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| <header> | | <header> |
| <h2>Solutions of Savchenko Physics Textbook</h2> | | <h2>Solutions of Savchenko Physics Textbook</h2> |
| <p class="author"> | | <p class="author"> |
| Aliaksandr Melnichenka <br/> | | Aliaksandr Melnichenka <br/> |
| October 2023 | | October 2023 |
| </p> | | </p> |
| </header> | | </header> |
| | | |
| <main> | | <main> |
| <article> | | <article> |
| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> |
| | | |
| <h4>Statement</h4> | | <h4>Statement</h4> |
| <p> | | <p> |
| $6.5.1$ | | $6.5.1$ |
| a) The surface densities of charges in two planes are $\pm \sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? | | a) The surface densities of charges in two planes are $\pm \sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes? | | b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes? |
| </p> | | </p> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge. | | c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge. |
| </p> | | </p> |
| <h4>Solution</h4> | | <h4>Solution</h4> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is | | a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is |
| </p> | | </p> |
| <center> | | <center> |
| $d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$ | | $d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$ |
| </center> | | </center> |
| <p calss="TxtSolutions"> | | <p calss="TxtSolutions"> |
| As $dq = \pm\sigma~dS$, | | As $dq = \pm\sigma~dS$, |
| </p> | | </p> |
| <center> | | <center> |
| $dF = \frac{\sigma^2~dS}{2\varepsilon_0}$ | | $dF = \frac{\sigma^2~dS}{2\varepsilon_0}$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| and moving $dS$ to another member of equation, | | and moving $dS$ to another member of equation, |
| </p> | | </p> |
| <center> | | <center> |
| $\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$ | | $\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| b) Let's see following figure | | b) Let's see following figure |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="drawb.png" | | <img src="drawb.png" |
| loading="lazy" alt="6.5.1" width="300" /> | | loading="lazy" alt="6.5.1" width="300" /> |
| <figcaption> | | <figcaption> |
| Electric field of plates | | Electric field of plates |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is | | According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is |
| </p> | | </p> |
| <center> | | <center> |
| $\vec{E} = \vec{E_1} + \vec{E_2}$ | | $\vec{E} = \vec{E_1} + \vec{E_2}$ |
| </center> | | </center> |
| <center> | | <center> |
| $E = \frac{\sigma}{\varepsilon_0}$ | | $E = \frac{\sigma}{\varepsilon_0}$ |
| </center> | | </center> |
| <center> | | <center> |
| $\boxed{\sigma = \varepsilon_0 E} \;(1)$ | | $\boxed{\sigma = \varepsilon_0 E} \;(1)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so, | | The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so, |
| </p> | | </p> |
| <center> | | <center> |
| $dF = \frac{E}{2}dq = \frac{E}{2}\sigma~dS$ | | $dF = \frac{E}{2}dq = \frac{E}{2}\sigma~dS$ |
| </center> | | </center> |
| <p class=" TxtSolutions"> | | <p class=" TxtSolutions"> |
| Hence, | | Hence, |
| </p> | | </p> |
| <center> | | <center> |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| c) See following figure | | c) See following figure |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="drawc.png" | | <img src="drawc.png" |
| loading="lazy" alt="6.5.1" width="300" /> | | loading="lazy" alt="6.5.1" width="300" /> |
| <figcaption> | | <figcaption> |
| Electric field of plates | | Electric field of plates |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so, | | Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so, |
| </p> | | </p> |
| <center> | | <center> |
| $\vec{E} = 2\vec{E_1} = 2\vec{E_2}$ | | $\vec{E} = 2\vec{E_1} = 2\vec{E_2}$ |
| </center> | | </center> |
| <center> | | <center> |
| $E = \frac{\sigma}{\varepsilon_0}$ | | $E = \frac{\sigma}{\varepsilon_0}$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m | | with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m |
| </p> | | </p> |
| <center> | | <center> |
| $\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$ | | $\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$ |
| </center> | | </center> |
| <p class="TxtSolutions"> | | <p class="TxtSolutions"> |
| Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$ | | Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$ |
| </p> | | </p> |
| <center> | | <center> |
| $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$ | | $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$ |
| </center> | | </center> |
| <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> | | <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;"> |
| BSc. Luis Daniel Fernández Quintana<br> | | BSc. Luis Daniel Fernández Quintana<br> |
| Physics Department (FCNE)<br> | | Physics Department (FCNE)<br> |
| Universidad de Oriente, Cuba<br> | | Universidad de Oriente, Cuba<br> |
| </p> | | </p> |
| </article> | | </article> |
| </main> | | </main> |
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