Electric charge on hollow spherical conducting shells Gauss's Law(?)

In summary, the conversation discusses two spherical hollow conducting shells, one inside the other, with the outer shell having a radius four times that of the inner shell. Both shells have a positive charge, with the switch connecting them being open. When the switch is closed, all the charge moves to the outer shell due to the fact that the electric field inside a conductor is zero. This is because connecting the two shells creates a strange inner cavity, resulting in all the positive charges being pushed to the outer shell in order to maintain a zero electric field inside the whole conductor.
  • #1
offbeatjumi
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There are two spherical hollow conducting shells, one inside the other. The outer shell has a radius 4 times that of the inner shell. There is a switch connecting the two shells, which is for the moment, open. Both shells have a positive charge (inner shell = +20nC and outer shell = +60nC.

When the switch is closed... can someone explain why all the charge moves to the outer shell? The fact that there is a conducting wire (switch) connecting the two spheres makes it confusing for me to apply a Gaussian surface.

Thanks =)
 
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  • #2
The electric field inside a conductor is zero. Basically what you are doing by connecting the outer sphere and the inner sphere is just making a really strange inner cavity. To maintain the fact that the field inside the whole conductor, the two shells, is zero all the positive charges will have to go to the outer shell. This is kind of a radical example, but as I said, just think of connecting them as making a big conductor with a really weird cavity.
 

1. What is Gauss's Law and how does it relate to electric charge on hollow spherical conducting shells?

Gauss's Law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed by that surface. In the case of hollow spherical conducting shells, Gauss's Law states that the electric field inside the shell is zero, and the electric field outside the shell is the same as if all the charge were concentrated at the center of the sphere.

2. How can we calculate the electric field inside and outside of a hollow spherical conducting shell using Gauss's Law?

To calculate the electric field inside the shell, we use Gauss's Law and the fact that the electric field is zero inside the shell. To calculate the electric field outside the shell, we use Gauss's Law and the fact that the electric field is the same as if all the charge were concentrated at the center of the sphere.

3. What is the significance of a hollow spherical conducting shell in the context of electric charge and Gauss's Law?

A hollow spherical conducting shell is a special case that allows us to simplify the calculation of the electric field using Gauss's Law. This is because the conducting shell ensures that the electric field inside the shell is zero, and the electric field outside the shell follows the same pattern as if all the charge were concentrated at the center of the sphere.

4. Can Gauss's Law be applied to non-spherical shapes or objects made of materials other than conducting shells?

Yes, Gauss's Law can be applied to any closed surface, regardless of shape or material. However, the calculation of the electric field may be more complex for non-spherical shapes or objects made of materials other than conducting shells.

5. How does the electric charge distribution on a hollow spherical conducting shell affect the electric field inside and outside the shell?

The electric charge distribution on a hollow spherical conducting shell does not affect the electric field inside the shell, as it is always zero. However, the distribution of charge on the shell will affect the strength of the electric field outside the shell, as the electric field is stronger closer to areas with higher charge density.

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