A thin electron beam accelerated by the potential difference V enters a flat capacitor parallel to its plates. Determine the angular spread of electrons if a voltage V_0 sin wt is applied to the capacitor plates. The distance between the plates of the capacitor d is much smaller than its length l.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#7.3">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $7.3.9^*.$ A thin electron beam accelerated by the potential difference $V$ enters a flat capacitor parallel to its plates. Determine the angular spread of electrons if a voltage $V_0$ sin wt is applied to the capacitor plates. The distance between the plates of the capacitor $d$ is much smaller than its length $l$.

    <h3>Solution</h3>
    <p>
        

From the drawing
$$\tan \alpha = \frac{v_y}{v_x}$$
Law of conservation of energy
$$\frac{mv^2_x}{2}=eU$$
From where
$$v_x=\sqrt{\frac{2eU}{m}}$$
Force $\vec{F}$ acting on the particle:
$$F=eU=e\frac{U_0}{d}\sin\omega t$$
Second Newton's Laws
$$ma=\frac{eU_0}{d}\sin\omega t$$
By the definition of acceleration $a = \frac{dv}{dt}$
$$\frac{dv}{dt}=\frac{eU_0}{md}\sin\omega t$$
Let's regroup and integrate
$$\int _0^{v_y}dv=\frac{eU_0}{md}\int_0^t\sin\omega t\,dt$$
$$v_y=\frac{eU_0}{md\omega}(1-\cos\omega t)$$
Time for which the particle will move horizontally by the value $l$
$$t=\frac{l}{v_x}=l\sqrt{\frac{m}{2eU}}$$
Find the angle of velocity to the horizontal
$$\tan\theta = \frac{v_y}{v_x}=\frac{eU_0}{m\omega d}\sqrt{\frac{m}{2eU}} \left( 1-\cos\omega l \sqrt{\frac{m}{2eU}}\right)$$
Whence the angle $\angle \theta$
$$\boxed{\theta=\operatorname{arctg} \left( \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] \right)}$$
A little extra. Given that $d \ll l$ we can use the approximation $\tan x \approx x$
$${\theta= \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$
Find the required scattering angle
$${\Delta \alpha=2\theta= \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$

    <h4>Answer</h4>
    <p>
        $$\Delta\alpha=\pm\operatorname{arctg}\bigg\{\frac{V_{0}}{d\omega}\sqrt{\frac{2e}{m_{e}V}}\bigg[1-\operatorname{cos}\bigg(\omega l\sqrt{\frac{m_{e}}{2eV}}\bigg)\bigg]\bigg\}$$
    </p>
    <p style="text-align: right; font-style: italic; font-size: 14;">
      Lutfulloyev Shukurullo<br>
    </p>



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