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Charge on semicircle

  1. May 10, 2014 #1
    1. The problem statement, all variables and given/known data
    This is not a homework problem so there is no problem statement. More of a conceptual question.

    Consider a charge on a ring:
    eler3.gif

    If this was a semicircle opposed to a ring, how would this change the equation since there is not an opposing dEr force.

    [tex]

    \boldsymbol{E} = \boldsymbol{\hat{z}} \frac{\rho_l R(-\hat{\boldsymbol{r}}R + \hat{\boldsymbol{z}Z})}{4 \pi \varepsilon_0 (R^2 + Z^2)^{3/2}}\int_{0}^{\pi }d\phi

    [/tex]
     
  2. jcsd
  3. May 10, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    The vertical component is still easy to calculate (as you just have a smaller part of the ring contributing to it), but the other component (due to the broken symmetry it does not vanish any more) will need its own integral. No idea how easy/messy that gets.
     
  4. May 10, 2014 #3
    So I would have to do it twice.

    dE = dEr + dEz
     
  5. May 10, 2014 #4

    mfb

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    Staff: Mentor

    The integrals are different, but yes.
     
  6. May 10, 2014 #5
    Does this look reasonable:

    [tex]

    \frac{Q}{4\pi \varepsilon_0(R^2 + Z^2)^{3/2} }(Z\hat{z}-R\hat{r})

    [/tex]
     
  7. May 11, 2014 #6

    mfb

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    Staff: Mentor

    Z and z are the same?

    That would mean the electric field points in the same direction as the "r" line in the sketch. No, that cannot work.
    It would be easier to find the error if you show your whole work.
     
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