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Find the flux through the uncharged disc of radius R due to point charge '+q' at a distance x from the center of disc and on its axis.

diagram

So I tried to solve this by gauss law. First I started by connecting the points on the periphery of the disc to the point charge so as to obtain a cone (with α as semi vertical angle) . Now to apply gauss law I considered a gaussian sphere with +q charge at its center. Now charge q acts for whole sphere but I only need the part of the sphere where disk is present.

But here is the problem as you can see in the diagram that the disc and curved surface of area are not coinciding as there is some part left. (Pls correct me If I am visualizing wrongly)

So my question is can I consider that part to be negligible and move on taking solid angle as 2π(1-cosα)? or else I wont be able to solve this by gauss law.

Frobenius
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HelpAyoungPhysicist
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2 Answers2

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Consider the following gaussian surface: the curved part of the sphere$(S_1)$ attached to the flat disc at end of cone$(S_2)$.

enter image description here

What is the flux through this surface? Zero (Why?). Now, the surface integral of flux through this surface can be written in the following way:

$$ \int_{S_1} \vec{E} \cdot\vec{ dS }+\int_{S_2} \vec{E} \cdot \vec{dS}=0$$

In the above considered surface, the flat disc has a normal pointing towards the charge, to flip this add a negative sign into the equation(look at the picture). Finally the result is that the flat disc and curved surface has equal flux passing through them.

tryst with freedom
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  • Why is the electric flux through the surface zero? Gauss' Law seems to very explicitly say otherwise. – Bob May 11 '21 at 20:13
  • @Bob How so? The surface I took is charge free – tryst with freedom May 11 '21 at 20:17
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    Oh, I think I got the problem. When I said curved part of sphere, I meant the part which is attached to the cone. I have now added a picture in case that clarifies – tryst with freedom May 11 '21 at 20:18
  • Can we also directly say that the flux entering and flux leaving are equal and opposite in magnitude, so they cancel each other and we get zero ? as flux outflow is negative and inflow is positive. – HelpAyoungPhysicist May 12 '21 at 10:22
  • Hmm, what's wrong with the mathematical expression? I don't like the imagination of thinking flux as the 'flow' because it leads to some serious problems. – tryst with freedom May 12 '21 at 10:24
  • no i got your idea by integral as angle btw the vectors is acute and obtuse we get that they are opposite in sign – HelpAyoungPhysicist May 12 '21 at 10:25
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    Oh, I get the problem , the idea is that in surface integrals, we always take that the normal of the surface points outside from the volume enclosed by the surface. That's why the integral got a negative sign. It's a bit difficult to explain unless you have actually computed a surfac eintegral. Try learn some double integrals and go for the second chapter of div, grad curl and all that. That should clear your doubts. @HelpAyoungPhysicist – tryst with freedom May 12 '21 at 10:28
  • Actually by flow I meant entering and exiting a closed surface that's it! – HelpAyoungPhysicist May 12 '21 at 10:28
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    Believe me, trying to interpret flux as a flow is a bad idea see this question I had posted – tryst with freedom May 12 '21 at 10:28
  • The entering and exiting is a part of your imagination i.e: how you took the normal. Depending on how you take the surface normal, a same area can have positive flux or negative flux. – tryst with freedom May 12 '21 at 10:30
  • @Buraian ok looks like I misinterpret that but anyway thanks I will look into it. – HelpAyoungPhysicist May 12 '21 at 10:32
  • Yeah, don't worry about it. The flowing makes sense only in certain contexts, since it doesn't work always, it's better not to think that way at all @HelpAyoungPhysicist – tryst with freedom May 12 '21 at 10:34
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If the volume between the flat surface and the curved surface is charge free, then any flux which enters through the flat surface must leave through the curved surface.

R.W. Bird
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  • So we can say that same flux passes through disc then through the curved surface so if we find the flux for that curved surface the same flux must have come through the disc. That's it right? – HelpAyoungPhysicist May 13 '21 at 09:41
  • That would be true as long as the boundaries of the disk are the same as those of the curved surface. – R.W. Bird May 13 '21 at 14:06