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| + | <header style="text-align:center;"> | ||
| <h2>Solutions of Savchenko Physics Textbook</h2> | |||
| <p class="author"> | |||
| Aliaksandr Melnichenka <br/> | |||
| October 2023 | |||
| </p> | |||
| </header> | |||
| − | | ||
| − | <article> | ||
| − | <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | ||
| − | |||
| − | <h4>Statement</h4> | ||
| + | <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | ||
| + | |||
| + | <h3> Statement </h3> | ||
| <p> | |||
| $6.5.1$ | |||
| 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? | |||
| </p> | |||
| <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? | |||
| </p> | |||
| @@ -42,135 +53,77 @@ | |||
| <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. | |||
| </p> | |||
| − | <h | ||
| − | | ||
| − | 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> | ||
| − | <center> | ||
| − | $d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$ | ||
| − | </center> | ||
| − | <p calss="TxtSolutions"> | ||
| − | As $dq = \pm\sigma~dS$, | ||
| − | </p> | ||
| − | <center> | ||
| − | $dF = \frac{\sigma^2~dS}{2\varepsilon_0}$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | and moving $dS$ to another member of equation, | ||
| − | </p> | ||
| − | <center> | ||
| − | $\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | b) Let's see following figure | ||
| − | </p> | ||
| − | <center> | ||
| + | |||
| + | <h3>Solution</h3> | ||
| + | <p> | ||
| + | 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 | ||
| + | $$d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$$ | ||
| + | As $dq = \pm\sigma~dS$, | ||
| + | $$dF = \frac{\sigma^2~dS}{2\varepsilon_0}$$ | ||
| + | and moving $dS$ to another member of equation, | ||
| + | $$\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$$ | ||
| + | b) Let's see following figure | ||
| + | </p> | ||
| + | <center> | ||
| <figure> | |||
| <img src="drawb.png" | |||
| loading="lazy" alt="6.5.1" width="300" /> | |||
| <figcaption> | |||
| Electric field of plates | |||
| </figcaption> | |||
| </figure> | |||
| − | </center> | ||
| − | | ||
| − | 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> | ||
| − | | ||
| − | | ||
| − | </center> | ||
| − | <center> | ||
| − | | ||
| − | </center> | ||
| − | <center> | ||
| − | | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | | ||
| − | | ||
| − | <center> | ||
| − | | ||
| − | | ||
| − | | ||
| − | | ||
| − | </p> | ||
| − | <center> | ||
| − | $P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \;(2)$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | Putting $(1)$ into $(2)$ | ||
| − | </p> | ||
| − | <center> | ||
| − | $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}}$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | c) See following figure | ||
| − | </p> | ||
| − | <center> | ||
| + | </center> | ||
| + | <p> | ||
| + | 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 | ||
| + | $$\vec{E} = \vec{E_1} + \vec{E_2}$$ | ||
| + | $E = \frac{\sigma}{\varepsilon_0}$ | ||
| + | $$\boxed{\sigma = \varepsilon_0 E} \;(1)$$ | ||
| + | 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, | ||
| + | $$dF = \frac{E}{2}dq = \frac{E}{2}\sigma~dS$$ | ||
| + | Hence, | ||
| + | $$P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \;(2)$$ | ||
| + | Putting $(1)$ into $(2)$ | ||
| + | $$\boxed{P_e = \frac{\varepsilon_0~E^2}{2}}$$ | ||
| + | c) See following figure | ||
| + | </p> | ||
| + | <center> | ||
| <figure> | |||
| <img src="drawc.png" | |||
| loading="lazy" alt="6.5.1" width="300" /> | |||
| <figcaption> | |||
| Electric field of plates | |||
| </figcaption> | |||
| </figure> | |||
| − | </center> | ||
| − | | ||
| − | 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, | ||
| − | | ||
| − | <center> | ||
| − | | ||
| − | </center> | ||
| − | <center> | ||
| − | $E = \frac{\sigma}{\varepsilon_0}$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m | ||
| − | </p> | ||
| − | <center> | ||
| − | $\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions"> | ||
| − | Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$ | ||
| − | </p> | ||
| − | <center> | ||
| − | $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$ | ||
| − | </center> | ||
| − | <p class="TxtSolutions" 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> | ||
| + | </center> | ||
| + | <p> | ||
| + | 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, | ||
| + | $$\vec{E} = 2\vec{E_1} = 2\vec{E_2}$$ | ||
| + | $$E = \frac{\sigma}{\varepsilon_0}$$ | ||
| + | with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m | ||
| + | <div class="scroll-wrapper"> | ||
| + | $$\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$$ | ||
| + | </div> | ||
| + | Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$ | ||
| + | $$\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$$ | ||
| </p> | |||
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| + | <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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<header style="text-align:center;"> | ||
| <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> | ||
| |
<h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | ||
| <article> | |||
| <h3 id="back-link"><a href="../">$\leftarrow$Back</a></h3> | <h3> Statement </h3> | ||
| <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> | ||
| @@ -42,135 +53,77 @@ | |||
| <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> | ||
| <h |
|||
| |
<h3>Solution</h3> | ||
| 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> | 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 | ||
| <center> | $$d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$$ | ||
| $d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq~\hat{E}$ | As $dq = \pm\sigma~dS$, | ||
| </center> | $$dF = \frac{\sigma^2~dS}{2\varepsilon_0}$$ | ||
| <p calss="TxtSolutions"> | and moving $dS$ to another member of equation, | ||
| As $dq = \pm\sigma~dS$, | $$\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$$ | ||
| </p> | b) Let's see following figure | ||
| <center> | </p> | ||
| $dF = \frac{\sigma^2~dS}{2\varepsilon_0}$ | <center> | ||
| </center> | |||
| <p class="TxtSolutions"> | |||
| and moving $dS$ to another member of equation, | |||
| </p> | |||
| <center> | |||
| $\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$ | |||
| </center> | |||
| <p class="TxtSolutions"> | |||
| b) Let's see following figure | |||
| </p> | |||
| <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> | ||
| 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> | $$\vec{E} = \vec{E_1} + \vec{E_2}$$ | ||
| |
$E = \frac{\sigma}{\varepsilon_0}$ | ||
| |
$$\boxed{\sigma = \varepsilon_0 E} \;(1)$$ | ||
| </center> | 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, | ||
| <center> | $$dF = \frac{E}{2}dq = \frac{E}{2}\sigma~dS$$ | ||
| |
Hence, | ||
| </center> | $$P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \;(2)$$ | ||
| <center> | Putting $(1)$ into $(2)$ | ||
| |
$$\boxed{P_e = \frac{\varepsilon_0~E^2}{2}}$$ | ||
| </center> | c) See following figure | ||
| <p class="TxtSolutions"> | </p> | ||
| |
<center> | ||
| |
|||
| <center> | |||
| |
|||
| |
|||
| |
|||
| |
|||
| </p> | |||
| <center> | |||
| $P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \;(2)$ | |||
| </center> | |||
| <p class="TxtSolutions"> | |||
| Putting $(1)$ into $(2)$ | |||
| </p> | |||
| <center> | |||
| $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}}$ | |||
| </center> | |||
| <p class="TxtSolutions"> | |||
| c) See following figure | |||
| </p> | |||
| <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> | ||
| 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, | ||
| |
$$\vec{E} = 2\vec{E_1} = 2\vec{E_2}$$ | ||
| <center> | $$E = \frac{\sigma}{\varepsilon_0}$$ | ||
| |
with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m | ||
| </center> | <div class="scroll-wrapper"> | ||
| <center> | $$\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$$ | ||
| $E = \frac{\sigma}{\varepsilon_0}$ | </div> | ||
| </center> | Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$ | ||
| <p class="TxtSolutions"> | $$\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$$ | ||
| with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m | |||
| </p> | |||
| <center> | |||
| $\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$ | |||
| </center> | |||
| <p class="TxtSolutions"> | |||
| Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$ | |||
| </p> | |||
| <center> | |||
| $\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$ | |||
| </center> | |||
| <p class="TxtSolutions" 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> | </p> | ||
| </article> | |||
| </main> | |||
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| MathJax = { | BSc. Luis Daniel Fernández Quintana<br> | ||
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| inlineMath: [['$', '$'],], | Universidad de Oriente, Cuba<br> | ||
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