# What is the dielectric constant of a pure conductor?

Dielectric constant is the ratio of permittivity of a medium to the permittivity of free space. How to find dielectric constant of a conductor?

• I'm voting to close this question as off-topic because it shows no research efforts.
– user36790
Commented Feb 18, 2016 at 19:42
• The first result in Google for "permittivity of conductors" brought me there. Commented May 11, 2017 at 10:13
• The answer by @SambeetPanigrahi is the only one which does not assume that we are dealing with a static field (which perhaps was implied, but was not actually stated in the question). Commented Nov 25, 2021 at 11:08
• I'm not sure why this is attracting close votes as "homework-like". Even if it's a problematic question for other reasons, it's definitely not "homework-like". Commented Nov 27, 2021 at 13:43

For static electric fields the permittivity of a perfect conductor is infinite.

Let the value of an external electric field in free space (relative permittivity = 1) be $$E$$.

If this is applied to a material of relative permittivity $$\epsilon_r$$ then the electric field in the material is $$\dfrac {E}{\epsilon_r}$$

Inside a conductor the electric field is zero hence its relative permittivity is infinite.

• This is true only for static electric field. Oscillating field can exist in conductor and we can define non-infinite permittivity. Commented Jul 4, 2021 at 12:24
• Only the permittivity of a perfect conductor is infinite. In such a case $\epsilon_r=-1$ giving $E=0$ inside the conductor. Commented Nov 25, 2021 at 10:42
• @my2cts $\epsilon_{\rm r} \ne -1$ rather $\epsilon_{\rm r} = \infty$. Commented Mar 6, 2023 at 12:10
• There is a muddle here which I have attempted to clear up in my answer. Commented Jun 1 at 12:35

Inside a metal, there is no formation of dipoles, hence there is no polarization as such. We have free electrons in metals not bound like that of a dielectric. Hence we can argue its electric susceptibility $$\chi$$ = 0. We know $$\epsilon_r = 1 + \chi$$, so it can be said that its relative permittivity($$\epsilon_r$$) is 1, considering electrostatics problems. For time varying fields, i.e. electrodynamics, we define complex permittivity as $$\hat{\epsilon}=\epsilon \times (1+\sigma/i\omega\epsilon)$$, where for metals we can have the imaginary part $$\sigma/\omega\epsilon >> 1$$. Thus for metals $$\hat{\epsilon}=i\sigma/\omega)$$, which is a large imaginary value considering high conductivity of metals.

Though, not a source for this answer,the basic idea was gained by observing the value of $$\epsilon = \epsilon_0$$ being used in Introduction to Electrodynamics by David Griffiths, in problems (see Chapter 9, Problem 20,bit (b)).

• What is $\epsilon$ in your expression for errr... $\epsilon$ for a conductor? Commented Nov 25, 2021 at 13:12
• What is the argument for $\chi = 0$? $\chi$ is defined as the constant in the law $\mathbf P = \chi \mathbf E$. When both $\mathbf P$ and $\mathbf E$ vanish, $\chi$ can have any non-zero value. Commented Jun 1 at 13:03

Value $k$ gives an idea of how it isolates the charges. Insulators are used for this purpose so the conductors do less in this regard.

Drude formula is not so bad for real metals:

$$\varepsilon(\lambda) = 1 - \frac{1}{\frac{\lambda_p}{\lambda}(\frac{\lambda_p}{\lambda} + i \gamma)}$$, where $$\lambda$$ is the wavelength in the vacuum.

For gold in the infrared region, $$\lambda_p$$ is about 190 nm, and $$\gamma$$ is about 0.005.

Bit what is a "pure" conductor? If you want to remove the losses, put $$\gamma = 0$$. $$\lambda_p$$ term is instrinsic to the mass and density of electrons.

For a perfect conductor, $$\frac{1}{\varepsilon} = 0$$. For a real good conductor, $$\frac{1}{\varepsilon}$$ is a small negative number in the optical domain and a small imaginary number in RF or DC domain.

The muddle arises when dealing with the response of a conductor to an oscillating field. There are two different ways in which the calculation can be set up.

Method 1. The response is accounted for by saying that there is a total current $${\bf j} = \sigma {\bf E}$$ where $$\sigma$$ is the conductivity. In this approach there is no polarization and $$\epsilon_r = 1. \tag{1}$$

Method 2. The response is accounted for by saying that there is an oscillation in the polarization. The oscillating polarization is $${\bf P} = i {\bf j}/\omega = i \sigma {\bf E}/\omega$$ (for a field going as $$\exp(-i \omega t)$$) so we get $$\epsilon_r = 1 + \frac{i \sigma}{\epsilon_0 \omega} . \tag{2}$$

How can there be two answers (1) and (2)? It is because the movement of charge can be interpreted in two ways: either as contributing to movement of free charge in the total current density $$\bf j$$ or as an oscillation of polarization. In the second case we must set the remaining "conduction current" $${\bf j}_c = 0 .$$

In method (1) you must use the Maxwell equations in terms of just $$\bf E$$ and $$\bf B$$ and the total charge and current. In method (2) you are implicitly adopting the Maxwell equations in terms of $$\bf E$$, $$\bf B$$, $$\bf D$$ and $$\bf H$$ and then there appears in the equations only the part of the charge and current densities not already ascribed to polarization and magnetization. Both methods of calculation are correct and consistent with themselves, but they must not be muddled together.

In my opinion the first method is clearer in the case of conductors (the second method is useful for dielectrics but that is another story).

• Both accounts are problematic. 1) assumes there is only conduction current, thus no current due to motion of bound charge is allowed; but in general, both $\mathbf E,\mathbf P$ may be non-zero in conductor, and when $\mathbf P$ changes due to motion of bound charge, there should be additional current. This can be fixed by allowing polarization current, so net current is $\mathbf J = \sigma \mathbf E + \partial_t \mathbf P$. 2) is much worse, because it allows conduction current in definition of $\mathbf P,\chi,\epsilon$, which is contrary to how these concepts are motivated in dielectrics. Commented Jun 1 at 13:21
• General medium both polarizes and is somewhat conductive with some free charge, so I think the relevant concepts should be taught in such a way that they apply to such general medium, not just to idealized dielectrics with zero free charge and idealized conductors that can't polarize. Commented Jun 1 at 13:25
• I agree the first comment. For the second, I would say that in a teaching context I would introduce the idealized cases first, before going to the general case. Commented Jun 1 at 13:54
• I agree with teaching idealized cases first, but preferably, if possible, teaching the general case in a way that won't contradict the concepts already taught (here, polarization being due to state of bound charge only). Commented Jun 1 at 13:58
• The muddle arises when dealing with the response of a conductor to an oscillating field. The OP had an electrostatics tag associated with the question. Commented Jun 1 at 21:13

Dielectric constant is proportional to the ratio of polarization density (P) and electric field (E) which means dielectric constant is inversely proportional to electric field.

The Electric field (E) inside a conductor is always zero under the static situation so the dielectric constant for conductor is infinite.

If permittivity is infinite inside conductor, then won't it mean that for a given time varying electric field (say E(t)=250sin wt), the displacement current density is very high inside conductor? Theoretically, it should be zero though.
And electric field is zero inside conductor, because of the effect of induced electric field inside the conductor, in direction opposite to external field. Induced field adds vectorially to external field and cancels its effect.

The net or resultant electric field inside in a conductor is zero therefore dividing electric field by zero we get infinity

The more in the value of dielectric constant the more is the nonconducting property. If dielectric constant of a conductor is infinite, it will be a perfect insulator. Electric field inside a conductor in electrostatic condition is because of induced electric filed of induced charges, opposing external electric field and it is not due to infinite value of dielectric constant or permittivity of the conductor. And the displacement current inside a conductor with time varying electric field like a.c voltage across a conductor, is zero because pemitivity of the conductor is zero.

• isnt the dielectric of metals infinity? Commented Nov 25, 2021 at 8:10

A pure conductor which is placed between potential difference it strats conducting. So dielectric constant is zero

• The permittivity of a conductor is infinite says the accepted answer. You should argue why it is the oposite. Whan means strats ? Commented Apr 18, 2017 at 10:27