Solution

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\title{Savchenko 6.5.3}

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\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.
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