Правка разделов «Solution», «Answer»
en/6.5.3.md
+4 −16
| ### Statement | |||
| $6.5.3.$ [Insert the problem statement] | |||
| @@ -4,19 +4,6 @@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: | |||
| @@ -91,7 +78,7 @@Solution | |||
| \begin{equation} | |||
| − | \fbox{$\sigma=-3\epsilon_0E$} | ||
| + | \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} | |||
| @@ -113,9 +100,10 @@Solution | |||
| \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] | ||
| + | a)$\sigma=\varepsilon_0E$;$P=\frac{3 \varepsilon_0 E^2}{2}$ b)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$ | ||
| ### Statement | ### Statement | ||
| $6.5.3.$ [Insert the problem statement] | $6.5.3.$ [Insert the problem statement] | ||
| @@ -4,19 +4,6 @@Statement | |||
| ### Solution | ### Solution | ||
| \documentclass{article} | |||
| \usepackage{graphicx} % Required for inserting images | |||
| \title{Savchenko 6.5.3} | |||
| \begin{document} | |||
| \section{Solution} | |||
| Boundary conditions of two fields: | Boundary conditions of two fields: | ||
| \begin{equation} | \begin{equation} | ||
| D_{2n}-D_{1n}=\sigma | D_{2n}-D_{1n}=\sigma | ||
| \end{equation} | \end{equation} | ||
| where $\sigma$ is a free charge at the boundary. | where $\sigma$ is a free charge at the boundary. | ||
| $D_{1n}$ and $D_{2n}$ are normal components of electric flux density which are equal: | $D_{1n}$ and $D_{2n}$ are normal components of electric flux density which are equal: | ||
| \begin{equation} | \begin{equation} | ||
| D_{1n}=\varepsilon_0E_{1n}=\varepsilon_0E | D_{1n}=\varepsilon_0E_{1n}=\varepsilon_0E | ||
| \end{equation} | \end{equation} | ||
| \begin{equation} | \begin{equation} | ||
| D_{2n}=\varepsilon_0E_{2n}=2\varepsilon_0E | D_{2n}=\varepsilon_0E_{2n}=2\varepsilon_0E | ||
| \end{equation} | \end{equation} | ||
| Using last two equations we get our surface charge density: | Using last two equations we get our surface charge density: | ||
| \begin{equation} | \begin{equation} | ||
| \fbox{$\sigma=\varepsilon_0E$} | \fbox{$\sigma=\varepsilon_0E$} | ||
| \end{equation} | \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: | 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} | \begin{equation} | ||
| 2E=E_0+E_0' | 2E=E_0+E_0' | ||
| \end{equation} | \end{equation} | ||
| \begin{equation} | \begin{equation} | ||
| E=E_0'-E_0 | E=E_0'-E_0 | ||
| \end{equation} | \end{equation} | ||
| Adding eq(5) and eq(6) we get: | Adding eq(5) and eq(6) we get: | ||
| \begin{equation} | \begin{equation} | ||
| E_0'=\frac{3}{2}E | E_0'=\frac{3}{2}E | ||
| \end{equation} | \end{equation} | ||
| The total force acting on the portion is: | The total force acting on the portion is: | ||
| \begin{equation} | \begin{equation} | ||
| \Delta F=\sigma \Delta SE_0'=\frac{3\varepsilon_0E^2}{2}\Delta S | \Delta F=\sigma \Delta SE_0'=\frac{3\varepsilon_0E^2}{2}\Delta S | ||
| \end{equation} | \end{equation} | ||
| So the pressure at the interface between two media is: | So the pressure at the interface between two media is: | ||
| \begin{equation} | \begin{equation} | ||
| \fbox{$P=\frac{3\varepsilon_0E^2}{2}$} | \fbox{$P=\frac{3\varepsilon_0E^2}{2}$} | ||
| \end{equation} | \end{equation} | ||
| Similarly, for the second case($E_1=E$;$E_2=-2E$) we have: | Similarly, for the second case($E_1=E$;$E_2=-2E$) we have: | ||
| @@ -91,7 +78,7 @@Solution | |||
| \begin{equation} | \begin{equation} | ||
| \fbox{$\sigma=-3\epsilon_0E$} | \fbox{$\sigma=-3\varepsilon_0E$} | ||
| \end{equation} | \end{equation} | ||
| \begin{equation} | \begin{equation} | ||
| E_0'=-\frac{1}{2}E | E_0'=-\frac{1}{2}E | ||
| \end{equation} | \end{equation} | ||
| \begin{equation} | \begin{equation} | ||
| \fbox{$P=\frac{3\varepsilon_0E^2}{2}$} | \fbox{$P=\frac{3\varepsilon_0E^2}{2}$} | ||
| \end{equation} | \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: | 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} | \begin{equation} | ||
| P=\frac{\varepsilon_0(E_2^2-E_1^2)}{2} | P=\frac{\varepsilon_0(E_2^2-E_1^2)}{2} | ||
| @@ -113,9 +100,10 @@Solution | |||
| \end{equation} | \end{equation} | ||
| From last equation we see that even if the sign of $E_2$ changes, the pressure will stay the same. | From last equation we see that even if the sign of $E_2$ changes, the pressure will stay the same. | ||
| \end{document} | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | a)$\sigma=\varepsilon_0E$;$P=\frac{3 \varepsilon_0 E^2}{2}$ b)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$ | ||