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en/6.5.3.md +121 −0
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+### Statement
+
+$6.5.3.$ [Insert the problem statement]
+
+### Solution
+
+\documentclass{article}
+\usepackage{graphicx} % Required for inserting images
+
+\title{Savchenko 6.5.3}
+
+\begin{document}
+
+
+\section{Solution}
+
+
+
+
+Boundary conditions of two fields:
+\begin{equation}
+D_{2n}-D_{1n}=\sigma
+\end{equation}
+
+
+where $\sigma$ is a free charge at the boundary.
+
+
+$D_{1n}$ and $D_{2n}$ are normal components of electric flux density which are equal:
+
+
+\begin{equation}
+D_{1n}=\varepsilon_0E_{1n}=\varepsilon_0E
+\end{equation}
+
+
+
+
+\begin{equation}
+D_{2n}=\varepsilon_0E_{2n}=2\varepsilon_0E
+\end{equation}
+
+
+Using last two equations we get our surface charge density:
+
+
+\begin{equation}
+\fbox{$\sigma=\varepsilon_0E$}
+\end{equation}
+
+
+To find the pressure we could consider a thin cylindrical shell at the boundary of two media.Consider a portion of a cylinder with charge $\sigma\Delta S$.Let the field of this part be $E_0$ and that of the remaining part be $E_0'$.From the superposition we know that:
+
+
+
+\begin{equation}
+2E=E_0+E_0'
+\end{equation}
+
+\begin{equation}
+E=E_0'-E_0
+\end{equation}
+
+
+
+Adding eq(5) and eq(6) we get:
+
+
+\begin{equation}
+E_0'=\frac{3}{2}E
+\end{equation}
+
+
+
+The total force acting on the portion is:
+
+
+\begin{equation}
+\Delta F=\sigma \Delta SE_0'=\frac{3\varepsilon_0E^2}{2}\Delta S
+\end{equation}
+
+
+So the pressure at the interface between two media is:
+
+\begin{equation}
+\fbox{$P=\frac{3\varepsilon_0E^2}{2}$}
+\end{equation}
+
+
+Similarly, for the second case($E_1=E$;$E_2=-2E$) we have:
+
+
+\begin{equation}
+\fbox{$\sigma=-3\epsilon_0E$}
+\end{equation}
+
+\begin{equation}
+E_0'=-\frac{1}{2}E
+\end{equation}
+
+
+\begin{equation}
+\fbox{$P=\frac{3\varepsilon_0E^2}{2}$}
+\end{equation}
+
+
+
+As we obtained the pressure at the interface stayed the same.In fact,we could obtain a general for $P$ using analogous reasoning:
+
+
+\begin{equation}
+P=\frac{\varepsilon_0(E_2^2-E_1^2)}{2}
+\end{equation}
+
+From last equation we see that even if the sign of $E_2$ changes, the pressure will stay the same.
+\end{document}
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]