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Electric Field due to a Semicircle of charge

  1. Sep 15, 2014 #1
    1. The problem statement, all variables and given/known data

    A semicircle of radius a is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge -Q is distributed uniformly around the right half of the semicircle.
    What is the magnitude and direction of the net electric field at the origin produced by this distribution of charge?
    Express your answer in terms of Q, a, and constants.

    2. Relevant equations

    dq=λdL
    dE=kdq/a^2

    3. The attempt at a solution

    I picked some random section (Θ up from the negative x-axis) on the left half of the semicircle and labeled it as dq with the length adΘ. I set dq=QadΘ and then dE=kQdΘ/a. Since the vertical components will all cancel, I focused in on the horizontal component and said dE_x=(kQdΘ/a)cosΘ. Since cosΘ=x/a at my point dq, I said dE_x=kQxdΘ/a^2.
    I know I'll have to pull out the constants and integrate dΘ over the interval ∏/2 to ∏, but how to I get rid of the x? Integrate it over 0 to a? Wouldn't I also need a dx? I originally thought it was a constant, but I can't leave my final answer in terms of x.
     

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  3. Sep 15, 2014 #2
    If you use polar coordinates (r, Θ) note that r and x are related. In a more strict manner the formula would be dE=kdq/r^2, where r^2 = x^2 + y^2. So you get rid of the x by keeping in mind that you are working in polar coordinates in a problem with a symmetry that allows you to cancel out the vertical component.

    PS. I think you dragged an error writing dE=kQdΘ/r instead of dE=kλdΘ/r (?)
     
  4. Sep 15, 2014 #3
    I didn't even think to use polar coordinates since we haven't yet in class, but it makes perfect sense.

    I substituted λ for Q because the charge of the quarter-circle was Q. If it is supposed to be λ in that equation then how will I eventually get rid of that variable?
     
  5. Sep 15, 2014 #4

    gneill

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    Hi physics24sam, Welcome to Physics Forums.

    Symmetry can be your friend :smile:

    If you simultaneously choose two dq's appropriately, one from each quadrant, you can guarantee that the resultant field dE at the origin has only an x-component as the y-components cancel. That should simplify things a bit. (Pay attention to your integration limits).

    Note that the Q charges are spread over the entire arc in a quadrant. You may want to rethink what comprises a dq. Hint: What's the linear charge density along an arc?

    You might want to sketch a diagram showing the vector sum of the complimentary dE's corresponding to a complementary pair of dq's.
     
  6. Sep 15, 2014 #5

    collinsmark

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    Hello physics24sam,

    Welcome to Physics Forums! :smile:

    This sort of goes along with what cwasdqwe said in the previous post.

    There is an error with your "dq=QadΘ" statement.

    It might help to first establish the line charge density, λ.

    [tex] \lambda = \frac{\mathrm{Charge}}{\mathrm{Length}} [/tex]

    Starting with what we know, that's

    [tex] \lambda = \frac{Q}{\mathrm{Arc \ length \ of \ a \ quarter \ circle \ of \ radius} \ a} [/tex]

    Now you can find, [itex] dq = \lambda (a d\theta). [/itex]

    For the rest of the problem, note that:

    [tex] \vec {dE} = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{r^2} \hat r [/tex]

    You should be able to make an easy substitution for [itex] r^2 [/itex] for this problem. (Hint: this radius, [itex] r [/itex], is a constant for this problem. It doesn't change. It's a trivial substitution in this case.)

    That unit (direction) vector [itex] \hat r [/itex] is the key here. You were right to leverage symmetry. Just make sure you do it correctly.

    Hint:

    [tex] \hat r = \cos \theta \ \hat x + \sin \theta \ \hat y [/tex]

    [Edit: I type slow. Didn't see all the other responses.]
     
  7. Sep 15, 2014 #6
    I knew already that the y-components would cancel and I would only have to deal with x-components. Would that change my limits somehow?

    I know for a line dq=λdL. I assumed that λ=Q, but now I am rethinking that since λ is charge per length and Q is just charge... Would dq=(Q/.5∏a)dL → dq=(Q/.5∏a)adΘ → dq=QdΘ/.5∏?
    I said that the length dL of dq would be adθ since the circumference of a circle is 2∏r. Was I also wrong in thinking that?
     
  8. Sep 15, 2014 #7

    I meant it is always better to write variables until the end, and only then substitute if required. In this case λ can be +q/dl or -q/dl. (!)
     
  9. Sep 15, 2014 #8
    So I was right in thinking that my dL=adΘ! That makes me feel a little better...

    I have already decided that I was originally wrong and that λ=Q/.5∏a since the arc length of a quarter-circle is ∏r/2.

    dE=kdq/r^2 → dE=k(Q/.5∏a)(adΘ)/r^2 → dE=kQdΘ/.5∏r^2

    To get the x-component of that do I just multiply it by cosΘ? This is where I was running into problems earlier. I substitute cosΘ for x/a, but now I have an x in my problem that I don't know how to get rid of. Should I just leave the cosΘ as is and integrate cosΘdΘ over the interval ∏/2 to ∏ rather than just dΘ? That will get me just -1.
     
  10. Sep 15, 2014 #9

    gneill

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    If you know that the two x-components of your dE are equal and in the same direction, you only have to deal with one of them and double the value. That means you only need to consider one of the arcs in one quadrant. That fixes your integration limits accordingly.

    Regarding dq, you have come to the right conclusion. For esthetic reasons you might want to make the 0.5 in the denominator a 2 in the numerator :smile:
     
  11. Sep 15, 2014 #10
    I thought I was already taking that approach... My integration limits are ∏/2 to ∏. Wouldn't that only cover the second quadrant?
     
  12. Sep 15, 2014 #11

    gneill

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    Okay, that's fine. You could also use the first quadrant to make your limits 0 to ##\pi/2##.
     
  13. Sep 15, 2014 #12
    Isn't the answer for the integral going to be -1 either way?
     
  14. Sep 15, 2014 #13

    gneill

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    Well, I agree that its magnitude will be 1. We may differ on the sign...

    Sketch the setup with the two arcs of charge and draw in the vectors representing the dE's due to charge elements from each arc. What direction do the dE vectors point?
     
  15. Sep 15, 2014 #14
    I understand that the dE_x vectors both point in the +x-direction, but if I put in the integral of cosΘdΘ over the interval ∏/2 to ∏ on my calculator it is giving me the answer of -1. The answer is positive one for an interval of 0 to ∏/2, but I've been looking at the second quadrant for the entirety of the problem so far and would like to stay consistent.
     
  16. Sep 15, 2014 #15

    gneill

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    Take a look at the area under the curve of cos(θ) from ##\pi / 2## to ##\pi##. Is it positive or negative?

    attachment.php?attachmentid=73175&stc=1&d=1410834513.gif

    A sneaky fix to change the sign is to exchange the two limits of integration: integrate from ##\pi## to ##\pi / 2##.
     

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  17. Sep 15, 2014 #16
    Okay, so I got E_net=4kQ/∏a^2 as my final magnitude and it is pointing in the +x-direction.
     
  18. Sep 15, 2014 #17

    gneill

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    Yup. Looks good.
     
  19. Sep 10, 2017 #18

    Fab

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    How did you get the 4? Originally, I had Ex = \int (dEcosΘ) = (kQ/∏a^2) \int (from 0 to ∏) cosΘdΘ. But since the integral of cosΘ from 0 to ∏ is 0 (making Ex = 0), I changed the limits of integration from 0 to ∏/2 and got Ex = 2(kQ/∏a^2), but the correct answer is 4(kQ/∏a^2)..What did I do wrong?
     
  20. Sep 10, 2017 #19

    collinsmark

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    Hello @Fab,

    Welcome to Physics Forums (PF)!

    I suggest starting a new thread. Go ahead and rewrite the problem statement and follow the template, showing your work as you go. In these situations it is preferable to start a new thread rather than ask new questions in somebody else's [old] thread. With a new thread you are more likely to get help pointing you in the right direction.
     
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