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We know that the electric field inside and outside of a spherical conductor is given by $0$ and $\frac{kq}{r^2}$ respectively. The 2nd formula is derived using Gauss's law by creating a spherical gaussian surface at the distance we are interested in. But Gauss's law states that the flux is $\frac{q_{\mathrm{inside}}}{\epsilon_0}$. But if we consider the sphere itself as the gaussian surface forthe distance $r=R$,we are assuming that a charge $q$ is inside the surface whereas the charge $q$ is on the surface. So no $q_{\mathrm{inside}}$ exists in this case. How are we then still being able to apply the Gauss law?

Below I have attached a picture of an article claiming that the result is actually $\frac{kq}{2R^2}$. Now I am really confused since untill this day I knew it to be $\frac{kq}{R^2}$ but didn't know the proof. Kindly clear this confusion.

enter image description here

David Bailey
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madness
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1 Answers1

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Where does "the surface" starts and ends going from the center to the outside? Does a shell of zero thickness really exists? So, I think you have to consider a "limit" condition with r > R and tends to R (R+). I also read that article, basically if it works or not :) it's a useful exercise of logic. In the real life you can deal with potential and electric field very very close to the surface, but not exactly at distance 0 from the shell of 0 thickness. :) [EDITED] Also, if we consider the "surface" like a shell of some Angstrom thickness, we should assume that for a "point" test charge the Electric field right oh that shell is very variable due to the distance among the test charge position and effectiive charge particle on the shell: in that way we sould consider an "average value" on the entire surface.

Antonio de Lucia
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