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Nonuniform Semicircle of Charge problem

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data
    A non-uniformly charged semicircle of radius R=28.0 cm lies in the xy plane, centered at the origin, as shown. The charge density varies as the angle θ (in radians) according to λ = + 5.8 θ, where λ has units of μC/m.
    (picture included)
    What is the y - component of the electric field at the origin?

    2. Relevant equations

    dE=[tex]\frac{dq}{4\pi\epsilon r^{2}}sin\theta[/tex]

    where [tex]\epsilon[/tex] = 8.85*10^{-12}

    [tex]\theta r = s[/tex]

    [tex]\lambda=\frac{q}{s}[/tex]
    -------------------------------------------

    3. The attempt at a solution

    First I derived [tex](5.8*10^{-6})\theta^{2} r=q[/tex]

    and took the derivative [tex]2(5.8*10^{-6}) r \theta d\theta=dq[/tex]

    substituted into equation 1
    dE=[tex]\frac{2(5.8*10^{-6})}{4\pi\epsilon r}\theta sin(\theta)d\theta[/tex]

    Integrated from 0 to pi.
    E=[tex]\frac{(5.8*10^{-6})}{2\pi\epsilon r}\int\theta sin(\theta)d\theta[/tex]

    The whole integral simplifies down to pi,which cancels out and leaves
    E=[tex]\frac{(5.8*10^{-6})}{2\epsilon r}[/tex]

    I am getting 1.17*10^{6} N/C for an answer, but it's not correct.
    Thanks for the help in advance.
     

    Attached Files:

    Last edited: Jan 15, 2010
  2. jcsd
  3. Jan 15, 2010 #2

    ideasrule

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    It's supposed to be [tex](5.8*10^{-6}/2)\theta^{2} r=q[/tex]. A less roundabout way of doing this is to note that dq=λ*dL and dL=Rd(theta), so dq=λRd(theta).
     
  4. Jan 15, 2010 #3
    why is that? I'm not seeing where that 2 came from. Does the 2 come in before the derivative because it's only half a circle?
     
  5. Jan 15, 2010 #4

    ideasrule

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    A better question would be how you got [tex](5.8*10^{-6})\theta^{2} r=q[/tex] in the first place. Didn't you use dq/dl=λ, dq=λ*dl=λ*R*dθ?
     
  6. Jan 15, 2010 #5
    I used [tex]\theta r = L[/tex] and [tex]\lambda=\frac{q}{L}[/tex] to get [tex]\lambda=\frac{q}{\theta r}[/tex]. I substituted [tex]5.8\theta[/tex] for lambda and rearranged to solve for q. I didn't know how else to take the nonuniform charge into account.

    Again, thanks for the help.
     
  7. Jan 15, 2010 #6

    ideasrule

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    Lambda does not equal q/L because the charge density varies across the circle. Try using the method I described: dq=λ*dL and dL=Rd(theta), so dq=λRd(theta).
     
  8. Jan 15, 2010 #7
    Okay, I understand why I was wrong before. Is it correct to go from dq=λRd(theta) to dq =5.8(theta)Rd(theta)?
     
  9. Jan 15, 2010 #8

    ideasrule

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    Yup, that's correct.
     
  10. Jan 15, 2010 #9
    Integrated from 0 to pi.
    E=[tex]
    \frac{(5.8*10^{-6})}{4\pi\epsilon r}\int sin(\theta)d\theta
    [/tex]

    =>(skipping a few steps) =>
    E=[tex]
    \frac{(5.8*10^{-6})}{4\pi\epsilon r}(2)[/tex]
    I believe that would be correct.
    And it should also be positive because it points towards the negative charge, right?
     
  11. Jan 15, 2010 #10

    ideasrule

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    Yeah, it's positive.
     
  12. Jan 15, 2010 #11
    thanks so much for the help!
     
  13. Jan 25, 2010 #12
    How did you get rid of the
    [tex]
    (\theta) from \int (\theta)sin(\theta) to \int sin(\theta)
    [\tex]
     
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