# Homework Help: Relative permittivity of a conductor

1. Aug 1, 2009

### ananthu

1. The problem statement, all variables and given/known data

Is the relative permittivity (ie.dielectric constant) of a conductor (metal) zero,or,infinity? Prof. Jaiswal in his golden physics (for XII std. CBSE) states that it is "zero". Another author Prof. Mohindroo in his book based on the same XII std. CBSE physics syllabus, says it is "infinity".
In the series " Interactive physics" by the authors R.Ravi, P.Kamaraj, R. Ranganathan published by Learning media (P)Ltd. (India), they say again it is "infinity". Which is the authentic statement? Interestingly, all these people are all well informed scholars and they have their own logical reasons. But, science does not permit ambiguity, I believe.
Will some one throw light on this tricky concept with proper reasoning, please?

2. Relevant equations
Coulomb's law in electrostatics

3. The attempt at a solution
It is a concept based question

2. Aug 1, 2009

My reasoning may be simplistic here but if it is correct to define relative permittivity as the ratio of the capacitance of a given capacitor with metal as the dielectric compared to the capacitance with vacuum as the dielectric then the answer is zero.If the space between a capacitor plates is filled with a metal it becomes one large conductor and the capacitance becomes zero(very nearly)

3. Oct 28, 2009

### mmiguel1

I think it is infinity.
The dielectric constant is related to the electronic susceptance in an isotropic material.
The susceptance is basically the ratio of polarization to applied electric field.
You can think about a conductor as having "bound" electrons in that they cannot leave the entire material, but are free to polarize across the entire length of a conductor.
When you apply an external electric field to a conductor, you polarize the entire conductor, such that the polarization causes the electric field inside the conductor to be zero (electrostatic equilibrium). In a normal dielectric, the bound electrons cannot move as far as in a conductor, they have a much smaller polarization.
The definition of a dipole moment is charge separation times separation distance.
In a conductor, the induced dipoles have distances of the magnitude of the size of the macroscopic object, which is much much larger than the dipole distances of an insulating dielectric (atomic scale).
Hence, the polarization vectors in a conductor are nearly infinite compared to the polarization vectors of a dielectric (whose order of magnitude we are accustomed to dealing with). The susceptance is therefore near infinite, and so is the relative permittivity.

Last edited: Oct 28, 2009
4. Oct 28, 2009

### mmiguel1

I believe I can reconcile the two "proofs" for this problem.
The capacitance proof takes advantage of the fact that capacitance is proportional to permittivity. You can construct a capacitor with a specific dielectric, and compare its capacitance to a capacitor with identical geometry, but no dielectric.
Let C_0 be the capacitance with no dielectric and the C be the capacitance with a dielectric.
You therefore have
C / C_0 = e / e_0, I use e instead of epsilon.
C / C_0 = e_r = relative permittivity
For the purpose of this discussion, let us assume we are working with parallel plate capacitors.
The capacitance argument goes: If you replace the dielectric material with a conductor, you will measure no capacitance. Therefore C = 0 and e_r = 0/C_0 = 0.

Here is my rebuttal.
This sort of test is invalid when determining the permittivity properties of a conductor.
The reason is that the wires and parallel plates are made of the same material as the "dielectric" which in this case is a conductor.
You would not trust the results of this capacitance test if the wires and the plates were made of the same material as an insulating dielectric. You no longer have a capacitor, but merely a piece of bulk material with a strange shape.
The key thing when dealing with permittivity is the effects of polarization in the presence of an external electric field.
Since we are interested in measuring the corresponding property for a conductor, we must construct a test in which the only differences in the result are due to the polarization differences between materials.
We can find a sufficient test by assuming that no charge transfer can occur between the dielectric conductor and the wires/parallel plates.
Assume you have constructed some ingenious way to prevent charge transfer from occurring between the dielectric conductor and the wire/parallel plates.
Now, apply an ideal voltage source with voltage V_0 to our "capacitor".
This potential difference will create an electric field inside of our conductor.
The typical response of a conductor to an electric field is the near instantaneous movement of charge in order to cancel out the applied field and reach static equilibrium (electric field reaching zero inside the conductor). This typically happens in less then a femtosecond.
That is the typical case. In our case, the conductor will experience charges migrating due to the electric field that try to cancel out the field. If we truly have an ideal voltage source however, the electric field inside the conductor will persist no matter how many charges move within the conductor. These charges cannot escape from the dielectric conductor either because of our earlier assumption. In this ideal model, the charges will build up on each side of the dielectric conductor indefinitely. In other words, Q approaches infinity. The capacitance by definition is Q / V_0 and since Q is infinite, so is the capacitance.
Hence by this modified capacitance test, we have C/C_0 = infinity = e_r.
The relative permittivity is infinite as well.
I think this should settle the debate.

5. Oct 30, 2009

### ananthu

Thank you all. The reply of mmiguel 1 is more convincing than that given by Dadface. I too think that the infinity should be logically correct.
Regarding the very name given to this property of the material ie."the permittivity", I have my own reservations.
when the relative permittivity is more for a substance, it allows or permits only less electric field to pass through it.For example. the relative permittivity for water is 80. So the electric flux passing through it is 80 times less than that passes through air. So to speak, the greater the permittivity, lesser its ability to permit the lines of force. Then how come this property got its present name ie.relative permittivity? Or is any historical reason for this name? (But in the case of a magnetic substance the word "permeability" conveys its meaning correctly, I think!)

6. Oct 30, 2009

### Born2bwire

I have alway dealt with complex permittivity. In that context, the imaginary part of the permittivity is infinite for a perfect conductor as it is directly related to the conductivity of the material. The real part is not considered by us, we keep it finite though, usually just choosing the relative real part of the permittivity to be unity.

In a mathematical sense I would prefer infinity over zero, if I were to apply Gauss' law in a homogeneous perfect conductor, I would avoid the singularity of the 1/0 on the right-hand side should I expess the left-hand side solely from the electric field. Still, for the most part I would consider this to be a matter of little consequence. Making the loss to be infinite fully retains the qualities of the perfect conductor for electromagnetic waves and setting the real part to be non-zero and finite helps keep the mathematics (especially in a computer program) well mannered.

7. Oct 30, 2009

Hello everybody.Would it be fair to say that as far as conductors are concerned the equation relative permittivity=Cr/Co does not apply or is lacking in some respect?

8. Oct 30, 2009

### Born2bwire

The relative permittivity is an indication of the polarization of the medium. An applied electric field in a material will induce dipole moments in the materials atoms or molecules. These dipole moments will create a net reduction in the total electric field (since the moment created has its own secondary electric field that opposes the applied field). However, the dipole moments store energy which gives an indication on why a dielectric filled capacitor can store more energy. Now with a metal, the ability for the polarization of the metal can't be infinite, and being zero would be unphysical. If it was infinite, that would imply that you could keep dumping energy into the metal and it would be stored. However, the metal does absorb incident power, but it does not store it, it is dissipated by induced currents from the incident fields. The means for this to occur is by the conduction current, which is described by the imaginary part of the permittivity. So the the real part should be finite, the necessary behavior will be encompassed in the imaginary part.

Now some of this behavior would be encompassed by making the real part infinite, but I think it would cause problems when you introduce time-varying fields. For example, the wavelength of the electromagnetic waves would be related to the real part of the square root of the product of the permittivity and permeability. An infinite real part would imply an infinite wavelength, but this is not true if we take a good conductor (good, not perfect) as an indication of the limiting case.

9. Oct 30, 2009

Would the equation apply when the fields are non varying that is when the capacitors are fully charged?

10. Oct 30, 2009

### Born2bwire

Yeah, the general wave equations are valid down to DC where they decouple into the static equations, just that we normally make approximations when we go down to low frequency cases (like the quasi-static case).

11. Oct 30, 2009

Thanks born2bwire,it makes sense now.I just googled and discovered that the equation refers not to the relative permittivity but to the relative static permittivity.Perhaps that was what Prof Jauswal was referring to.

12. Nov 1, 2009

### ananthu

Friends! Your discussions are going very acadamic and intellectual indeed. Congrats! But please clear a layman's doubt I raised above.Especially when You introduce this concpt of permittiivity to a higher secondary student how will you convincingly explain the logic of the term? ie. "permittivity" means "what does it exactly permit?"

13. Nov 1, 2009

### rl.bhat

The purpose of capacitor is to store the charges. The substance which permits to store more charge in the capacitor has more permittivity.

14. Feb 8, 2011

### sanjivmarathe

Coming back to the original question - Electric field is inversely proportional to permittivity. Field inside a conductor placed in an electric field is zero. Hence permittivity is almost infinite.

15. Aug 23, 2012

### DavidPun

Actually, all the above appears to be wrong. The relative permittivity of any material is composed of a real part and an imaginary part ( εr + jεim) A metal at the approrpriate point relative to its plasmon resonance (where it behaves like the metals we all know and love) has a negative real part and positive imaginary part which is quite large compared to a dielectric material. The definition of a perfect metal is one where the imaginary part goes to zero and it only has a negative real part. To determine how this affects the propagation of a field through a metal, you have to convert this to the usual (n+ j k ) since n purely affects the propagation of the field in term of phase and k affects only the absorption of the field i.e. loss. When you use the dielectric constants or relative permittivity however, it is essential for a "lossless" metal to have a zero imaginary part, but in real metals when it is not zero, both the real and the imaginary part contribute to n and k.
If you end up with an infinity, something has gone wrong

16. Aug 27, 2012

### Born2bwire

Considering that this is a very old thread, this is not the place to discuss this but I'd rather not leave it on this point. However, your assertions are incorrect. A perfectly conducting metal has an infinite imaginary part as it is considered to be infinitely lossy (infinite conductivity). We rarely consider metals to have a negative permittivity (excepting bulk structures that are designed to have a resonance) as that occurs above the plasma frequency which is generally in frequencies outside of the regime for classical electromagnetics. The metals that "we all know and love" as you put it are behaving below the plasma resonance where the real part is generally taken around unity and the imaginary part blows up. Take a look at the permittivity model governed by a simple conductivity and the permittivity model for a single plasma resonance. Below the plasma frequency, the real part is approximately unity and the imaginary part blows up as the frequency decreases for a given finite conductivity.

17. Apr 15, 2013

### rf7627

The value should be 0 but it is infinity.

what i dont understand about the permitivity of metals is that if it is a conductor, the charge will never remain at a place and will be transferred from particle to particle through conduction via the electrons. thus, the charge at a point at a point will always be zero. thus, the relative permitivity of a conductor should always be zero. but, t is infinity. HOW?

18. Aug 8, 2013

### virpe

Dear colleagues

When we try to solve Maxwell's equations analytically, we come to the equation σ/(ωε) >> 1 for good conductor σ/(ωε) << 1 for good insulator. Therefore, for the good conductor it's not possible to obtain >>1 if the permittivity ε is infinity (this sets the majority of the literature but very very unclear based on no strict mathematical proofs and derivations), because the equation would be in this case zero but that means good insulator. In my opinion, the permittivity approaches to zero, that means "more zero" than in case of insulator. Don't forget, ε0=8.85x10-12 As/Vm -this value is already close to zero (ε=ε0εr).

Vice versa logic is not relevant because σ/(ωε) >> 1 has no sense in the case of having permittivity close to infinity because it reduces the equation to zero, but the "task of conductor conductivity σ" is to increase the result (σ has never no chance in comparison to ε :-) when permittiviy is travelling to zero).

A agree the most with the opinion of rf7627.

19. Aug 8, 2013

### rude man

Static permittivity of a conductor is the same as that of vacuum.

20. Aug 8, 2013

### vanhees71

From a general point of view there is not so much formal difference between the permittivity of a dielectric and conductance of a metal. For the latter, a crude classical model is that for a quasi-free gas of conduction electrons, moving in the external electric field and subject to friction from collisions. The equation of motion for such a conduction electron reads (in non-relativistic approximation, which is sufficient for any practical purposes)
$$m \dot{\vec{v}}=-m \gamma \vec{v}+q\vec{E}.$$
Neglecting the spatial variation of $\vec{E}$ along the relevant distances of the electron's motion, we find by finding the retarded Green's function of the operator $\partial_t+\gamma$
$$\vec{v}(t)=\int_{-\infty}^{\infty} \mathrm{d} t' \Theta(t-t') \exp[-\gamma(t-t')]\frac{q}{m} \vec{E}(t').$$
The current is given by
$$\vec{j}(t)=n q \vec{v} = \frac{n q^2}{m}\mathrm{d} t' \Theta(t-t') \exp[-\gamma(t-t')]\frac{q}{m} \vec{E}(t'),$$
where $n$ is the conduction-electron density, which we also consider as spatially homogeneous.

Writing the electric field as Fourier transform,
$$\vec{E}(t,\vec{x})=\int_{-\infty}^{\infty} \frac{\mathrm{d} \omega}{2 \pi} \exp(-\mathrm{i} \omega t) \tilde{\vec{E}}(\omega,\vec{x})$$
and also
$$\vec{j}(t,\vec{x})=\int_{-\infty}^{\infty} \frac{\mathrm{d} \omega}{2 \pi} exp(-\mathrm{i} \omega t) \tilde{\vec{j}}(\omega,\vec{x}),$$
we find in the frequency domain
$$\tilde{\vec{j}}(\omega,\vec{x})=\sigma(\omega) \tilde{\vec{E}}(\omega,\vec{x})$$
with
$$\sigma(\omega)=\frac{n q^2}{m} \frac{1}{\gamma-\mathrm{i} \omega}.$$

The only difference in the case of a non-conducting dieelectric is that here all the electrons are bound and in linear-response theory can be described as moving in a harmonic potential with frequency $\omega_0$ and some friction coefficient$\gamma$. The only difference here is that we express the response to a (weak) electric field in terms of the polarization
$$\vec{P}=n q \vec{x}$$
In Fourier space the Green's function gives the electric susceptibility,
$$\chi_e(\omega)=\frac{n q^2}{m} \frac{1}{\omega_0^2-\mathrm{i} \gamma \omega-\omega^2},$$
$$\tilde{\vec{P}}(\omega,\vec{x})=\chi_e(\omega) \tilde{\vec{E}}(\omega,\vec{x}).$$
In Heaviside-Lorentz units one thus has
$$\tilde{\vec{D}}=\tilde{\vec{E}}+\tilde{\vec{P}}=(1+\chi_e) \tilde{\vec{E}},$$
and thus
$$\epsilon(\omega)=1+\chi_e(\omega).$$
In a real material you have several eigen frequencies, corresponding to the different quantum-theoretical bound states of the electrons to their ions, and also in a conductor you usually have some bound electrons, so that $\sigma$ and $\chi_e$ are given by the sum of the corresponding terms.

That there is not so much difference in the two cases, because on a microscopic level the total current is given by the conduction current and the current due to the motion of the bound electrons. The latter obviously is given by
$$\vec{j}_{\text{bound}}=\partial_t \vec{P}$$
or, in frequency space,
$$\tilde{\vec{j}}_{\text{bound}}=-\mathrm{i} \omega \tilde{\vec{P}}.$$
Thus one finds the conductivity for $\omega_0=0$ for the dielectric case via this relation
$$\sigma=\left .-\mathrm{i} \chi_e \right|_{\omega_0=0}.$$