[NOTES/EM-03010]-Electric Field Inside an Empty Cavity in a Conductor

As an application of the Maxwell's second equation \begin{equation} \nabla \times \vec{E} =0. \end{equation} we will prove that the electric field inside an empty cavity in a conductor is zero. To prove the above statement, let us assume that the electric field can be nonzero and derive a contradiction. Let us consider the integral of ${\rm curl}\,\vec{E}$ along a closed path consisting of $\Gamma_1$ along an electric line of force inside the cavity and $\Gamma_2$ in the body of conductor. Then \begin{align}\label{eq003} \int_{\Gamma_1}\vec{E}\cdot\overrightarrow{dl} + \int_{\Gamma_2} \vec{E}\cdot\overrightarrow{dl}=\int_{\Gamma_1}\vec{E}\cdot \overrightarrow{dl}~\ne~0 \end{align} The right hand side is positive because $\Gamma_2$ runs parallel to the field and $\vec{E}\cdot\overrightarrow{dl}=\vec{E}\cdot\overrightarrow{dl}$ is positive all along the path $\Gamma_1$. To get a contradiction, we note that the left hand side must vanish because Stokes theorem gives \begin{align}\label{eq004} \int_{\Gamma_1}\vec{E}\cdot\overrightarrow{dl} + \int_{\Gamma_2}\vec{E}\cdot\overrightarrow{dl} = \iint_S (\nabla\times\vec{E})\cdot dS=0\,\,. \end{align} Hence the field inside an empty cavity must vanish.