Electric Field of a Spherical Conducting Shell

In summary, a spherical shell of inner radius a and outer radius b carries a total charge of +Q distributed on the surface of a conducting shell. 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?
  • #1
johnnyies
93
0
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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  • #2
The easy answer is Guass' 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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  • #3
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.
 

Related to Electric Field of a Spherical Conducting Shell

1. What is the formula for the electric field of a spherical conducting shell?

The formula for the electric field of a spherical conducting shell is E = Q/(4πε0R2), where Q is the total charge on the shell, ε0 is the permittivity of free space, and R is the radius of the shell.

2. How does the electric field of a conducting shell differ from that of a point charge?

The electric field of a spherical conducting shell is constant on the surface of the shell and inside the shell, while the electric field of a point charge follows an inverse-square law, becoming weaker as the distance from the charge increases.

3. What happens to the electric field inside a conducting shell if a point charge is placed inside it?

If a point charge is placed inside a conducting shell, the electric field inside the shell will be zero. This is because the charges on the shell will rearrange themselves in such a way that the electric field inside the shell cancels out the electric field of the point charge.

4. How does the electric field of a conducting shell change if the shell is not spherical?

If the conducting shell is not spherical, the formula for the electric field will change. Instead of R being the radius of the shell, it will be the distance from the center of the shell to the point where the electric field is being measured. The rest of the formula remains the same.

5. Can the electric field of a conducting shell be negative?

No, the electric field of a conducting shell is always positive or zero. This is because the electric field is a vector quantity that points in the direction of the force on a positive test charge. Since the charges on a conducting shell are always positive and repel each other, the electric field will always point away from the shell, making it positive.

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