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.
@@ -34,9 +30,7 @@Solution
#### 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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c. Permittivity increases with pressure (the quantity $\varepsilon - 1$ is directly proportional to the pressure).
@@ -1,14 +1,10 @@
### Statement
### Statement
$6.6.1.$ [Insert the problem statement]
### Solution
### Statement
$6.6.1.$
$6.6.1.$
a. What explains the decrease in the electric field strength in a substance?
a. What explains the decrease in the electric field strength in a substance?
b. What is the permittivity (dielectric constant) of a substance?
b. What is the permittivity (dielectric constant) of a substance?
c. How does the permittivity of a gas depend on its pressure?
c. How does the permittivity of a gas depend on its pressure?
### Solution
### Solution
<b>a.</b> The decrease in the electric field strength in a substance is explained by the <b>polarization of the dielectric</b>.
<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.
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:
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:
$$\vec{E} = \vec{E}_0 + \vec{E}_{\text{pol}}$$
$$\vec{E} = \vec{E}_0 + \vec{E}_{\text{pol}}$$
In scalar form (for magnitudes, since they are directed oppositely):
In scalar form (for magnitudes, since they are directed oppositely):
$$E = E_0 - E_{\text{pol}}$$
$$E = E_0 - E_{\text{pol}}$$
Consequently, the resulting field inside the dielectric is weaker than the external one.
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):
<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):
$$\varepsilon = \frac{E_0}{E}$$
$$\varepsilon = \frac{E_0}{E}$$
For vacuum, $\varepsilon = 1$; for all dielectrics, $\varepsilon > 1$.
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 $n$:
<b>c.</b> For gases, the difference $(\varepsilon - 1)$ is directly proportional to the molecular concentration $n$:
$$\varepsilon - 1 \sim n$$
$$\varepsilon - 1 \sim n$$
From the ideal gas equation of state, at a constant temperature ($T = \text{const}$), the concentration of molecules is directly proportional to the pressure: $n = \frac{P}{kT}\implies n \sim P$.
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:
Therefore:
$$\varepsilon - 1 \sim P$$
$$\varepsilon - 1 \sim P$$
Thus, as the pressure increases, the permittivity of the gas increases, with the quantity $(\varepsilon - 1)$ growing linearly.
Thus, as the pressure increases, the permittivity of the gas increases, with the quantity $(\varepsilon - 1)$ growing linearly.
@@ -34,9 +30,7 @@Solution
#### Answer
#### Answer
a. By the polarization of the dielectric (the emergence of bound charges creating an opposing field).
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 ($\varepsilon = E_0 / E$).
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
c. Permittivity increases with pressure (the quantity $\varepsilon - 1$ is directly proportional to the pressure).
