Why the Dielectric strength decrease with the frequency?

Question

According to Wikipedia, the dielectric strength decrease with the frequency:

Is it possible to explain what is happening when the frequency increase ? I try to understand how to generate an arc electric without applying a high voltage in the air. (I cannot change the temperature or the humidity or use a gaz). Nevertheless, I ca increase the frequency of the AC voltage.

Answer 1

The following is from Chapter 7 of Jackson's E&M Book, Third Edition (i.e., blue cover)

If you start with a simple model for the material where there's some natural frequency of oscillation, $\omega_{o}$, and a damping rate, $\gamma$, then you can show the equation of motion for the electrons in the material is basically: $$ \ddot{\mathbf{x}} + \gamma \ \dot{\mathbf{x}} + \omega_{o}^{2} \ \mathbf{x} = - \frac{ e }{ m } \ \mathbf{E} \tag{0} $$ where $\mathbf{x}$ is the position of the electrons and the dots are total time derivatives, $\mathbf{E}$ the electric field, $e$ is the fundamental charge, and $m$ is the particle mass. If we assume all time-varying quantities are proportional to $e^{-i \ \omega \ t}$, then we can show that the electric dipole moment, $\mathbf{p}$, is given by: $$ \mathbf{p} = -e \ \mathbf{x} = \frac{ e^{2} }{ m } \left[ \omega_{o}^{2} - \omega^{2} - i \ \omega \ \gamma \right]^{-1} \ \mathbf{E} \tag{1} $$ We also know that the electric polarization, $\mathbf{P}$, is defined as follows: $$ \mathbf{P} = \varepsilon_{o} \ \chi_{e} \ \mathbf{E} = \left( \varepsilon - \varepsilon_{o} \right) \ \mathbf{E} \tag{2} $$ where $\varepsilon_{o}$ is the permittivity of free space, $\varepsilon$ is the dielectric permittivity, and $\chi_{e}$ is the electric susceptibility of the material, where $\left( 1 + \chi_{e} \right) = \tfrac{ \varepsilon }{ \varepsilon_{o} }$.

If we assume a pure substance with $n$ molecules per unit volume and $n \ Z$ electrons per unit volume, then the relative permittivity is given by: $$ \frac{ \varepsilon\left( \omega \right) }{ \varepsilon_{o} } = 1 + \frac{ n \ Z \ e^{2} }{ m \ \varepsilon_{o} } \sum_{j} \ \left[ \omega_{j}^{2} - \omega^{2} - i \ \omega \ \gamma_{j} \right]^{-1} \tag{3} $$ where $j$ is the j^th molecule, $\omega_{j}$ is the oscillation or binding frequency of the j^th molecule, and $\gamma_{j}$ is a damping constant for the j^th molecule. Note that the term in front of the sum is sometimes called the electron plasma frequency, $\omega_{pe} = \sqrt{ \tfrac{ n \ Z \ e^{2} }{ m \ \varepsilon_{o} } }$.

Is it possible to explain what is happening when the frequency increase ?

Yes, if you look at Equation 3 you will see that as $\omega$ increases, the second term on the right-hand side decreases because $\omega_{j}$ and $\gamma_{j}$ are independent of $\omega$. In the high frequency limit, Equation 3 reduces to: $$ \lim_{\omega \gg \omega_{j}} \frac{ \varepsilon\left( \omega \right) }{ \varepsilon_{o} } \simeq 1 - \left( \frac{ \omega_{pe} }{ \omega } \right)^{2} \tag{4} $$ which is the dispersion relation for free electromagnetic waves in a plasma.