relationship between those two angles. Notice that θ is a fixed angle while β varies to cover all the charge elements. Your diagram looks good. How would you project ##dE_\perp## onto the the vertical dotted line?
     
  15. Feb 25, 2015 #14
    Multiplying it with [itex]cos \beta[/itex]?
    If I am right, then I need limits for the integral. Would it be 2*[from pi/2 to zero], or 2*[zero to pi/2]?
     
  16. Feb 25, 2015 #15

    TSny

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    I would use twice the integral from 0 to pi/2.
     
  17. Feb 25, 2015 #16
    Thank you so much, I've managed to get the correct answer.
    Just one more thing. Should the final result's sign always obey the direction of the vector?
     
    Last edited: Feb 25, 2015
  18. Feb 25, 2015 #17

    TSny

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    Yes. If you are asked for Ey, then the answer should be negative. If you are asked for the magnitude of the y component, then that generally means just the absolute value.
     
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