- #1

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- Thread starter yanyin
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- #1

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- #2

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What is the expression for Potential due to a Spherical shell

In General

[tex] V=\frac{Q}{4\pi \epsilon_0 R} **&\mbox{for}** r \leq R[/tex]

AND

[tex] V=\frac{Q}{4\pi \epsilon_0 r} **&\mbox{for}** r \geq R[/tex]

In General

[tex] V=\frac{Q}{4\pi \epsilon_0 R} **&\mbox{for}** r \leq R[/tex]

AND

[tex] V=\frac{Q}{4\pi \epsilon_0 r} **&\mbox{for}** r \geq R[/tex]

Last edited:

- #3

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So if you're inside the shell, you don't see it; if you're outside the shell then (again by an application of Gauss' law) it affects you in the same way that a concentric point charge of the same total charge would, i.e., a one over R potential.

That should get you started; you treat two shells with the superposition principle (i.e., add the two potentials together.)

P

- #4

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=> 0 which is not true.so if you're inside the shell, you don't see it

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