4
$\begingroup$

If there's no material in vacuum, then why is there a non-zero vacuum permittivity, $\epsilon_0$, and what is the cause of the vacuum polarized?

$\endgroup$
1
  • 2
    $\begingroup$ what's your source/ where are you reading this? $\endgroup$
    – pentane
    Commented Aug 2, 2018 at 2:36

2 Answers 2

3
$\begingroup$

The permititvity of the vacuum is not related to polarization. It signifies the strength by which a charge affects another through vacuum (Gauss's law), and in vaccum it influences the strength by which a change in the electric field affects the magnetic field (Maxwell's correction to Ampere's law).

Polarization only becomes relevant when averages over matter, to reach equations that are only true for these averaged quantities rather than the microscopic fields. Then one defines polarization and you can reach equations in matter (Maxwell's equations in matter) that are similar to Maxwell's equations in a vacuum. But this is only a similarity of mathematical form. The microscopic equations, that hold always and not just for the averages, have only the vacuum permittivity by itself, with no polarization.

$\endgroup$
1
$\begingroup$

if no material exist in vacuum, then why does vacuum has permittivity epsilon naught

In this link:

In a dielectric material, the presence of an electric field E causes the bound charges in the material (atomic nuclei and their electrons) to slightly separate, inducing a local electric dipole moment. The electric displacement field D is defined as

$D=ε_0E + P$

where P is the polarization density in the medium. As the vacuum has not charges, this should be zero. D is proportional to E for vacuum, and carries the units necessary for consistency.

This is a design for an experiment to measure the permittivity of vacuum.

The vacuum is not polarized as there are no charges to get an induced polarization due to the field E.

$\endgroup$
2
  • $\begingroup$ There are virtual charged particles in vacuum. $\endgroup$ Commented Sep 7, 2021 at 15:05
  • 1
    $\begingroup$ @DavidJonsson Permittivity is classical electrodynamics. virtual particles belong to quantum electrodynamics (QED). The quantum vacuum is a different story en.wikipedia.org/wiki/QED_vacuum "The electrical permittivity of quantum electrodynamic vacuum can be calculated, and it differs slightly from the simple ε0 of the classical vacuum." $\endgroup$
    – anna v
    Commented Sep 7, 2021 at 18:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.