Motion of charged particles in an electric field > Motion in an alternating electric field > Problem 7.3.10

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

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

a)$v=\frac{\omega l}{2\pi n}$ b)$d_B-d_A=\frac{4\pi eV_0 n}{m\omega^2d}$ $n$-integer