# Confusion in understanding the relation between permittivity and capacitance

The insulating medium between the capacitor plates is called dielectric. This appears in the Coulomb's law:

$$F = \frac{q_1 q_2}{4 \pi ɛ r^2}$$

ɛ is the "permittivity" constant which depends on the medium. This can be written as:

ɛ = $ɛ_0×ɛ_r$

where ɛr is also called "dielectric constant" and ɛ0 is the vacuum/space permittivity.

So when I look at the Coulomb's law equation I can see that as ɛ decreases which means as ɛr decreases the force increases and this means the electric field increases.

So the equation tells me that if the relative permittivity decreases the electric field increases. It also means if I use a more insulating material I would obtain a stronger electric field.

Isn't "permittivity" a counterintuitive term here? So "permittivity increases" should mean "the material should permit more". But in this case increasing the permittivity decreases the Force and the electric field E.

Now let's look at another equation(capacitance equation) where ɛ appears:

C = ɛ×(A/d)

Now if the relative permittivity increases the capacitance increases. But this is also counterintuitive. Metals have almost infinite permittivity but in capacitors they use insulating material which means low relative permittivity. I would think to increase C the material between the capacitor plates would be more insulating(more insulating material, smaller ɛ). But the equation says the opposite.

Where am I getting things wrong here? I would appreciate any clarification.

I'm not sure that "permittivity" is entirely such a good term. Or a bad one. It just is. And we live with it. That said, here's a discussion that may help a little:

The net electric field inside the dielectric is the sum of the field due to the capacitor's plates and the field that is due to the induced dipoles within the dielectric. (Those dipoles align in a direction that directly opposes the capacitor's electric field.)

The potential difference (voltage) between the plates of a capacitor depends entirely upon the electric field in the region that is between the conducting plates.

Since the dipoles act to reduce the net electric field's magnitude, it follows that the required voltage for a given charge on the plates is less than it would be without the electric dipoles in the dielectric material. In effect, it takes less energy per unit added-charge.

The dielectric constant, $K$ ($\varepsilon_r$), is defined as:

$$\vec{E}_{insulator}=\frac{\vec{E}_{applied}}{K}$$

And it represents the factor by which the net electric field is weakened inside the insulating dielectric.

For a capacitor with plate area $A$ and charge $Q$, the applied field is simply $\frac{Q}{A}\varepsilon_0$ and the net field inside the capacitor (and throughout the assumed uniform dielectric insulating medium) is $\frac{1}{K}\frac{Q}{A}\varepsilon_0$. (The dielectric constant is related to atomic polarizability. But that's a more complex subject than I want to address here.)

A way I like to visualize this is to imagine that energy can only be stored in the vacuum of space itself. (As I imagine it here, this also applies equally to magnetic energy as well as for electric energy.) But as a dielectric sets up electric dipoles to oppose the applied field, these dipoles act as short circuits that bridge over a small bit of the physical plate separation. (Energy, as I'm imagining it here, cannot be stored in the dipoles themselves, as only vacuum can do so.) In between the dipoles where no other dipoles yet exist, is more vacuum space, of course. (Or, if filled with matter it is filled with matter that does not set up a dipole.) But if you imagine the physical plate separation as being partially bridged by these dipole short circuits then you can "see" that there is less effective vacuum distance left over, within which energy may be stored.

In short, the plate separation has been reduced by these intervening electric field short circuit elements. This effectively means the gap is reduced. And therefore the capacitance is increased.

I mentioned that the voltage (potential difference) is about the intervening electric field and that the plates themselves must have an electric field of zero (within the conducting plates themselves.) Physicists take the field potential to be derived by looking at one plate from an infinitely far away place (because the Einstein's theory of relativity changes the idea of a field from a matter of simple convenience now into a required idea), then looking at the other plate also from an infinitely far away place, so that the result is something like this:

\begin{align*} V_A&=-\int_\infty^A \vec{E}\bullet \textrm{d} \vec{l}\\\\ V_B&=-\int_\infty^B \vec{E}\bullet \textrm{d} \vec{l}\\\\ \therefore \Delta V &= V_B - V_A \end{align*}

Since the molecular dipoles consist of stationary point charges, the round trip path integral of the electric field due to the dipoles must sum to zero. (It's impossible to be non-zero.) This fact allows us, without getting into the precise atomic details of how, to conclude that the direction of the average field in the insulator must point oppositely to the applied field.

• I have to read this several times and make some research. But at least for now I understood the answer of my question is not that simple. I was thinking I was making a very basic mistake like the definition of permittivity. But obviously its not the case... – atmnt Jun 3 '17 at 19:15
• good conductors have no dielectric permittivity , only insulator materials, it's that simple . but poor conductors have noteable small relative permittivity with low k. k=1 is the lowest value. Conducting charges and Blocking or resisting charge flow due to storage are opposite functions of resistors and capacitors. – Tony Stewart Sunnyskyguy EE75 Jun 3 '17 at 19:25
• @TonyStewart.EEsince'75 this video made me understood what permittivity is about basically: youtube.com/watch?v=5x8kj02ar34 – atmnt Jun 3 '17 at 19:41
• @TonyStewart.EEsince'75 I had started to discuss conductors and then removed it from my reply, as not being constructive to the OP's interest. Conductors are, in effect charged spheres, and the interior electric field is everywhere zero (but the potential relative to infinity is not, of course.) It just gets mired when I go that direction. So I deleted it. – jonk Jun 3 '17 at 19:45
• I remember some said everything is relative but conductors cannot create or discharge itself, but obvious it can store a charge of one polarity but a capacitor always has both polarities so @user134429 needs to absorb all the above because he is mainly a capacitor or a dielectric computer student – Tony Stewart Sunnyskyguy EE75 Jun 3 '17 at 19:53

I also had this question. This was my thought process:

So to construct a capacitor with a lot of capacitance, you want to separate a lot of charge (with an insulator, otherwise the charge will move to cancel the E field) and then minimize the distance between the plates to minimize the voltage. If your insulator has high permittivity from dipoles that reduces the E field and the voltage more, then you have more capacitance. The insulator can't have free charge like a conductor, because then it won't keep the charges on the plates separated - it has to have high permittivity from non-conducting dipoles. I'm imagining a deionized water capacitor (deionized water is non-conductive but H2O molecule is a dipole).

Permittivity usually is a term used with dielectrics and it helps to determine the the amount of force generated into the dielectric. As an example of a capacitor, there are two electrical fields E the first one is generated by the difference in charges between the two plates of the capacitor and let us call it the forward electrical field. The other one (which opposes the first one) is generated into the dielectric of the capacitor. The more permettivity the dielectric has the lower opposing electrical field is going to be generated thus bigger total elecatrical field is achieved.

E_total = E_forward - E_backward

So you can understand the permettivity as following, high permittivity materials permit or allow the forward electrical field to pass through more.

Regarding the second part,

Metals have almost infinite permittivity

According to my knowledge the real relative permittivity of materials is almost 1 (not infinity). So using a metal instead of a dielectric is going to decrease the capacitance.

This quote from wikipedia:

Permittivity is typically associated with dielectric materials, however metals are described as having an effective permittivity, with real relative permittivity equal to one.

For better understanding of permettivity I would recommend this video. It has a comprehensive explanation of what is going on inside a dielectric when exposed to an electrical field.

• what is being permitted here? electric flux? "permittivity is higher" means what? does that mean the same voltage creates more electric flux? if so why in Coulomb's law increasing the permittivity is decreasing the electric field? and why in capacitance equation decreasing the permittivity(putting more insulting material) decreasing the flux and so the capacitance? – atmnt Jun 3 '17 at 18:51
• The forward electrical field is being permitted. Permettivity is higher means better insulator or higher permitted forward electrical field . – Macit Jun 3 '17 at 19:05
• Have a look at the video I have shared in my answer. It will help you understand the meaning of permettivity. – Macit Jun 3 '17 at 19:14
• yes here is even a simpler nice explanation: youtube.com/watch?v=5x8kj02ar34 – atmnt Jun 3 '17 at 19:38

All insulators are capacitors and visa versa, although they do have leakage resistance. but sometimes on the surface called creepage and usually defined as homogeneous inside.

An ideal dielectric blocks all charge flow with infinite capacitance or permittivity. (not conductors) The perfect insulator.

Consider only 2 material conductors and insulators.

Conductors transfer charges easily so permittivity which resists charge conduction is meaningless. Permittivity is counter-intuitive and not measurable in good conductors and permitivity is usually very low, not infinite but measureable in large resistors values.

The permittivity is just then ratio air of resisting charge flow. The higher the $e_r$ or k factor the more charge per voltage can be stored and the greater insulation or resistance to charge flow. We know flow rates are defined by some initial charge voltage with a decay time slope starting at Rp*C and the greater the k or C value the greater the time constant and thus resistance charge flow.

We know Supercaps are big in hundred's of Farads with a finite series resistance ESR and a large shunt leakage resistance so they cannot be charged up quickly and the I^2ESR =Pd causes internal temp rise is the limiting rate.

Also we know batteries are really huge Capacitors except have a chemical cell voltage at 0 % State of Charge (SOC) and rises some 8% for lead acid to 20% for LiPo to 100% SOC. However, there is a charge voltage factor which must considered for each chemistry. Even Supercaps have some memory like Lead acid and NiCad but much less significant for LiPo.

LiPo batteries are in fact on the order of one hundred thousand Farads but have faster aging rates with charge cycles (300) than SuperCaps .(100k~1M cycles) and ESR's ranging from 5 to 35mOhms typ.Thus the RC discharge times usually rated in mAh are very long but depend on self heating from discharge rates and other factors as the chemistry ions change the dielectric properties and thus the k factor.

When liquid electrolytic caps or batteries age they rise in ESR sharply and drop in C sharply. Whereas rechargeable batteries drop so the same when depleted, the ESR and C value is restored upon recharging beyond 10%, so ESR is not always a good indicate or State of Charge unless <10% SoC or completely destroyed (usually open , sometimes short)

next semiconductors .. I just answered that for diode RC values, both nonlinear in another query.

The other factor is that body capacitance between resistor terminals can become lower AC impedance when R is very high.

e.g. if body is 5pF and R=10M what frequency becomes the breakpoint for current?

Then for insulators what properties do you expect for high voltage for k=1 a vacuum gap, is Ideal but in air particles and air pressure cause ion flow so breakdown effects occur rapidly above 1kV/mm for a sharp point to 3kV/mm for clean parallel surfaces. (or 1V/um to 3V/um thus in silicon only 15nm wide, the impurities are so low in the dielectric capacitance that they have very high V/um ratings (I think >> 100V/um).

This is the fundamental law of materials. Semiconductors have both insulators and doped ion insulators with conductor terminals to become voltage controlled resisitors.

All capacitors are DEFINED by their geometric ratio A/d and dielectic constant , k. Similarity all coax has a fixed A/d ratio and thus a fixed characteristic impedance with pF/m capacitance.

• what is k? btw most part of your answer are detalis which im not asking i dont know why you writing those. you are talking about ESR SuperCaps SOC. are you sure you are answering the right question. my question is simple. you know a lot but obviously im not asking what u are talking about. – atmnt Jun 3 '17 at 19:01
• read again I said $e_r$=k .. All you need , ehm, want to know then is the 4th paragraph. All pure materials are either a conductor with no dielectric, or it is an insulator with a dielectric with rel. permittivity and no resistance ideally in series or parallel. It is impurities that mix the two. and electrode area/distance. Pure Vacuum is the best insulator but in air, mica (but porcelain is cheaper) is used as the best k>1 insulator followed by polymer metal film caps like Polyurethane, polyester, teflon, etc – Tony Stewart Sunnyskyguy EE75 Jun 3 '17 at 19:14