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I know that the electric field intensity at the origin of a uniformly charged sphere is 0.

Question: Does this mean that the electric field intensity of a non-uniformly charged sphere at the origin is 0 as well? I have tried this and got 0 as a result, am I correct?

Edit:

Noticed this changes the question a lot.

$\rho_{v} = \alpha r$ , where $\alpha$ is constant

$a<r<b$ , $0<\theta <\pi$ , $0<\phi <2\pi$

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  • $\begingroup$ Trying to understand your edit: do you mean that spherical symmetry is preserved but each "shell" of the sphere has a different charge density? $\endgroup$ – Floris Jun 6 '17 at 16:18
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If you have a non-uniformly charged sphere, you have removed symmetry from the system. In the limiting case, "non uniform distribution of charge on a sphere" is two unequal point sources on opposite ends of the sphere. You can see trivially that that would give a non-zero field at the origin.

So either your derivation is wrong, or you are leaving some information out with regards to your setup / boundary conditions.

UPDATE

You edited the question since I first wrote my answer. If you can consider your sphere to be made up of many spherical shells, centered on the origin, and with different charge densities, then each of those shells has zero field at the center and their superposition will still give zero field at the center.

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Hint: Think about the symmetry of the charge distribution, especially along the radial direction.

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No, the electric field will not be zero at the origin of a non-uniform charged sphere. Explain: You can assume a uniform charged sphere consists of two hemispheres. The electric field due to two uniformly charged hemispheres will cancel each other resulting zero electric field at origin. But in case of non-uniform charged spheres, the net electric field is not zero at origin.

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