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| <h2>Solutions of Savchenko Problems in Physics</h2> | | <h2>Solutions of Savchenko Problems in Physics</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <h3 id="back-link"><a href="../../#7.3">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#7.3">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <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 $l$. | | $7.3.9^*.$ 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 $l$. |
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| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $7.3.9^*$ | | For problem $7.3.9^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| | | |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="draw.png" | | <img src="draw.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| Trajectory of a particle in an electric field | | Trajectory of a particle in an electric field |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| | | |
| From the drawing | | From the drawing |
| $$\tan \alpha = \frac{v_y}{v_x}$$ | | $$\tan \alpha = \frac{v_y}{v_x}$$ |
| Law of conservation of energy | | Law of conservation of energy |
| $$\frac{mv^2_x}{2}=eU$$ | | $$\frac{mv^2_x}{2}=eU$$ |
| From where | | From where |
| $$v_x=\sqrt{\frac{2eU}{m}}$$ | | $$v_x=\sqrt{\frac{2eU}{m}}$$ |
| Force $\vec{F}$ acting on the particle: | | Force $\vec{F}$ acting on the particle: |
| $$F=eU=e\frac{U_0}{d}\sin\omega t$$ | | $$F=eU=e\frac{U_0}{d}\sin\omega t$$ |
| Second Newton's Laws | | Second Newton's Laws |
| $$ma=\frac{eU_0}{d}\sin\omega t$$ | | $$ma=\frac{eU_0}{d}\sin\omega t$$ |
| By the definition of acceleration $a = \frac{dv}{dt}$ | | By the definition of acceleration $a = \frac{dv}{dt}$ |
| $$\frac{dv}{dt}=\frac{eU_0}{md}\sin\omega t$$ | | $$\frac{dv}{dt}=\frac{eU_0}{md}\sin\omega t$$ |
| Let's regroup and integrate | | Let's regroup and integrate |
| $$\int _0^{v_y}dv=\frac{eU_0}{md}\int_0^t\sin\omega t\,dt$$ | | $$\int _0^{v_y}dv=\frac{eU_0}{md}\int_0^t\sin\omega t\,dt$$ |
| $$v_y=\frac{eU_0}{md\omega}(1-\cos\omega t)$$ | | $$v_y=\frac{eU_0}{md\omega}(1-\cos\omega t)$$ |
| Time for which the particle will move horizontally by the value $l$ | | Time for which the particle will move horizontally by the value $l$ |
| $$t=\frac{l}{v_x}=l\sqrt{\frac{m}{2eU}}$$ | | $$t=\frac{l}{v_x}=l\sqrt{\frac{m}{2eU}}$$ |
| Find the angle of velocity to the horizontal | | Find the angle of velocity to the horizontal |
| $$\tan\theta = \frac{v_y}{v_x}=\frac{eU_0}{m\omega d}\sqrt{\frac{m}{2eU}} \left( 1-\cos\omega l \sqrt{\frac{m}{2eU}}\right)$$ | | $$\tan\theta = \frac{v_y}{v_x}=\frac{eU_0}{m\omega d}\sqrt{\frac{m}{2eU}} \left( 1-\cos\omega l \sqrt{\frac{m}{2eU}}\right)$$ |
| Whence the angle $\angle \theta$ | | Whence the angle $\angle \theta$ |
| $$\boxed{\theta=\operatorname{arctg} \left( \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] \right)}$$ | | $$\boxed{\theta=\operatorname{arctg} \left( \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] \right)}$$ |
| A little extra. Given that $d \ll l$ we can use the approximation $\tan x \approx x$ | | A little extra. Given that $d \ll l$ we can use the approximation $\tan x \approx x$ |
| $${\theta= \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$ | | $${\theta= \frac{U_0}{\omega d}\sqrt{\frac{e}{2mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$ |
| Find the required scattering angle | | Find the required scattering angle |
| $${\Delta \alpha=2\theta= \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$ | | $${\Delta \alpha=2\theta= \frac{U_0}{\omega d}\sqrt{\frac{2e}{mU}} \bigg[ 1-\cos\omega l \sqrt{\frac{m}{2eU}}\bigg] }$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$\Delta\alpha=\pm\operatorname{arctg}\bigg\{\frac{V_{0}}{d\omega}\sqrt{\frac{2e}{m_{e}V}}\bigg[1-\operatorname{cos}\bigg(\omega l\sqrt{\frac{m_{e}}{2eV}}\bigg)\bigg]\bigg\}$$ | | $$\Delta\alpha=\pm\operatorname{arctg}\bigg\{\frac{V_{0}}{d\omega}\sqrt{\frac{2e}{m_{e}V}}\bigg[1-\operatorname{cos}\bigg(\omega l\sqrt{\frac{m_{e}}{2eV}}\bigg)\bigg]\bigg\}$$ |
