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A small circular piece of radius b much smaller than a is cut from the surface of a spherical shell of radius a. What's the Electric Field at the center of the aperture (magnitude as well as direction)? This is not a homework question. But the reason for asking it is - The electric field at the center of a disk ("of infinitely small width ") is zero . But in the formula for the Electric Field due to disk as I put L=0 E= p/2e * (1- (L/(R^2 +L^2)^1/2) , I don't get E=0 ( where p is the charge density and e is epsilon) I searched about it on net I found something which I didn't understand wholly . I tried to attach the file but it isn't uploading. Secondly , I couldn't integrate (in my question) over the remaining charge distribution to sum the contribution of all the elements to find the field at the center of the aperture. I wasn't able to do it because the field strength was to be found out at the center of the aperture. The center of the aperture was troubling. The question also had a hint with it which said " Remembering the superposition principle , you can think about the effect of replacing the piece removed , which itself is a practically a little disc. " Does it mean that we simply find out the remaining charge on sphere (after subtracting the charge gone with the disc )? I suppose it shouldn't mean this. I would like to understand the answer using both the ways because I want to find the mistake in integration as well as understand the hint. The question figure is- enter image description here

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I'll only explain a couple of stuff and you'll do the rest. All you need to do here is find the electric field for a disc of radius r, and take the limit as r tends to zero. This limiting field wil NOT be zero because of the field produced by the point charges at a point on the conductor. Next subtract this field from the field of a conductor ($\sigma / \epsilon_0$) This will show you an interesting fact. For the spherical conductor, at any point $P$ of its surface, half the field is produced by the charges at all other points$P' \ne P$ on the surface, and the other half is produced by the charge at P. The superposition of these two contributors gives the familiar field just outside a conductor as $\sigma / \epsilon_0$.

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  • $\begingroup$ Thanks for the answer. I carried out the entire integration of the question and then did what you said. By both the ways I got the E =sigma/2*epsilon. The question is would the answer for the E remain the same ie sigma/2*epsilon had the disk taken out be not of a radius very small compared to the radius of the sphere. In other words if b wouldn't have been very small (in the question it is given that b is negligible wrt to a ) , would the answer remain the same ? Because the integration I carried out gives this answer if b is very very small. $\endgroup$
    – Shashaank
    Oct 25 '16 at 11:47
  • $\begingroup$ of course they wont be the same. The fact that you cannot neglect it anymore is proof of that alone. $\endgroup$
    – Lelouch
    Oct 25 '16 at 11:54
  • $\begingroup$ But what I could get out of books (which I know is wrong but don't know where I am getting it wrong ) is that the field anywhere just on the surface of the conductor is sigma/epsilon half of which is due to the charges at that patch and half due to the rest of the charges (as you said too) . So howsoever big a patch we choose or cut out the field at that patch before cutting was sigma/epsilon and after cutting it reduces to half of its previous value because only the field due to rest of the charges remains which as we know is half. It doesn't depend on the size of the patch cut. $\endgroup$
    – Shashaank
    Oct 25 '16 at 12:01
  • $\begingroup$ Is it due to the assumptions taken in integration because the element taken is a ring. ? But we wouldn't be able to take a ring if the patch cut was larger due to curvature effects . $\endgroup$
    – Shashaank
    Oct 25 '16 at 12:02
  • $\begingroup$ No, that is absolutely incorrect. The size of the patch has to be infinitesimally small. By the definition i gave, you can see that it will only be true if we consider the field at a point. The patch must also be point sized, denoting the charge at the point in consideration. $\endgroup$
    – Lelouch
    Oct 25 '16 at 12:20

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