1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Semi-circle linear charge Electric Field

  1. Aug 28, 2016 #1

    RJLiberator

    User Avatar
    Gold Member

    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]

    Now:

    [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

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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

    RJLiberator

    User Avatar
    Gold Member

    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

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    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

    RJLiberator

    User Avatar
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Semi-circle linear charge Electric Field
Loading...