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Positive charge "Q" is distributed uniformly around a semicircle

  1. Sep 14, 2014 #1

    squelch

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    1. The problem statement, all variables and given/known data

    Positive charge Q is distributed uniformly around a semicircle of radius a. Detirmine ##\vec{E}## around at the center of curvature P.

    2. Relevant equations

    $$\vec{E}=K\frac{Q}{r^2}\hat{r}$$

    3. The attempt at a solution

    Taking a small slide of length dl from the semicircle, the field ##d\vec{E}## due to the charge ##dQ## is ##d\vec{E}=K\frac{dQ}{a^2}\hat{r}##.

    Where I'm getting confused is the definition of dQ
    $$dQ=\rho*a*d\theta$$

    What is ##\rho## meant to represent?
     
  2. jcsd
  3. Sep 14, 2014 #2

    squelch

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    I still don't *completely* understand what was meant by ##\rho##, but I see now that ##\lambda## is used a lot in its place. I did the following procedure:

    The ##\vec{E}## due to an infinitesimal portion of the semicircle, ##dl=ad\theta## with charge ##dQ## can be shown as:
    ##d\vec{E}=K\frac{dQ}{a^2}\hat{r}## where ##dQ=\lambda dl=\lambda ad\theta##

    The semicircle extends from ##0→\pi## radians, so we can integrate ##d\vec{E}## as

    [tex]\vec E = \int_0^\pi {K\frac{{\lambda a}}{{{a^2}}}d\theta } = \frac{{K\lambda }}{a}\int_0^\pi {d\theta = } \frac{{K\lambda }}{a}[\theta ]_0^\pi = \frac{{K\pi \lambda }}{a}[/tex]

    Because ##\lambda=\frac{Q}{\theta a}## and in this case ##\theta = \pi## then ##\lambda=\frac{Q}{\pi a}## and:

    $$\vec{E}=\frac{KQ}{a^2}$$

    Is this procedure sensible?
     
  4. Sep 15, 2014 #3

    ehild

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    Do not forget that the electric field is a vector, and the contributions of the line elements are of different directions. One of the components cancel because of symmetry....

    ehild
     
  5. Sep 15, 2014 #4

    BiGyElLoWhAt

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    ##\rho## is rho and its normally a density of some kind. Here its a charge density and uniform or constant. It varies with theta as youve demonstrated. Lambda and rho are basically the same thing, just variables and can sometimes be used differently in different contexts.
     
  6. Sep 15, 2014 #5

    BiGyElLoWhAt

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    Also, where did ##a## come from?
     
  7. Sep 15, 2014 #6

    squelch

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    ##a=r##

    So, I had redone the procedure because my answer didn't match other expected answer. Please let me know if this corrected procedure makes sense:

    The ##\vec{E}## due to an infintesmal portion of the semicircle, ##dl=ad\theta## with charge ##dQ## is:
    ##d\vec{E}=K\frac{dQ}{a^2}\hat{a}## where ##dQ=\lambda dl=\lambda ad\theta##

    Its components are, therefore,
    In X:
    $$d\vec{E}_x = d\vec{E}cos\theta d\theta = K\frac{dQ}{i^2}$$
    $$\because dQ=\lambda ad\theta$$
    $$d\vec{E}_x = \frac{K\lambda}{a}cos\theta d\theta$$
    Integrating, we find that:
    [tex]{\vec E_x} = \frac{{K\lambda }}{a}\int_0^\pi {\cos \theta d\theta } = \frac{{K\lambda }}{a}[\sin \theta ]_0^\pi = 0\hat{i}[/tex]

    In Y:
    [tex]d{\vec E_y} = \frac{{K\lambda }}{a}\sin \theta d\theta [/tex]
    Integrating, we find that
    [tex]{\vec E_y} = \frac{{K\lambda }}{a}\int_0^\pi {\sin \theta d\theta } = \frac{{K\lambda }}{a}[ - \cos \theta ]_0^\pi = \frac{{K\lambda }}{a}[ - 1 - 1] = \frac{{ - 2K\lambda }}{a}[/tex]
    ##\because \lambda=\frac{Q}{\pi a}##
    $$\vec{E}_y=\frac{-2KQ}{\pi a^2}$$

    $$\therefore \vec{E}_{total}=\vec{E}_x\hat{i} + \vec{E}_y\hat{j} = \frac{-2KQ}{\pi a^2}\hat{j}$$
     
  8. Sep 15, 2014 #7

    ehild

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    It might be correct, but how are the coordinate axes related to the axis of the semicircle?

    ehild
     
  9. Sep 15, 2014 #8

    squelch

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    The image matches this diagram:

    YF-21-096.jpg
     
  10. Sep 15, 2014 #9

    ehild

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    It is correct. Nice work and nice figure!

    ehild
     
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