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en/6.6.1.md
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| + | ### Statement | ||
| + | |||
| + | $6.6.1.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $6.6.1.$ | ||
| + | a. What explains the decrease in the electric field strength in a substance? | ||
| + | b. What is the permittivity (dielectric constant) of a substance? | ||
| + | c. How does the permittivity of a gas depend on its pressure? | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | <b>a.</b> The decrease in the electric field strength in a substance is explained by the <b>polarization of the dielectric</b>. | ||
| + | When a dielectric is placed in an external electric field $\vec{E}_0$, the electric charges within its atoms and molecules shift (or the existing dipoles align). As a result, uncompensated macroscopic <b>bound (polarization) charges</b> appear on the surfaces of the dielectric. | ||
| + | These charges create their own internal electric field $\vec{E}_{\text{pol}}$, which is directed opposite to the external field. According to the superposition principle, the resulting field $\vec{E}$ inside the substance is: | ||
| + | $$\vec{E} = \vec{E}_0 + \vec{E}_{\text{pol}}$$ | ||
| + | In scalar form (for magnitudes, since they are directed oppositely): | ||
| + | $$E = E_0 - E_{\text{pol}}$$ | ||
| + | Consequently, the resulting field inside the dielectric is weaker than the external one. | ||
| + | |||
| + | <b>b.</b> <b>Permittivity</b> (or relative dielectric constant, $\varepsilon$) is a dimensionless physical quantity that shows how many times the magnitude of the electric field $E$ inside a homogeneous isotropic dielectric is less than the magnitude of the external field $E_0$ (in a vacuum): | ||
| + | $$\varepsilon = \frac{E_0}{E}$$ | ||
| + | For vacuum, $\varepsilon = 1$; for all dielectrics, $\varepsilon > 1$. | ||
| + | |||
| + | <b>c.</b> For gases, the difference $(\varepsilon - 1)$ is directly proportional to the molecular concentration $n$: | ||
| + | $$\varepsilon - 1 \sim n$$ | ||
| + | From the ideal gas equation of state, at a constant temperature ($T = \text{const}$), the concentration of molecules is directly proportional to the pressure: $n = \frac{P}{kT} \implies n \sim P$. | ||
| + | Therefore: | ||
| + | $$\varepsilon - 1 \sim P$$ | ||
| + | Thus, as the pressure increases, the permittivity of the gas increases, with the quantity $(\varepsilon - 1)$ growing linearly. | ||
| + | |||
| + | #### Answer | ||
| + | a. By the polarization of the dielectric (the emergence of bound charges creating an opposing field). | ||
| + | b. A quantity indicating how many times the field strength in a substance is less than in a vacuum ($\varepsilon = E_0 / E$). | ||
| + | c. Permittivity increases with pressure (the quantity $\varepsilon - 1$ is directly proportional to the pressure). | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $6.6.1.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $6.6.1.$ | |||
| a. What explains the decrease in the electric field strength in a substance? | |||
| b. What is the permittivity (dielectric constant) of a substance? | |||
| c. How does the permittivity of a gas depend on its pressure? | |||
| ### Solution | |||
| <b>a.</b> The decrease in the electric field strength in a substance is explained by the <b>polarization of the dielectric</b>. | |||
| When a dielectric is placed in an external electric field $\vec{E}_0$, the electric charges within its atoms and molecules shift (or the existing dipoles align). As a result, uncompensated macroscopic <b>bound (polarization) charges</b> appear on the surfaces of the dielectric. | |||
| These charges create their own internal electric field $\vec{E}_{\text{pol}}$, which is directed opposite to the external field. According to the superposition principle, the resulting field $\vec{E}$ inside the substance is: | |||
| $$\vec{E} = \vec{E}_0 + \vec{E}_{\text{pol}}$$ | |||
| In scalar form (for magnitudes, since they are directed oppositely): | |||
| $$E = E_0 - E_{\text{pol}}$$ | |||
| Consequently, the resulting field inside the dielectric is weaker than the external one. | |||
| <b>b.</b> <b>Permittivity</b> (or relative dielectric constant, $\varepsilon$) is a dimensionless physical quantity that shows how many times the magnitude of the electric field $E$ inside a homogeneous isotropic dielectric is less than the magnitude of the external field $E_0$ (in a vacuum): | |||
| $$\varepsilon = \frac{E_0}{E}$$ | |||
| For vacuum, $\varepsilon = 1$; for all dielectrics, $\varepsilon > 1$. | |||
| <b>c.</b> For gases, the difference $(\varepsilon - 1)$ is directly proportional to the molecular concentration $n$: | |||
| $$\varepsilon - 1 \sim n$$ | |||
| From the ideal gas equation of state, at a constant temperature ($T = \text{const}$), the concentration of molecules is directly proportional to the pressure: $n = \frac{P}{kT} \implies n \sim P$. | |||
| Therefore: | |||
| $$\varepsilon - 1 \sim P$$ | |||
| Thus, as the pressure increases, the permittivity of the gas increases, with the quantity $(\varepsilon - 1)$ growing linearly. | |||
| #### Answer | |||
| a. By the polarization of the dielectric (the emergence of bound charges creating an opposing field). | |||
| b. A quantity indicating how many times the field strength in a substance is less than in a vacuum ($\varepsilon = E_0 / E$). | |||
| c. Permittivity increases with pressure (the quantity $\varepsilon - 1$ is directly proportional to the pressure). | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||