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Upon trying to find the electric field using integration (as done below) we get the following result :

enter image description here

According to me there are two problems with this result :

  1. This function is not of odd parity. It should be of odd parity because the disc divides the space into two symmetric parts and therefore the value of electric field in the two parts must be of equal magnitudes and opposite sign (because of opposite direction).
  2. The value of electric field at center of disc is coming out to be finite according to the result which I feel intuitively must be zero.

So my question is :

Are these arguments right? If yes, where was the mistake done while deriving the result. If no, why?

Thank you very much :)

Image source : hyperphysics

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  • $\begingroup$ Related? with a nice diagram to show that the electric field is not zero at $z=0$.. $\endgroup$
    – Farcher
    Jan 13 '19 at 15:04
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The result is correct only for $z> 0$. If you follow the calculation carefully, you will see that it appears $\sqrt{{{z}^{2}}}=\left| z \right|$. If $z<0$, $\sqrt{{{z}^{2}}}=-z$ or $\frac{\sqrt{{{z}^{2}}}}{z}=\pm 1$

To be more precise : you have to make a change of variable : $z\left( \int\limits_{0}^{R}{\frac{rdr}{{{\left( {{r}^{2}}+{{z}^{2}} \right)}^{3/2}}}} \right)=z\left( \int\limits_{{{z}^{2}}}^{{{R}^{2}}+{{z}^{2}}}{\frac{1}{2}\frac{du}{{{\left( u \right)}^{3/2}}}} \right)=-\frac{z}{\sqrt{{{R}^{2}}+{{z}^{2}}}}+\frac{z}{\sqrt{{{z}^{2}}}}=\pm 1-\frac{z}{\sqrt{{{R}^{2}}+{{z}^{2}}}}$

It is well known that the electric field is discontinuous at the crossing of a charged sheet. Your symetry argument (value 0) would be valuable for a volumic distribution of charges. For a surfacic distribution, the electric field is not defined on the sheet. You should abandon the surfacic model to pass to a volumic model and the field would be zero in 0 if the symmetry is preserved.

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  • $\begingroup$ Could you please elaborate a bit more like where did we use If $z<0$, $\sqrt{{{z}^{2}}}=-z$ and how to justify it being non zero at center? $\endgroup$
    – user219667
    Jan 13 '19 at 14:36
  • $\begingroup$ I detailed my answer! $\endgroup$ Jan 13 '19 at 14:54
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You are wrong in the intergation result. Instead of 1 it should be $\text{sign}(z)$. The electric field projection $E_z$ is proportional to an odd function of $z$, so it is odd too which is visible from your first integral. It is also intuitively comprehensible. And it is zero when $z=0$. As often happens in Physics, we use some approximations, in particular, in your case the notion of the surface charge is governed by the inequality $r\gg a$, where $a$ is the plate thickness.

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  • $\begingroup$ By odd function I mean wikipedia and how is it 0? It is $k\sigma2\pi$ $\endgroup$
    – user219667
    Jan 13 '19 at 14:10
  • $\begingroup$ It is $2\pi\sigma k$ above the plate, and it is $-2\pi\sigma k$ below the plate. In the middle $E_z$ is zero due to absolute symmetry of the charge distribution. $\endgroup$ Jan 13 '19 at 14:30
  • $\begingroup$ Why is it $−2πσk$ below the plate? Sorry, I am new to electrostatics and slow to learn. $\endgroup$
    – user219667
    Jan 13 '19 at 14:34
  • $\begingroup$ @my2cts: I meant $E_z$ of course since the question is about it. I modified my answer too. $\endgroup$ Jan 13 '19 at 16:37
  • $\begingroup$ The OP is not wrong. He knows there is a mistake and he asks where. You don't even tell hem that. Ez is not proportional to z. This is not an answer, it contains an error and it is not "nice". $\endgroup$
    – my2cts
    Jan 13 '19 at 16:40

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