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@@ -4,106 +4,36 @@Statement
### Solution
−Boundary conditions of two fields:
−\begin{equation}
−D_{2n}-D_{1n}=\sigma
−\end{equation}
+Let us write the boundary condition for the normal components of the electric displacement field:
+$$D_{2n} - D_{1n} = \sigma$$
+where $\sigma$ is the free surface charge density at the boundary.
+Since $D = \varepsilon_0 E$, for the first case ($E_1 = E$, $E_2 = 2E$) we have:
+$$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (2E) - \varepsilon_0 E = \varepsilon_0 E$$
+To find the pressure, consider a thin cylindrical shell at the boundary of the two media. Let us isolate a small portion of the surface with charge $\Delta q = \sigma \Delta S$. Let the field generated by this specific portion be $E_0$, and the field generated by all other charges in the system (the external field) be $E_0'$.
+From the superposition principle, the total fields on either side of the boundary are:
+$$E_2 = E_0' + E_0 = 2E$$
+$$E_1 = E_0' - E_0 = E$$
+By adding these two equations, we can find the external field $E_0'$:
+$$2E_0' = 3E \implies E_0' = \frac{3}{2}E$$
+The total force acting on this portion is:
+$$\Delta F = \sigma \Delta S E_0'$$
+Thus, the electrostatic pressure at the interface is:
+$$P = \frac{\Delta F}{\Delta S} = \sigma E_0' = (\varepsilon_0 E)\left(\frac{3}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$
−where $\sigma$ is a free charge at the boundary.
+Similarly, for the second case ($E_1 = E$, $E_2 = -2E$), we have:
+$$\sigma = \varepsilon_0 E_2 - \varepsilon_0 E_1 = \varepsilon_0 (-2E) - \varepsilon_0 E = -3\varepsilon_0 E$$
+The external field in this case is:
+$$E_0' = \frac{E_1 + E_2}{2} = \frac{E - 2E}{2} = -\frac{1}{2}E$$
+And the pressure is:
+$$P = \sigma E_0' = (-3\varepsilon_0 E)\left(-\frac{1}{2}E\right) = \frac{3\varepsilon_0 E^2}{2}$$
+<b>Why is the pressure the same?</b>
+We can obtain a general formula for the pressure $P$ using analogous reasoning. By substituting $E_0' = \frac{E_1 + E_2}{2}$ and $\sigma = \varepsilon_0(E_2 - E_1)$:
+$$P = \sigma E_0' = \varepsilon_0(E_2 - E_1)\frac{E_1 + E_2}{2} = \frac{\varepsilon_0(E_2^2 - E_1^2)}{2}$$
+From this final equation, we see that the electrostatic pressure depends on the squares of the electric fields. Therefore, changing the sign of $E_2$ leaves the pressure unchanged.
−$D_{1n}$ 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) $\sigma = \varepsilon_0 E$; $P = \frac{3\varepsilon_0 E^2}{2}$
−a)$\sigma=\varepsilon_0E$;$P=\frac{3 \varepsilon_0 E^2}{2}$ b)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$
+b) $\sigma = -3\varepsilon_0 E$; $P = \frac{3\varepsilon_0 E^2}{2}$