Prove: "If a cavity surrounded by conducting material is itself empty of charge, then the field within the cavity is zero."

My attempt:

Take a Gaussian surface that lies completely inside the cavity. Since $Q_{enc}=0$, $\iint\mathbf{E}\cdot d\mathbf{a}=0$. But this doesn't necessarily mean that $\mathbf{E}=\mathbf{0}$ everywhere on the Gaussian surface; it could just be perpendicular to $d\mathbf{a}$ at some places on the surface. How do I prove that $\mathbf{E}=\mathbf{0}$ everywhere within the cavity?


closed as off-topic by David Z May 3 at 9:38

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  • $\begingroup$ I'm not sure why my question has been put on hold. I've posted my attempt at a solution and for the record, it isn't even homework -- I'm just working my way through Griffith's. $\endgroup$ – Thomas May 3 at 9:49
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    $\begingroup$ I think this is a perfectly conceptual question. It's not "help me find a formula and plug numbers into it". If this is too homework-like, every single question on the 'related' sidebar on the right should be closed too. $\endgroup$ – knzhou May 3 at 9:56
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    $\begingroup$ Ah, but it has to be perpendicular to every gaussian surface within the cavity, and E=0 is therefore at least one solution. But there is always only one solution; so E must be 0. $\endgroup$ – camel May 3 at 10:33