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I found the following explanation of Griffiths Introduction to Electrodynamics, chapter 2.5, page 100-101.

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So is it correct to state the following?

$$\oint_{\gamma} \bf{E} \cdot \bf{ds}=0 \,\,\,\,\, \forall\gamma \,\,\,\,\implies \,\,\,\,\bf{E}=0 \,\,\,\, \mathrm{everywhere \,\,\, inside \,\,\, the \,\,\, conductor}$$

Where $\gamma$ is any closed curve made as described in the text (i.e. composed of a curve joining two points of the cavity and another curve passing only inside the conductor).

Furthermore, is it possible to conclude from here that there cannot be any charge on the internal surface of the conductor? That is

$$\bf{E}=0 \,\,\,\, \mathrm{everywhere \,\,\, inside \,\,\, the \,\,\, cavity} \,\,\,\implies\,\,\,\mathrm{ q=0\,\,\,everywhere \,\,\, on \,\,\, the \,\,\, internal \,\,\,surface \,\,\, of \,\,\, the \,\,\, conductor}$$

If so, how to justify this last implication?

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If $$\int_{\gamma}\vec{E}.\mathrm{d}\vec{s}=0\quad \forall\gamma\\ \int_{\gamma}\vec{E}.\mathrm{d}\vec{s}=\iint_{\delta}\left(\nabla\times\vec{E}\right).\mathrm{d}\vec{A}=0\quad \forall \delta\quad \text{inside the surface}\\ \text{It means}\quad \nabla\times\vec{E}=0$$ This hints at $\vec{E}$ being a gradient of some sort, in fact we know that the electric field would be an electrostatic gradient $\vec{E}=-\nabla V$, where $V$ is the electrostatic potential function. So, we have Gauss' law which allows the field to diverge. $$\nabla.\vec{E}=\frac{\rho}{\epsilon}$$ but since inside the conductor no charge exists, hence, the charge density $\rho = 0$, thus we have $$\nabla\times\vec{E}=0\quad\text{and}\quad \nabla.\vec{E}=0$$ This implies $\vec{E}$ has to be a constant vector for sure, which is zero in this case.

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If E=0, there cannot be charge on the internal surface, because the presence of charge would yield a discontinuity in the E field perpendicular to the surface (by Gauss' law). But the field is zero on both sides, so no discontinuity.

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Relating to your first question, yes it true that if the line integral is zero for all paths (as defined in the textbook), then the electric field is zero everywhere inside the cavity. The textbook proves this by showing that if the electric field is not zero at some point inside the cavity, then you can find a path along which the line integral is non-zero.

Assume that the electric field is non-zero at any point inside the cavity. Then there has to be a field line passing through the point. Since, there are no charges inside the cavity, the field line has to go from a positive charge on the surface of the conductor to a negative charge on the surface of the conductor. Then choose $\gamma$ to be the path along the field line.
Let the electric field along the path be $E_l$. By the definition of the path $E_l> 0$ all along the path. Therefore, the line integral is non-zero, which is not possible.

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protected by Qmechanic Dec 30 '16 at 6:53

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