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Consider a cavity inside a conductor, in a static situation (electrostatic). Let's consider the case when the cavity is empty: the potential on the boundary of the cavity (the conductor's inner surface) must be constant, because we can always go from one point on the boundary to another point on the boundary through a path completely inside the conductor material. Here the electric field is zero, thus

$$V_B-V_A=\int_A^B \mathbf{E}\cdot d \mathbf{l}=0$$

We can also conclude that the electric field inside the cavity is zero, because such a solution satisfies the boundary conditions (constant potential on the boundary) and the Laplace equation, thus by unicity of the solution we know this is the only correct solution.

Suppose now we have an arbitrary number of charges and/or conductors, possibly charged, in various locations inside the cavity. They're not touching the cavity surface: maybe they are suspended by isolating strings, or whatever. What happens now? I think the potential on the boundary of the cavity is still constant, because I can always go from any point on the cavity surface to any other point through a path completely inside the conductor. On the other hand, we now have charges inside the cavity and a zero electric field everywhere won't satisfy Poisson's equation, thus we cannot say anything about the electric field inside the cavity. Right?

Ps this isn't self-study. I gave my Physics exam at the University many years ago: I'm only trying to refresh some old knowledge.

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When you introduce charges into the cavity, you'll still have a constant potential on the inside cavity surface which you can chose to be zero. Then you have to solve the classical Dirichlet boundary value problem for the Poisson equation with charges as sources in order to obtain the electrostatic potential (and thus electric field) inside the cavity. $\epsilon$ is the absolute permittivity inside the cavity. $$ \Delta \Phi=-\frac{\rho(\vec r)}{\epsilon}$$ $\Phi(\vec r)=0$ for all $\vec r$ on boundary surface. The electric field is obtained from the unique solution for the potential $\Phi(\vec r)$ by $$\vec E=-\nabla \Phi(\vec r)$$

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  • $\begingroup$ Great, thanks! Can you please confirm that my proof of the assertion that "the potential is constant on the inside cavity surface" is correct, or give the right proof in case mine is wrong? $\endgroup$ – DeltaIV Dec 31 '17 at 18:07
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    $\begingroup$ "DeltaIV - Your proof seems correct to me. It is based on the assumption that the electric field is zero in the metal, which is correct. You can also assume that the lateral (tangential to the surface, not normal) electric field on any path along the surface is zero. This follows from the fact that if there was a lateral field then, due to the conductivity of the metal, currents would flow until an equilibrium, static (surface) charge distribution arises where no currents are flowing. The latter is a condition for (thermodynamic) equilibrium. $\endgroup$ – freecharly Dec 31 '17 at 19:09

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