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Recently, I derived a formula for the surface charge density on the surface of a conductor( no specific shape) that is placed in an electric field from the poissons equation of electrostatics. The only condition that I considered while deriving the formula is that the outer surface of the conductor is equipotential. Now I experimented with the formula considering spheres and planes (as they are comparatively simple to deal with.). Now, I observed that the electric field outside the sphere is perfectly consistent with illustrations that I see on the internet. However, the electric field inside the sphere is fairly constant but NOT ZERO. This is an Image I found on the internet

enter image description here

enter image description here

This is what i got when I used the desmos vector field generator( I coudnt find anything better than that.). The radius of the circle is 5 units. The electric field inside the conductor isnt zero. I dont know why and how else I am supposed to solve this

EDIT:Derivation enter image description hereenter image description here

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  • $\begingroup$ If your surface is an equipotential, the field inside has to be zero. Can you share your formula? $\endgroup$ – Raziman T V Jan 13 '17 at 9:43
  • $\begingroup$ Just added the derivation $\endgroup$ – Chandrahas Jan 13 '17 at 10:28
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Laplacian of the potential on the surface is not zero because there is induced charge on the surface. Equipotential only means that the potential is constant if you move along the surface, it says nothing about the variation of the potential normal to the surface. Starting from there, the derivation is incorrect.

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  • $\begingroup$ Note that the laplacian of the potential is taken with respect to the parametric variables. Which means that there is no perpendicular component involved ( just along the surface). Hence it is zero $\endgroup$ – Chandrahas Jan 13 '17 at 11:20
  • $\begingroup$ You need a third parameter for the normal direction. As such the Laplacian is incomplete. Imagine your surface is flat and lies in the $xy$ plane, you would completely ignore the $z$ derivatives, but that is what the charge causes $\endgroup$ – Raziman T V Jan 13 '17 at 11:34
  • $\begingroup$ I didnt quite understand your last statement, can you elaborate? Anyway, isnt it just a laplacian in 2 dimensions? well, it explains the field outside the conductor pretty well. The problem is only with the interior. I was thinking if it is a problem of the epsilon. I considered it to be epsilon not even on the interior of the conductor. $\endgroup$ – Chandrahas Jan 13 '17 at 11:41
  • $\begingroup$ When you are using the Poisson equation you have to use the full 3-D Laplacian. You are setting the 2D Laplacian to zero which is fine but later using it with the Poisson equation which is incorrect. $\endgroup$ – Raziman T V Jan 13 '17 at 11:45
  • $\begingroup$ The charge density/epsilon that I substituted is the charge that exists solely on the surface( 2d surface). So, everything that I formulated is in 2 dimensional. At the end, we get a charge distribution as a function of s and t ( the parametric variables). $\endgroup$ – Chandrahas Jan 13 '17 at 11:57

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