Правка разделов «Statement», «Solution», «Answer»
en/7.3.10.md
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| @@ -1,10 +1,12 @@ | |||
| ### Statement | |||
| − | $7.3.10.$ [Insert the problem statement] | ||
| + | $7.3.10.$ A device designed to select electrons with a specific velocity from an electron beam consists of a parallel-plate capacitor of length $l$, shielded on both sides by screens. The first screen has an input aperture $A$, and the second has a long output channel $B$. An alternating voltage with frequency $\omega$ and amplitude $V_0$ is applied to the plates. The distance between the plates is $d$. | ||
| + | a. What is the velocity of the electrons selected by the device from a beam entering parallel to the plates? | ||
| + | |||
| + | b.* By how much must aperture $A$ be narrower than channel $B$ to ensure that the selected group of electrons passes through the channel? | ||
| ### Solution | |||
| − | \section{Solution} | ||
| Let $V(t)=V_0\sin{\omega t}$.This voltage creates an alternating field which equals: | |||
| \begin{equation} | |||
| E(t)=\frac{V_0}{d}\sin{\omega t} | |||
| \end{equation} | |||
| Since there are no any forces in the direction of initial speed,the projection of the speed to that direction will stay the same.But,there is a vertical force acting on electrons which causes vertical displacement.By applying Newton's second law to $y$ axis we get: | |||
| \begin{equation} | |||
| m\frac{dv_y}{dt}=-e\frac{V_0}{d}\sin{\omega t} | |||
| \end{equation} | |||
| By multiplying both sides by $dt$ and integrating we get: | |||
| \begin{equation} | |||
| v_y(t)=\frac{eV_0}{m\omega d}(1-\cos{\omega t}) | |||
| \end{equation} | |||
| For the beam to exit through the channel, its vertical velocity component must be zero when passing through the outlet channel B.The time to pass the capacitor is: | |||
| \begin{equation} | |||
| t=\frac{l}{v_0} | |||
| \end{equation} | |||
| where $v_0$ is an initial speed of the beam. | |||
| Using that $v_y(\frac{l}{v_0})=0$ we get the following: | |||
| \begin{equation} | |||
| \cos{\frac{\omega l}{v_0}}=1 | |||
| \end{equation} | |||
| Solving last equation we get our initial speed: | |||
| \begin{equation} | |||
| v_0=\frac{\omega l}{2\pi n} | |||
| \end{equation} | |||
| where $n$ is an integer.The speed of the electrons when exiting the device will be exactly equal to $v_0$ as the vertical velocity becomes zero: | |||
| \begin{equation} | |||
| \fbox{$v=v_0=\frac{\omega l}{2\pi n} $} | |||
| \end{equation} | |||
| Let widths of channels A and B are $d_A$ and $d_B$ respectively. | |||
| Let us consider the outermost electron relative to the point of origin. If this electron can pass through the channel, then the entire beam will also pass through. | |||
| \begin{equation} | |||
| \Delta y=\frac{d_B-d_A}{2}=\int_{0}^{\frac{l}{v_0}}\frac{eV_0}{m\omega d}(1-\cos{\omega t})dt | |||
| \end{equation} | |||
| By integrating last equation and using obtained eq(7) we get: | |||
| \begin{equation} | |||
| \fbox{$d_B-d_A=\frac{4\pi eV_0 n}{m\omega ^2d} $} | |||
| \end{equation} | |||
| @@ -82,4 +84,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | a)$v=\frac{\omega l}{2\pi n}$ b)$d_B-d_A=\frac{4\pi eV_0 n}{m\omega^2d}$ $n$-integer | ||
| @@ -1,10 +1,12 @@ | |||
| ### Statement | ### Statement | ||
| $7.3.10.$ [Insert the problem statement] | $7.3.10.$ A device designed to select electrons with a specific velocity from an electron beam consists of a parallel-plate capacitor of length $l$, shielded on both sides by screens. The first screen has an input aperture $A$, and the second has a long output channel $B$. An alternating voltage with frequency $\omega$ and amplitude $V_0$ is applied to the plates. The distance between the plates is $d$. | ||
| a. What is the velocity of the electrons selected by the device from a beam entering parallel to the plates? | |||
| b.* By how much must aperture $A$ be narrower than channel $B$ to ensure that the selected group of electrons passes through the channel? | |||
| ### Solution | ### Solution | ||
| \section{Solution} | |||
| Let $V(t)=V_0\sin{\omega t}$.This voltage creates an alternating field which equals: | Let $V(t)=V_0\sin{\omega t}$.This voltage creates an alternating field which equals: | ||
| \begin{equation} | \begin{equation} | ||
| E(t)=\frac{V_0}{d}\sin{\omega t} | E(t)=\frac{V_0}{d}\sin{\omega t} | ||
| \end{equation} | \end{equation} | ||
| Since there are no any forces in the direction of initial speed,the projection of the speed to that direction will stay the same.But,there is a vertical force acting on electrons which causes vertical displacement.By applying Newton's second law to $y$ axis we get: | Since there are no any forces in the direction of initial speed,the projection of the speed to that direction will stay the same.But,there is a vertical force acting on electrons which causes vertical displacement.By applying Newton's second law to $y$ axis we get: | ||
| \begin{equation} | \begin{equation} | ||
| m\frac{dv_y}{dt}=-e\frac{V_0}{d}\sin{\omega t} | m\frac{dv_y}{dt}=-e\frac{V_0}{d}\sin{\omega t} | ||
| \end{equation} | \end{equation} | ||
| By multiplying both sides by $dt$ and integrating we get: | By multiplying both sides by $dt$ and integrating we get: | ||
| \begin{equation} | \begin{equation} | ||
| v_y(t)=\frac{eV_0}{m\omega d}(1-\cos{\omega t}) | v_y(t)=\frac{eV_0}{m\omega d}(1-\cos{\omega t}) | ||
| \end{equation} | \end{equation} | ||
| For the beam to exit through the channel, its vertical velocity component must be zero when passing through the outlet channel B.The time to pass the capacitor is: | For the beam to exit through the channel, its vertical velocity component must be zero when passing through the outlet channel B.The time to pass the capacitor is: | ||
| \begin{equation} | \begin{equation} | ||
| t=\frac{l}{v_0} | t=\frac{l}{v_0} | ||
| \end{equation} | \end{equation} | ||
| where $v_0$ is an initial speed of the beam. | where $v_0$ is an initial speed of the beam. | ||
| Using that $v_y(\frac{l}{v_0})=0$ we get the following: | Using that $v_y(\frac{l}{v_0})=0$ we get the following: | ||
| \begin{equation} | \begin{equation} | ||
| \cos{\frac{\omega l}{v_0}}=1 | \cos{\frac{\omega l}{v_0}}=1 | ||
| \end{equation} | \end{equation} | ||
| Solving last equation we get our initial speed: | Solving last equation we get our initial speed: | ||
| \begin{equation} | \begin{equation} | ||
| v_0=\frac{\omega l}{2\pi n} | v_0=\frac{\omega l}{2\pi n} | ||
| \end{equation} | \end{equation} | ||
| where $n$ is an integer.The speed of the electrons when exiting the device will be exactly equal to $v_0$ as the vertical velocity becomes zero: | where $n$ is an integer.The speed of the electrons when exiting the device will be exactly equal to $v_0$ as the vertical velocity becomes zero: | ||
| \begin{equation} | \begin{equation} | ||
| \fbox{$v=v_0=\frac{\omega l}{2\pi n} $} | \fbox{$v=v_0=\frac{\omega l}{2\pi n} $} | ||
| \end{equation} | \end{equation} | ||
| Let widths of channels A and B are $d_A$ and $d_B$ respectively. | Let widths of channels A and B are $d_A$ and $d_B$ respectively. | ||
| Let us consider the outermost electron relative to the point of origin. If this electron can pass through the channel, then the entire beam will also pass through. | Let us consider the outermost electron relative to the point of origin. If this electron can pass through the channel, then the entire beam will also pass through. | ||
| \begin{equation} | \begin{equation} | ||
| \Delta y=\frac{d_B-d_A}{2}=\int_{0}^{\frac{l}{v_0}}\frac{eV_0}{m\omega d}(1-\cos{\omega t})dt | \Delta y=\frac{d_B-d_A}{2}=\int_{0}^{\frac{l}{v_0}}\frac{eV_0}{m\omega d}(1-\cos{\omega t})dt | ||
| \end{equation} | \end{equation} | ||
| By integrating last equation and using obtained eq(7) we get: | By integrating last equation and using obtained eq(7) we get: | ||
| \begin{equation} | \begin{equation} | ||
| \fbox{$d_B-d_A=\frac{4\pi eV_0 n}{m\omega ^2d} $} | \fbox{$d_B-d_A=\frac{4\pi eV_0 n}{m\omega ^2d} $} | ||
| \end{equation} | \end{equation} | ||
| @@ -82,4 +84,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | a)$v=\frac{\omega l}{2\pi n}$ b)$d_B-d_A=\frac{4\pi eV_0 n}{m\omega^2d}$ $n$-integer | ||