Electrostatics > Electrical pressure. The energy of an electric field > Problem 6.5.3

Statement

$6.5.3.$ What are the surface charge density and the electrostatic pressure at the boundary between two fields with magnitudes $E$ and $2E$? What about $E$ and $-2E$? In the second case, the surface charge density is three times larger. Why then is the electrostatic pressure the same in both cases?

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{$\sigma=-3\varepsilon_0E$}
\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.

Answer

a)$\sigma=\varepsilon_0E$;$P=\frac{3 \varepsilon_0 E^2}{2}$ b)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$