Added Luis's English solution of 6.2.10
en/6.2.10.md
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| + | <meta name="date" content="2023-10" scheme="YYYY-MM"> | ||
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| + | <title>Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is \rho, and the second plate -\rho. Find the maximum electric field intensity.</title> | ||
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| + | <header style="text-align:center;"> | ||
| + | <div id = "logo"> | ||
| + | <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | ||
| + | </div> | ||
| + | <p class="author"> | ||
| + | Solutions of Savchenko Problems in Physics <br> | ||
| + | <i><b>knowledge must be free</b></i> | ||
| + | </p> | ||
| + | </header> | ||
| + | |||
| + | <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| + | <p> | ||
| + | $6.2.10.$ Two infinite planes intersecting at an angle $\alpha$ divide space into four regions. What is the electric field strength in regions 1 and 2 if the surface charge density of the planes is $\pm \sigma$? | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="statement.png" | ||
| + | loading="lazy" width="350" /> | ||
| + | <figcaption> | ||
| + | For problem $6.2.10$ | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | Consider the following figure... | ||
| + | </p> | ||
| + | <center> | ||
| + | <figure> | ||
| + | <img src="figure.png" | ||
| + | loading="lazy" width="350" /> | ||
| + | <figcaption> | ||
| + | Field Analysis | ||
| + | </figcaption> | ||
| + | </figure> | ||
| + | </center> | ||
| + | <p> | ||
| + | It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$. | ||
| + | </p> | ||
| + | <p> | ||
| + | For region 1 (see part $a$ of above figure): | ||
| + | </p> | ||
| + | <p> | ||
| + | $$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$ | ||
| + | </p> | ||
| + | <p> | ||
| + | and for $x$-direction, $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$. So, as $$E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2}$$, and taking in account that $$\sin{\alpha} = \sqrt{\frac{1-\cos{\alpha}}{2}}$$, it is obtained | ||
| + | </p> | ||
| + | <h4>Answer 1</h4> | ||
| + | <p> | ||
| + | $$E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}$$ | ||
| + | </p> | ||
| + | <p> | ||
| + | For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$. | ||
| + | </p> | ||
| + | <p> | ||
| + | Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$ | ||
| + | </p> | ||
| + | |||
| + | <h4>Answer 2</h4> | ||
| + | <p> | ||
| + | $$E_{R2} = \frac{\sigma}{\varepsilon_0}\cos{\frac{\alpha}{2}}$$ | ||
| + | </p> | ||
| + | |||
| + | <p style="text-align: right; font-style: italic; font-size: 14;"> | ||
| + | BSc. Luis Daniel Fernández Quintana<br> | ||
| + | Physics Department (FCNE)<br> | ||
| + | Universidad de Oriente, Cuba<br> | ||
| + | </p> | ||
| + | |||
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| <!DOCTYPE html> | |||
| <html lang="en"> | |||
| <head> | |||
| <meta charset="utf-8"> | |||
| <meta name="viewport" content="width=device-width, initial-scale=1.0"> | |||
| <meta http-equiv="content-language" content="en"> | |||
| <meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda"> | |||
| <meta name="description" content="Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is \rho, and the second plate -\rho. Find the maximum electric field intensity."> | |||
| <meta name="author" content="Aliaksandr Melnichenka"> | |||
| <meta name="date" content="2023-10" scheme="YYYY-MM"> | |||
| <meta property="og:title" content="Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is \rho, and the second plate -\rho. Find the maximum electric field intensity."> | |||
| <meta property="og:image" content="img/logo.png"> | |||
| <meta property="og:description" content="Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is \rho, and the second plate -\rho. Find the maximum electric field intensity."> | |||
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| <title>Two infinite plates of thickness h are charged uniformly in volume and stacked together. The bulk charge density of the first plate is \rho, and the second plate -\rho. Find the maximum electric field intensity.</title> | |||
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| }); | |||
| </script> | |||
| </head> | |||
| <body style=""> | |||
| <header style="text-align:center;"> | |||
| <div id = "logo"> | |||
| <span><img src = "../../img/book.png"><span><span>Savchenko Solutions</span> | |||
| </div> | |||
| <p class="author"> | |||
| Solutions of Savchenko Problems in Physics <br> | |||
| <i><b>knowledge must be free</b></i> | |||
| </p> | |||
| </header> | |||
| <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3> | |||
| <h3> Statement </h3> | |||
| <p> | |||
| $6.2.10.$ Two infinite planes intersecting at an angle $\alpha$ divide space into four regions. What is the electric field strength in regions 1 and 2 if the surface charge density of the planes is $\pm \sigma$? | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="statement.png" | |||
| loading="lazy" width="350" /> | |||
| <figcaption> | |||
| For problem $6.2.10$ | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <h3>Solution</h3> | |||
| <p> | |||
| Consider the following figure... | |||
| </p> | |||
| <center> | |||
| <figure> | |||
| <img src="figure.png" | |||
| loading="lazy" width="350" /> | |||
| <figcaption> | |||
| Field Analysis | |||
| </figcaption> | |||
| </figure> | |||
| </center> | |||
| <p> | |||
| It's known that the electric field intensity for one of the faces of a infinite plane with surface charge density is $E = \frac{\sigma}{2\varepsilon_0}$. | |||
| </p> | |||
| <p> | |||
| For region 1 (see part $a$ of above figure): | |||
| </p> | |||
| <p> | |||
| $$E_{R1y} = E - E \cos{\alpha} = \frac{\sigma}{2\varepsilon_0}(1-\cos{\alpha})$$ | |||
| </p> | |||
| <p> | |||
| and for $x$-direction, $$E_{R1x} = E \sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$. So, as $$E_{R1} = \sqrt{E_{R1x}^2+E_{R1y}^2}$$, and taking in account that $$\sin{\alpha} = \sqrt{\frac{1-\cos{\alpha}}{2}}$$, it is obtained | |||
| </p> | |||
| <h4>Answer 1</h4> | |||
| <p> | |||
| $$E_{R1} = \frac{\sigma}{\varepsilon_0}\sin{\frac{\alpha}{2}}$$ | |||
| </p> | |||
| <p> | |||
| For region 2, for $y$-direction: $$E_{R2y} = E(1+\cos{\alpha})=\frac{\sigma}{2\varepsilon_0}(1+\cos{\alpha})$$ and for $x$-axis, $$E_{R2x} = E\sin{\alpha} = \frac{\sigma}{2\varepsilon_0}\sin{\alpha}$$. | |||
| </p> | |||
| <p> | |||
| Again, $$E_{R2} = \sqrt{E_{R2x}^2+E_{R2y}^2}$$ and taking in account that $$\cos{\frac{\alpha}{2}} = \sqrt{\frac{1+\cos{\alpha}}{2}}$$ | |||
| </p> | |||
| <h4>Answer 2</h4> | |||
| <p> | |||
| $$E_{R2} = \frac{\sigma}{\varepsilon_0}\cos{\frac{\alpha}{2}}$$ | |||
| </p> | |||
| <p style="text-align: right; font-style: italic; font-size: 14;"> | |||
| BSc. Luis Daniel Fernández Quintana<br> | |||
| Physics Department (FCNE)<br> | |||
| Universidad de Oriente, Cuba<br> | |||
| </p> | |||
| <footer class="row container"> | |||
| <br> | |||
| <p> | |||
| <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small> | |||
| </p> | |||
| <p> | |||
| <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small> | |||
| </p> | |||
| </footer> | |||
| </body> | |||
| </html> | |||