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

a. The decrease in the electric field strength in a substance is explained by the polarization 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 bound (polarization) charges 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. Permittivity (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, $\varepsilon > 1$.

c. 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.