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Conductors with off-centered cavity |
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| Feb19-11, 10:57 PM | #1 |
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Conductors with off-centered cavity
Suppose we had an uncharged solid spherical conductor with radius R, and some spherical cavity of radius a that is located b units above R, where a+b<R. If we place a charge inside the cavity at the center, how would you quantitatively express the E-field lines outside of the conductor?
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| Feb19-11, 11:08 PM | #2 |
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This is NOT a homework. Concentric cavities are too easy. I'm wondering if we can somehow apply uniqueness here to show that it actually doesn't matter where the cavity is...
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| Feb20-11, 12:11 AM | #3 |
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The E field is easy to find using Gauss's law . And it doesn't matter where the charge is inside the sphere or the shape of the cavity.
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| Feb20-11, 12:37 AM | #4 |
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Conductors with off-centered cavityYou can use Gauss' Law to prove every part of that by setting up surfaces just outside the cavity and just outside the sphere and using the relation: [tex]\Phi_E = \int\vec{E}\cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}[/tex] Where you know [tex]\vec{E} = 0[/tex] inside the conductor. |
| Feb20-11, 01:07 AM | #5 |
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True. But to solve that surface integral simply one must apply a symmetry argument. How do you arrive about that symmetry? The induced charge will not be evenly distributed on the outside of the sphere, but is there a way to apply uniqueness to arguing that?
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| Feb20-11, 06:02 AM | #6 |
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Recognitions:
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The outer surface of the conductor will be an equipotential surface of radius R. An outside potential phi=Q/r satisfies this boundary condition. It is thus the unique potential outside the sphere.
It doesn't matter what goes on in cavities within the conduciting sphere. |
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