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Semi-circle linear charge Electric Field

  1. Aug 28, 2016 #1


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

    A semi-circle of radius R has a charge Q uniformly distributed over its length, which provides a line charge density λ. Determine E at the origin.

    2. Relevant equations

    3. The attempt at a solution

    View attachment 105239

    I can tell by argument of symmetry that the Electric field will be pointing in +y direction.

    If we take a sliver of the charge, call it dq, we will calculate the Electric field.

    [tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{dq}{R^2}\right)sin(θ) [/tex]

    Also: [tex] \left(\frac{dq}{dθ}\right) = \left(\frac{Q}{π}\right) [/tex]

    so: [tex] dq = \left(\frac{Qdθ}{π}\right) [/tex]


    [tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{Qsin(θ)}{πR^2}\right)dθ [/tex]

    After calculating the integral from θ = 0 to θ = π I get the following answer:

    [tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y} [/tex]

    Does this work appear to be correct? I want to make sure I get this easy material down before I go to the more difficult material in electromagnetism. If so, is it fair to say that the linear charge density typically symbolized as λ is equal to Q/π in this problem. Q is the charge in coulombs and pi is the length of the semicircle. So I could represent my answer as:

    [tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R^2}\right)\hat{y} [/tex]
  2. jcsd
  3. Aug 28, 2016 #2


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    I cannot open the file.
    Regardless of the missing picture, your final answer is correct.
    No, that cannot be a linear charge density since its dimension is the same as that of the charge. Note that the circumference of a semicircle must contain its radius.
    Last edited by a moderator: May 8, 2017
  4. Aug 28, 2016 #3


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    The radius is not 1, is it?

    The charge is spread over the semicircle. What is 1/2 the circumference of a circle of radius, R ?

    Also, missing from the problem statement is the location and orientation of the semi-circle. (I could not access the image.)
  5. Aug 29, 2016 #4


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    Screen Shot 2016-08-29 at 6.55.37 AM.png

    Hey guys,

    Sorry about the image not showing -- it was an accidental double post.

    I see your point on the radius. I watched a video on this problem and did not take into consideration that the radius of that video = 1, while my radius = R. So the Linear charge density would be Q/(pi*R), is this now correct?

    Due to this fact, I will have to change my answer slightly to:

    [tex]E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^3}\right)\hat{y}[/tex]
  6. Aug 29, 2016 #5


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    It's now correct.
    Why would there be an extra ##R##? Your previous answer (2nd equation from the last in post#1) is already correct.
  7. Aug 29, 2016 #6


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    If linear charge density
    [tex] λ = \left(\frac{Q}{πR}\right) [/tex]

    Then the final answer :

    [tex]E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y} [/tex]

    which is:
    [tex]E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R}\right)\hat{y} [/tex]

    Thank you for the help.
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