Новое решение

Arman правка от
Посмотреть решение История
правка #17917 позже →
en/7.3.10.md +85 −0
@@ -0,0 +1,85 @@
+### Statement
+
+$7.3.10.$ [Insert the problem statement]
+
+### 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}
+
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]