Putting a point charge Q inside the thin metallic shell but not the center. How to calculate the electric field outside the shell? Can I use Gauss's law to obtain it? Would the induced charge on the shell be zero because it is a "Thin" shell?
1 Answer
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Use Gauss's Law, and the existence and uniqueness theorem for Laplace's equation given the boundary condition.
No matter what charge distribution is inside the shell, it's an equipotential surface. Taking this as our boundary condition, the electric field outside the shell is in fact uniquely determined. Moreover, since this boundary condition is spherically symmetric, the field outside the shell will also be spherically symmetric.
Reasoning with Gauss's Law shows that the field outside the shell is dependent only on the total charge inside the shell, not its placement. So, the field outside the shell is the same as when the charge is placed at the center of the shell.
Regarding charge distribution on the shell, $-Q$ will be distributed on the inner surface of the shell in a way to cancel out the electric field inside the shell itself. Then, $+Q$ will be uniformly distributed on the outer surface of the shell.
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1$\begingroup$ That is to say, although the inner surface would be nonuniform, the outer surface would maintain uniform distribution? $\endgroup$– ALLinCommented Nov 15, 2021 at 9:10
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