Electric Field of a Spherical Conducting Shell

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SUMMARY

The discussion centers on the electric field behavior of a spherical conducting shell with inner radius 'a' and outer radius 'b' carrying a total charge of +Q. According to Gauss' Law, there is no electric field inside the shell because no charge is enclosed within a Gaussian surface placed inside it. The contributions to the electric field from various patches of charge on the shell cancel each other out, resulting in an electric field of zero inside the shell while an electric field exists outside the shell.

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  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric field concepts and behavior
  • Knowledge of Coulomb's Law and its applications
  • Basic principles of symmetry in physics
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  • Investigate the differences between electric fields generated by spherical shells and infinite sheets of charge
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johnnyies
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Not a homework question:

A spherical conducting shell of inner radius a and outer radius b carries a total charge of +Q distributed on the surface of a conducting shell.

http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoAndCaptions/detail/embed03.htm

If you watch prof. Lewin, he says that a uniformly distributed charge on a shell has an electric field outside the shell, but not inside. Can someone explain why so? I thought if we had a surface such as a plate, the E-field points out in both directions, but why doesn't it exist in a sphere?

sorry for no pics.
 
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The easy answer is Gauss' Law. Since we cannot arrange a Gaussian surface inside the shell that encloses any charge, then there is no electric flux and we can conclude by arguments of symmetry (by choosing the proper Gaussian surfaces) that there must be no electric field inside the spherical shell.

The hard way of proving this is to calculate it out by doing the integration of Coulomb's Law. Obviously though, should you do this you realize that because the electric field has an associated direction, that the contributions from say a patch of charge to the right of your observation point will be countered by a similar patch of charge to the left of your observation point. The shell ensures that you always have a volume of charge surrounding you, so electric field contributions will cancel out. This is the primary geometrical difference in comparison with a sheet of charge.
 
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Hi.
Let us consider simpler but essentially similar case, a spherical shell charged +Q. Inside the shell E=0. Gauss's law is the simplest way of explanation as Born2bwire said.

To supplement it, consider inward electric field caused by the opposite side of the shell. It reduces inward electric field generated at this side. And it strengthen outward electric field.

Opposite shell 0 inside ←←← Q →→→ Outside of shell

Considering charges on shell in opposite side
  
Opposite shell Q inside __ ← Q →→→→→ outside of shell

In this way all the other part of charged spherical shell cancel the part here in inward electric field generation.
Regards.
 

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