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Electric field above a circular loop

  1. Jul 12, 2014 #1

    Radarithm

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

    Find the electric field a distance z above the center of a circular loop of radius r that carries a uniform line charge λ.

    2. Relevant equations

    $$E=E_r\hat{r}+E_z\hat{z}$$
    $$E_r=\frac{\lambda}{4\pi\epsilon_0}\int_0^r\frac{1}{\mathcal{R}^2}\sin{\theta}\,dr$$
    $$E_z=\frac{\lambda}{4\pi\epsilon_0}\int_0^r\frac{1}{\mathcal{R}^2}\cos{\theta}\,dr$$
    $$\sin{\theta}=\frac{r}{\mathcal{R}}$$
    $$\cos{\theta}=\frac{z}{\mathcal{R}}$$
    $$\mathcal{R}=\sqrt{r^2+z^2}$$

    3. The attempt at a solution
    This question is rather simple but I still got it wrong (I checked the solutions manual and it had a different answer which I will post below).

    Carrying out the integrations for ##E_r## and ##E_z##:
    $$E_r=-\frac{\lambda}{4\pi\epsilon_0}\frac{1}{\sqrt{r^2+z^2}}\hat{r}$$
    $$E_z=\frac{\lambda}{4\pi\epsilon_0}\frac{r}{z\sqrt{z^2+r^2}}\hat{z}$$
    Therefore the electric field a distance z above a circular loop is:
    $$E=-\frac{\lambda}{4\pi\epsilon_0}\frac{1}{\sqrt{r^2+z^2}}\hat{r}+\frac{\lambda}{4\pi\epsilon_0}\frac{r}{z\sqrt{z^2+r^2}}\hat{z}$$
    The solution however is:http://i.gyazo.com/e5aba5c7109b6c119b36ad299003b1bc.png

    edit: Nevermind, caught my mistake. Can a mentor delete this thread?
     
    Last edited: Jul 12, 2014
  2. jcsd
  3. Jul 13, 2014 #2

    Simon Bridge

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    Or you can just post the mistake and so help someone else who does the same?
     
  4. Jul 13, 2014 #3

    Radarithm

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    The radial component of the electric field cancels out at every point due to the symmetry of the circle and the fact that the electric field arises from a line charge. This leaves us with the z component of the electric field, which can be calculated by carrying out the following integral (is it not a line integral?):
    $$\frac{1}{4\pi\epsilon_o}\int\frac{\lambda}{\mathcal{R}^2}\, dl$$
    where ##\mathcal{R}## is the hypotenuse of the right triangle formed by the radial distance + height from the line charge. To get the z component we need to take the cosine of this right triangle so ##\cos{\theta}## would be equal to ##z/r##. ##dl## would be equal to the circumference of the circle, ##2\pi r##.
    $$\frac{\lambda}{4\pi\epsilon_0}\int\frac{2\pi r}{z^2+r^2}\frac{z}{r}$$
    Which results in the following function:
    $$E=\frac{\lambda rz}{2\epsilon_0(z^2+r^2)^{3/2}}$$

    I incorrectly assumed that the radial part of the electric field does not cancel out which I can understand conceptually. Incorrect assumptions were made.
     
    Last edited: Jul 13, 2014
  5. Jul 13, 2014 #4

    Simon Bridge

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    Not bad - and yes it works out to a line integral.
    Just some pointers:

    It makes no sense to take the cosine of a right triangle.

    If we put P=(0,0,z), and Q is the location of a line element length dl on the loop (radius r) then the cosine of angle QPO is ##z/R: R^2=r^2+z^2##. Was that the angle you meant?

    dl is not going to be equal to the circumference of the circle - it is an infinitesimal line interval ##dl=r\;d\theta##.

    But you appear to have arrived at the correct answer anyway.
    I think you just mixed up r and R.

    Since most online sources use R for the circle and r for the hypotenuse, someone marking your work is likely to suspect that you have just copied the solution without looking at it properly. Take care.
     
  6. Jul 13, 2014 #5

    vela

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    The integral isn't really correct either because you're supposed to sum the electric field due to each infinitesimal line element, but your integral is summing the magnitudes of the contributions. In other words, you can't say
    $$\vec{E} = \frac{1}{4\pi\epsilon_o}\int\frac{\lambda}{\mathcal{R}^2}\, dl$$ obviously because the LHS is a vector and the RHS isn't, nor can you say
    $$\|\vec{E}\| = \frac{1}{4\pi\epsilon_o}\int\frac{\lambda}{\mathcal{R}^2}\, dl$$ because the correct expression is
    $$\|\vec{E}\| = \frac{1}{4\pi\epsilon_o}\int\frac{\lambda}{\mathcal{R}^2}\frac{z}{R}\, dl.$$
     
  7. Jul 13, 2014 #6

    Radarithm

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    Yes, that was the angle I meant.
    That makes sense. I guess I'll have to be careful about my notation; I should have also included the step where the line integral transforms into an integral for ##d\theta##. Thanks for the help!
     
  8. Mar 4, 2015 #7
    Cheers! and thank you Radarithm for coming back to explain your mistake, it helped me quite a bit.
     
  9. Mar 4, 2015 #8
    Could someone help me understand how the symmetry is expressed mathematically? I completely understand the symmetry of the situation conceptually but cannot see how to show it mathematically, I have reduced the electric field vector into its radial and axial components as in the following equation
    [tex]E=\frac{\lambda r}{4\pi\epsilon_0}(\frac{z}{(z^2+r^2)^\frac{3}{2}}\hat{z}-\frac{r}{(z^2+r^2)^\frac{3}{2}}\hat{r})(2\pi)[/tex]
    How would I proceed to cancel the radial component? I know the answer is the above equation with the radial component being 0 but I cant see how to show it.
     
  10. Mar 4, 2015 #9

    Simon Bridge

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    What you are doing is just that you know the result of the integration ... basically
    The charge element on the ring between ##\theta## and ##\theta+d\theta## gives rise to an electric field that can be expressed as ##\vec E(\theta) = E_z(\theta)\hat k + \vec E_r(\theta)## and you notice that ##\vec E_r(\theta+\pi) = -\vec E_r(\theta)## ... i.e. the radial vector on the opposite side of the loop has the same magnitude but points in the opposite direction.
     
    Last edited: Mar 4, 2015
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