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+ <title>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.</title>
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+ <header style="text-align:center;">
+ <h2>Solutions of Savchenko Problems in Physics</h2>
+ <p class="author">
+ Aliaksandr Melnichenka <br/>
+ October 2023
+ </p>
+ </header>
+
+ <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 $l$.
+
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ For problem $7.3.9^*$
+ </figcaption>
+ </figure>
+</center>
+<p>
+ </p>
+
+ <h3>Solution</h3>
+ <p>
+
+</p>
+<center>
+ <figure>
+ <img src="draw.png"
+ loading="lazy" width="230" />
+ <figcaption>
+ Trajectory of a particle in an electric field
+ </figcaption>
+ </figure>
+</center>
+<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
+$${\Delta \alpha=2\theta= \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$
+ </p>
+
+ <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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