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Say I try to find the magnitude of the electric field at any point within an insulating solid sphere. I know that in the case of a conductor, the electric field within it is 0. However, I have not learned anything about an insulator, so I assume that it would not be 0.

I used Gauss' Law and calculated the charge of the volume within the Gaussian surface, the radius of which is equal to the distance between the point of interest to the center of the sphere. So I got the right answer, but I want to know the physics behind it. Why does the remaining volume of the insulating sphere, which is just right outside the Gaussian surface, have no effect on the electric field at that point? Even to me, my question sounds flawed as I am pretty much asking why an insulator has no effect on an electric field. However, I just don't think it would be that simple.

Qmechanic
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Skipher
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3 Answers3

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This is somewhat similar to why the rest of the earth doesn't influence the gravitational field inside it. By the same logic, the net electric force of all of the charges on 1 half of the outer side cancel each other due to the presence of corresponding charges on the other half, resulting in no net field due to the outer shell charges.

Tamoghna Chowdhury
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There certainly does seem to be confusion here. It is a question of symmetry, but not just the symmetry of the shape and Gaussian surface.

It is electric charges that are responsible for electric fields, whether you are talking about insulators or conductors. You don't say, but I am assuming you are dealing with an insulator that has a certain amount of charge distributed uniformly within its volume? If so, then its spherical shape ensures the symmetry which enables you to use Gauss's law to solve the problem. If the shape was non-spherical or the charge was not uniformly distributed, then although Gauss's law is of course still completely applicable, in practice it is very difficult to apply in a simple analytical way.

A "Gaussian surface" that you construct should have the electric field lines either perpendicular or parallel to the surface. This is so that you can easily calculate the electric flux on the LHS of Gauss's law without having to worry about scalar products or resolving the electric field in the direction of the surface. If the charge is uniformly distributed and the insulator is spherical, then you can be sure that the electric field also has spherical symmetry and that the electric field is parallel to the surface vector of a spherical Gaussian surface with the same central point. In this case, the electric fields produced by the charges distributed in the material outside your Gaussian surface can be shown to cancel out inside that surface. This is often called the shell theorem. However, if the electric field was not spherically symmetric, because perhaps the charge was not uniformly distributed in the sphere, then a simple application of Gauss's law would not be possible and the fields due to charges outside a spherical surface would not cancel inside that surface.

In other words, the symmetry that you exploit to solve this problem is a symmetry of the chosen surface you construct and an assumption that the electric field is also symmetric.

The crucial difference between insulators and conductors is that in conductors the charges can rearrange themselves in response to asymmetries in the electric field in a way that can remove these asymmetries.

ProfRob
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Who says the outside field doesn't effect.The Gauss's law gives the net field due to entire charges inside or outside the Gaussian surface only the charge taken is what inside.

Aman
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