7
$\begingroup$

Permittivity is the measure determining the electric field produced by charge in a particular medium.

Now electric field, $E$ increases as ε (permittivity) decrease, and E decreases as ε increases, due to inverse proportionality of E to ε.

Talking in material (practical) terms, the permittivity - that is how much E field would be allowed in a medium- is due to the material of medium. For example, water medium has water molecules, so when two charges are placed in water, the Field from the two charges are resisted by water molecules, and so less NET field would be produced by the charges(as compared to when the two charges would have been placed in vacuum), and there would be less force between them.

In vacuum, there is no such mass or material object. So it should have permittivity approaching 0(and in fact 0 itself). But permittivity of free space (free space means- no electromagnetic waves, no particles, no charges, nothing in space, only absolute space) is 8.85×10-¹² F m-¹.

It is though a fact, that if ε of vacuum (free space) be 0, then there would be infinite force between two objects kept in free space, and it is physically not possible. But hypothetically it is possible. (Or is this hypothesis wrong?).

What makes the vacuum not have 0 permittivity?

$\endgroup$
3
  • $\begingroup$ Welcome to Physics SE. I did not downvote. Your thoughts led to the definition of a permittivity equals 1. $\endgroup$ Oct 5, 2017 at 10:47
  • 1
    $\begingroup$ @StefanBischof Haha. Do not worry about down vote. ;). Well, the link provided by you talks about Relative permittivity. So definitely for vacuum, it is 1. But in the question it is being asked why is permittivity of vacuum not 0, and not about the relative permittivity. $\endgroup$
    – user171164
    Oct 5, 2017 at 11:30
  • 2
    $\begingroup$ Keep in mind that empty space isn't empty space. It's full of quantum fluctuations. $\endgroup$
    – Brad S
    Oct 5, 2017 at 18:13

8 Answers 8

6
$\begingroup$

Both the previous answers (though correct) are somewhat misleading. What $\epsilon_0$ is measuring is the strength of the electric force. The force between two point charges is stated by Coulombs law, which states

$F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1q_2}{r^2}$, where q represents their charges and r is the distance between them. Electric forces exist everywhere in the universe, and $\epsilon_0$ is just a fundamental constant.

However, in any non-vacuum material, there are atoms composed of charges particles present all around. Because there can be movement of charges in a medium (specifically, electrons) the resulting forces and fields can completely change.

The term "permittivity of free space" is misleading, because is suggests that the vacuum is just another material. Electric fields and forces exist in a vacuum - materials simply react and can hence change the field.

The variable $\epsilon$ is specifically used to describe a particular type of material, a linear dielectric. This is a material where the polarization of the material is proportional to the applied electric field. This is described by the variable $\epsilon = \epsilon_0 \epsilon_r$, where $\epsilon_r$ is the relative permittivity, which is one or greater.

It is important to realize that there are many other possible equations of state that relate the movement of charges to an applied electric field. Linear dielectrics (insulators) behave differently from conductors, and other non-linear materials.

Note: I would like to apologize for my previous answer, which incorrectly stated that electric fields increase the attraction between two charges. This is not true, and in fact dielectrics usually decrease net forces between charges (although sometimes they do not). For further reading, I suggest introduction to electrodynamics by Griffiths.

$\endgroup$
3
  • $\begingroup$ Could you, please explain how is the force increased? If you replace the permittivity e0 by e0er for a material, where "er>1", F will decrease. $\endgroup$ Aug 24, 2022 at 14:03
  • $\begingroup$ Ok, I added to my explanation. Now it should be more clear that the force "increases" because it is added onto by forces between the material and the plates. $\endgroup$
    – Armaan
    Aug 29, 2022 at 3:37
  • $\begingroup$ This answer is incorrect. The permittivity is on the bottom of the Coulomb's law equation. If you increase the permittivity ie. by putting the charges in something that is not a vacuum, then the force will decrease. $\endgroup$
    – PhysMs
    Oct 9, 2022 at 14:26
5
$\begingroup$

Vacuum permittivity $\epsilon_0$ is defined by nature of light. In vacuum electromagnetic waves (light) propagates with speed of light $c_0$ in vacuum. By definition

$$\epsilon_0 = \frac{1}{µ_0\cdot {c_0}^2}$$

Let be $µ_0 = 4\pi\cdot 10^{-7}\frac{H}{m}$ in vacuum. Since speed of light is not infinite $\epsilon_0$ will not be 0.

$\endgroup$
1
$\begingroup$

In matter, due to partial screening of a charge $q$ by dipoles sticking on its surface, it's effective charge becomes $$q_{\text{e}}=q\frac{\epsilon_0}{\epsilon}$$

This is the definition of $\epsilon$.

In vacuum, there is no screening, and hence by definition, $\epsilon=\epsilon_0$.

$\endgroup$
1
$\begingroup$

I will tell you why it shouldn't be $0$. First of all, speed of light would become infinite since it's defined as

$$c=\frac{1}{\sqrt{\varepsilon_{0}\mu_{0}}}$$

this is not true, we know it from different experiments that speed of light is finite. In addition to that, magnetic field produced by current carrying wire would be $0$ everywhere

$$\textbf{B} = \frac{\mu_{0}}{4\pi}\int_{C} \frac{I \textbf{dl}\times \textbf{r'}}{\textbf{|r'|}^{3}}$$

Electric force exerted on charged particles would become infinite

$$\textbf{|F|} = \frac{1}{4\pi \varepsilon_{0}}\frac{|q_{1}q_{2}|}{r^2}$$

From mass-energy equivalence $E = \sqrt{(m_{0}c^2)^2+(pc)^2}$ , energy of a particle when $p=0$ will tend to infinite and relativistic mass tends to rest mass $m = \frac{m_{0}}{\sqrt{1-\frac{v^2}{c^2}}}$.

$\endgroup$
0
$\begingroup$

Another possible way to think about this, very similar to answers above. Imagine a charged particle (Q). By definition, flux taken through some surface which the field cuts through is given as, $$\Phi = \int{\vec{E}\cdot d\vec{A}}$$ Inverse square law associated with the electric field source is, $$\vec{E}=\frac{k_e Q}{r^2}\hat{r}$$ Then we can take the surface integral anywhere outside the source, let's make it an enclosing sphere, $$\Phi = \int^{\phi=2\pi}_{\phi=0}\int_{\theta=0}^{\theta=\pi}{\frac{k_e Q}{r^2}\hat{r} \cdot r^2 sin\theta\ d\phi\ d\theta\ \hat{r}}$$ $$\Phi = 4\pi k_e Q$$ Where, $k_e = 1/4\pi\epsilon_0$ $$\Phi = Q/\epsilon_0$$

For any finite charge enclosed the flux must be both non-zero and non-infinite, ruling out the possibility that the field constant of proportionality ($k_e$) is either zero or infinite.

$\endgroup$
0
$\begingroup$

It was already stated but I couldn't comment as a reply.

The simple answer is this: permittivity is inversely proportional to the (the square of) the speed of light. It takes time for electromagnetic interactions to propagate through space. if EM radiation moved infinitely fast through space, then permittivity would be zero.

So it's not that space itself causes the presence of permittivity, but instead that EM interactions take time to propagate.

in fact, photons are a type of virtual particle known as a Boson which "communicates", among other things, electric charge, and more specifically a charge differential (between two different charges including a charge and lack of charge)

$\endgroup$
0
0
$\begingroup$

"Permittivity" is a bad name, I'd think, for just this reason - intuitively, you'd imagine a material which is less "permittive" of electric fields to penetrate it to weaken the electric forces between charges, when in fact exactly the opposite is true. Although, its magnetic analogue, permeability, does in fact behave in the expected way. (Another weird one is that when one talks of permittivity in a medium, written $\epsilon$, it is not the same kind of quantity as $\epsilon_0$ but instead a multiplier thereof!)

Nonetheless, if we use the quantity as it should be phrased, what you'd really be asking is why does the vacuum permit an electric field at all, i.e. why is $\frac{1}{\epsilon_0} > 0$? Well, this constant actually is a dressed up version of the much more fundamental constant $\alpha$, which only emerges when you use the natural quantum of charge $e$, the electron charge. If you use $e$ as your measure, and in addition set some other constants to natural units, so that altogether $e = \hbar = c = 1$ (basically, your natural unit of speed is the universe's natural unit and your natural unit of phase space resolution is the universe's natural unit), Coulomb's law becomes

$$F_E = \alpha \frac{q_1 q_2}{r^2}$$

and $\alpha$ is a well-known mystery constant famously with value a bit smaller than $\frac{1}{137}$. It's not known if it "just is" that value, or if there is some way to derive it from a deeper form of physics not yet discovered. This constant has long been remarked upon for exactly such reason.

$\endgroup$
-1
$\begingroup$

Electromagnetic waves traveling through space perhaps is a incorrect view of thinking. Perhaps, space itself is something because it limits the light velocity for it has a permittivity and permeability. If space where nothing it wouldn't have the ability to limit velocities. Perhaps a more accurate view is that electromagnetic waves traveling through space is simply a rippling of the space or ether. In other words, space itself is part of the creation, not a simple void.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.