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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="../../#7.1">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#7.1">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $7.1.12^*.$ Determine what the accelerating potential difference $V$ should be in order for the electrons to follow the path shown in the figure. Radii of cylindrical capacitor plates $R_1$ and $R_2$. Potential difference between the plates $V_0$. | | $7.1.12^*.$ Determine what the accelerating potential difference $V$ should be in order for the electrons to follow the path shown in the figure. Radii of cylindrical capacitor plates $R_1$ and $R_2$. Potential difference between the plates $V_0$. |
| </p> | | </p> |
| <center> | | <center> |
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| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="230" /> | | loading="lazy" width="230" /> |
| <figcaption> | | <figcaption> |
| For problem $7.1.12^*$ | | For problem $7.1.12^*$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| </p> | | </p> |
| | | |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Gauss theorem | | Gauss theorem |
| $$\oint_{2\pi r}E\,dl=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{q}{2\pi \varepsilon_0 r}$$ | | $$\oint_{2\pi r}E\,dl=\frac{q}{\varepsilon_0}\Rightarrow E=\frac{q}{2\pi \varepsilon_0 r}$$ |
| Let's write down the small change of the electrostatic field and then integrate it | | Let's write down the small change of the electrostatic field and then integrate it |
| $$dU=E \,dr\Rightarrow \int dU=\int E \,dr$$ | | $$dU=E \,dr\Rightarrow \int dU=\int E \,dr$$ |
| Integrate from $R_1$ to $R_2$ | | Integrate from $R_1$ to $R_2$ |
| $$U_0=\frac{q}{2\pi \varepsilon_0 r}\int_{R_1}^{R_2}\frac{dr}{r}=\frac{q}{2\pi \varepsilon_0 r}\ln\frac{R_2}{R_1}$$ | | $$U_0=\frac{q}{2\pi \varepsilon_0 r}\int_{R_1}^{R_2}\frac{dr}{r}=\frac{q}{2\pi \varepsilon_0 r}\ln\frac{R_2}{R_1}$$ |
| Law of conservation of energy | | Law of conservation of energy |
| $$\frac{mv^2}{2}=eU$$ | | $$\frac{mv^2}{2}=eU$$ |
| $$e\frac{q}{2\pi \varepsilon_0 \frac{R_1+R_2}{2}}=\frac{mv^2}{(\frac{R_1+R_2}{2})}$$ | | $$e\frac{q}{2\pi \varepsilon_0 \frac{R_1+R_2}{2}}=\frac{mv^2}{(\frac{R_1+R_2}{2})}$$ |
| From where we get | | From where we get |
| $$\frac{q}{2\pi \varepsilon_0}=\frac{mv^2}{e}$$ | | $$\frac{q}{2\pi \varepsilon_0}=\frac{mv^2}{e}$$ |
| Substituting the previously obtained values | | Substituting the previously obtained values |
| $$\frac{U_0}{\ln\frac{R_2}{R_1}}=\frac{2eU}{e}$$ | | $$\frac{U_0}{\ln\frac{R_2}{R_1}}=\frac{2eU}{e}$$ |
| From where | | From where |
| $$U=\frac{U_0}{2\ln\frac{R_2}{R_1}}$$ | | $$U=\frac{U_0}{2\ln\frac{R_2}{R_1}}$$ |
| </p> | | </p> |
| | | |
| <h4>Answer</h4> | | <h4>Answer</h4> |
