Consider a uncharged metallic sphere with a charge inside of it but not in the center, e.g. $0.5$radius upwards. What does this electric field look like? Without the sphere it is easily calculated but with the metallic sphere and the off-centering confuses me. My guess is that the charge cannot "induce" another net charge on the metallic sphere (since it's not grounded) but only reorganize the configuration of inherent charges. In the literature it seems like the strength of the electric field is proportional only to the charge which makes sense, but it radiates out from the center instea of from the "off-center". How can this be?
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3 Answers
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This is about the Green function of the sphere, basically. Check out the related question Question about the Green's function for a conducting sphere. The best visualisation I could find with a quick Google was: https://www.researchgate.net/figure/Image-charge-method-relating-to-a-point-charge-and-a-grounded-conductive-sphere-a_fig3_302981294
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This problem is solved by images. The equation for a point charge inside is the same as for a charge outside. The field will be the sum of the coulomb fields of the two charges.
answered Apr 11, 2022 at 16:07
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Due to Gauss' theorem we know that the electric field on some imaginary sphere is described entirely by the charges within that sphere. If the charge is exactly in the centre of the imaginary sphere, from symmetry we know that the electric field has only a radial component and the field is constant on the entire surface. From this, you can directly obtain Coulumb's law, since the area of a sphere is $4\pi r^2$ where $r$ is the radius of this imaginary sphere. The electric field is then computed from applying Stokes theorem to $\partial_iE_i=\epsilon_0\rho$. We obtain $$E_r 4\pi r^2 = \epsilon_0 q$$
The metal sphere that is around that charge, will not change the total charge. The radial component of electric field on the sphere that is centred around the charge will thus not change. Since the electric charge does not coincide with the centre of the metal sphere, the electric field will thus not be constant on the surface of the metal sphere and as you pointed out, will induce a displacement of charges.
This displacement of charges is only allowed to change the tangential component of the electric field of the imaginary sphere centred around the charge due to Gauss' theorem. Note however, that the electric field is no longer constant on the imaginary sphere centred around the charge!
Now, what is important to note, is that for conducting metal surfaces there is the boundary conditions that the electric field on the surface is purely radial. From this one should be able to solve for the distribution of charges on the sphere. From an outside perspective, due to this boundary condition, the field looks as if it is radiated from the centre of the sphere. (Recall that the normal vector of the sphere points to the centre). One should thus expected that behaviour!
It would be interesting to think about whether the electric field (integrated over the metal sphere) depends on the distance of the charge to the surface.
