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Quarter circle and electric field

  1. Feb 10, 2015 #1
    1. The problem statement, all variables and given/known data
    So a quarter circle has a total charge of 25μC and has a 1m radius.
    upload_2015-2-10_15-54-17.png
    I need to find the resultant electric field at point P.

    2. Relevant equations

    The Y components will cancel here due to symmetry so we are left only caring about the x components.

    dE= (kdq)/(r^2) rˆ

    dq= λdl

    3. The attempt at a solution

    I don't have a value for lambda, so...

    dq= λdl
    Q= ∫λdl from 0 to (pi * r / 2) because 1/4 of 2*pi*r
    Q= λ∫dl
    25uC = λ*pi*1/4
    λ= 31.83uC/m

    k/r^2 * q = E
    (9*10^9)/(1^2) * 31.83 = 2.86*10^11

    The answer is really E = 2.03*10^5 N/C

    I know I'm going wrong at the integration but I don't know how to solve it.



     
    Last edited: Feb 10, 2015
  2. jcsd
  3. Feb 10, 2015 #2

    Nathanael

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    is the total charge 25μC or -12μC? Also, pi/4 is not the correct arc length.

    [[Edit: The total charge is 25μC (from the right answer) but you said it was -12μC at the beginning]]

    You said yourself, only the horizontal component matters. Integrate only the horizontal component.
     
  4. Feb 10, 2015 #3
    The total charge of the quarter circle is 25μC.

    So I just did

    Q = λ ∫cosΘdl from 0 to pi/2
    Q= λ ∫r*cosΘdΘ from 0 to pi/2
    Q= λr * (sin(pi/2) - sin(0)
    Q= λ = 25μC/m

    which means....

    (9*10^9) / 1^2 * 25μC/m = 2.25*/10^11

    still the wrong answer.
     
  5. Feb 10, 2015 #4

    Nathanael

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    Why are you integrating from 0 to pi/2?

    Also, multiply your answer by 10^-6 because you want the answer in C not in μC
     
  6. Feb 10, 2015 #5
    Well, the quarter circle extends over an angle of 90 degrees, so I thought going from 0 to pi/2 would simplify it.

    I just tried integrating from (3pi/8 to 5pi/8) and the answer to the integral is 0, so that definitely doesn't help.
     
  7. Feb 10, 2015 #6

    Nathanael

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    Yes the angle of the quarter circle is 90°, but there are infinite limits you can have that will cover 90°

    The "cosθ" term is to account for only the horizontal component, right? What is the minimum horizontal component? (Is it zero, like your limit of pi/2 implies?) Over what angles does the cosθ factor range?
     
  8. Feb 10, 2015 #7
    What do you mean by this?
     
  9. Feb 10, 2015 #8

    Nathanael

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    I was trying to think of a better way to explain it, sorry.

    You're integrating Ecosθ from 0 to pi/2... That means you're saying the horizontal component starts at the full field strength, and then decreases all the way to zero. Does it seem like this is true? Is there any piece of the quarter circle which contributes zero force in the horizontal direction?

    Furthermore, the reason you're only integrating the horizontal component is because of symmetry. What that means is that if we were to integrate Esinθ we should get zero (it should all cancel out).
    Does [itex]\int\limits_0^{\pi/2}\sin\theta d\theta=0[/itex] ??
     
    Last edited: Feb 10, 2015
  10. Feb 10, 2015 #9
    Well, the field wouldn't go from full strength to zero, I imagine. 25μC is uniformly distributed across the rod, so of course EsinΘ would end up by cancelling out. I essentially just need to find the sum of the electric field throughout the rod and I understand that, but I'm having trouble simply putting it down on paper. My bounds are screwed up, but I tried doing it from 3pi/8 to 5pi/8 because of the orientation of the quarter circle, but it didn't help at all.
     
  11. Feb 10, 2015 #10

    Nathanael

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    For which integration limits does ∫Esinθdθ equal zero? (The difference of the upper and lower limits has to equal pi/2)

    The picture really does give the limits away. Look how θ is defined in the picture. What is the maximum θ? What is the minimum θ?
    (I don't know how you got 3pi/8 and 5pi/8)
     
  12. Feb 10, 2015 #11
    I could simply try integrating ∫dq because λ,k and r are all constant, but I don't know the bounds in that case, which makes it hard to carry on.
    I have to go to another class right now, so I'll be back in ~2 hrs. I really appreciate your help, though!
     
  13. Feb 10, 2015 #12
    The maximum θ is 45° and the minimum is -45°, no?
     
  14. Feb 10, 2015 #13

    Nathanael

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    Yes :)
     
  15. Feb 10, 2015 #14
    So from that point onwards, I have kλ/r * (sin45 - sin(-45)), but I'm still lacking λ. How do I calculate the value of lambda? Would I just divide Q by the lenght of the semicircle (2πr/4)?
     
  16. Feb 10, 2015 #15

    Nathanael

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    Yep
     
  17. Feb 10, 2015 #16
    I found it! Infinite thanks for your help and patience!
     
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