Правка раздела «Statement»

Arman правка от 19 апр. 2026 г., 06:14

правка #17916 предыдущая #17915 ← раньше позже →

en/6.5.3.md +1 −1

		@@ -1,6 +1,6 @@
			### Statement

		−	$6.5.3.$ [Insert the problem 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)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$
		ещё строк без изменений 103

	@@ -1,6 +1,6 @@
	### Statement		### Statement

	$6.5.3.$ [Insert the problem 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		### 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:


	\begin{equation}		\begin{equation}
	\fbox{$\sigma=-3\varepsilon_0E$}		\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}
	\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.




	#### Answer		#### Answer

	a)$\sigma=\varepsilon_0E$;$P=\frac{3 \varepsilon_0 E^2}{2}$ b)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$		a)$\sigma=\varepsilon_0E$;$P=\frac{3 \varepsilon_0 E^2}{2}$ b)$\sigma=-3\varepsilon_0E$;$P=\frac{3\varepsilon_0E^2}{2}$
	ещё строк без изменений 103