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I'm a bit confused about how to calculate the $E-\text{field}$ in a spherical capacitor.

As many sources state out, its calculated the same way as for a point charge e.g. for a charged sphere: $E = Q/(4\pi\epsilon r^2)$ for $r\geq\text{the radius of the sphere}$.

This is equal to the eletric field of the inner part of the capacitor, again only for $r\geq\text{ the radius of the inner part}$.

This leads me to the following question: What's the electric field "produced" by the outer part of the capacitor inside the capacitor?

Many sources in german language state out, that the electric field inside a conductive hollow sphere is zero. It seems logic to me that this is the case for the center of the sphere, but what about a point near the rim of the sphere?

Edit: I understand that field lines go from positive to negative charges. Since we have no negative charges inside a empty hollow sphere we have no field lines there ergo no electric field. But in case of a capacitor we put a charged sphere inside the the shell with opposite sign of the charge as the shell. Why do we then ignore the field from the shell?

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In a spherical capacitor you have two concentric conductor spheres. According to Gauss law, if the inner shell(radius $r_1$) has a charge $Q$, the electric field in the dielectric at a radius $r≧r_1$, will only be determined by this charge, there is no influence of a charge at the radius $r_2$ of the outer shell. This is even true if there is no outer shell at all.

You have no electric field inside a charged conductive sphere as long as you have no charge there. If, e.g. you insert a positive point charge inside a hollow conductive sphere (also away from the center) you will have an inside electric field distribution with field lines ending on induced charges on the inner surface of the conductor.

In the capacitor, there is no contribution to the electric field from the symmetric charges on the outer shell in the space between the conductive shells. This can be directly shown by adding the contributions of all charge elements on the outer shell using Coulomb's law. This has already been demonstrated by Isaac Newton with his Shell Theorem for his analogous gravitational law.

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In a spherical shell conductor, any field inside the shell will cause the electrons at the surface to rearrange so that the electric field inside will be 0.

For a thought experiment, think of a small electric field just inside the surface. that field would apply a force on the surface electrons and cause them to move and cancel out the force.

This is covered quite well in Classical Electrodynamics by Jackson. http://www.fisica.unlp.edu.ar/materias/electromagnetismo-licenciatura-en-fisica-medica/electromagnetismo-material-adicional/Jackson%20-%20Classical%20Electrodynamics%203rd%20edition.pdf/view

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  • $\begingroup$ Sure, in a conductor we can expect the field to become 0. But if we have a spherical capacitor with inner radius somewhat smaller as the outer, so that we can't assume the field to be homogenous. And lets say the outer one is positive charged and the inner negative. And if we put then a negative charge between the two. Then the force on this negative charge comes at least partially from the positive charges on the shell, seems as no? But why? if we have a charge inside a charged hollow sphere I can understand that the sum of all forces is 0, but near the rim this can't be true. $\endgroup$ – Felix Crazzolara Oct 19 '16 at 14:35

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