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i.e. What does $\epsilon_{0}$ mean in real life? Like pressure = force exerted over an area, momentum = mv which is some measure of the motion of a mass, acceleration = how quickly the velocity changes, etc. Is it a constant I just have to know, like $\pi = \frac{circumference}{diameter}$ or $c \approx 3\cdot10^{9} m/s$? Does it only have the units it does so that everything works out nicely in equations? Up until now I've just been treating it as part of $k=(4\pi \epsilon_{0})^{-1}$ but I ran into it in a different context and now I have no clue what it does.

Sorry if this question sounds like nonsense, my understanding of this is rudimentary so I can't tell if I'm using the right words in the right context. I know this question has been answered more than once but I can't make sense of or visualize the explanations I've already seen. Something about permitting electric field lines, but then someone else says that's wrong and it's about polarizability?

Also, if permittivity is impossible to explain in simple terms, I don't mind being told that it's beyond intuition or to wait until the concept suddenly makes sense one day. thanks

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What does ϵ0 mean in real life?

It simply means that you are using SI units. In SI units Coulomb's law is $$F = \frac{1}{4 \pi \epsilon_0}\frac{q_1 q_2}{r^2}$$ but in Gaussian units Coulomb's law is $$F=\frac{q_1 q_2}{r^2} $$ so $\epsilon_0$ doesn't even show up. That is because Gaussian units are dimensionally consistent in Maxwell's equations and Coulomb's law so they do not need additional conversion factors, whereas SI units are only dimensionally consistent for mechanical laws like Newton's 2nd law.

Something about permitting electric field lines, but then someone else says that's wrong and it's about polarizability?

Nothing like that, it is simply a unit conversion because you are using SI units. There is nothing special about $\epsilon_0$ itself, and it doesn’t tell you anything particular about the universe.

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    $\begingroup$ "...it doesn’t tell you anything particular about the universe."—Well, it tells you how strong the electrostatic force is. $\endgroup$
    – Buzz ♦
    Commented Sep 20, 2022 at 19:27
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    $\begingroup$ No, the strength of the electrostatic force doesn’t disappear in Gaussian units, like $\epsilon_0$ does. It does tell you how big the Newton is compared to the Coulomb. $\endgroup$
    – Dale
    Commented Sep 20, 2022 at 20:22
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    $\begingroup$ @Buzz Coulomb’s Law in Gaussian units tells you how strong the electrostatic force is, without using nonsense like the “permittivity of free space”. $\endgroup$
    – Ghoster
    Commented Sep 20, 2022 at 20:44
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    $\begingroup$ I think @Buzz would not be dissatisfied with your otherwise completely perfect answer if you had also said that unlike $\epsilon_0$ that is by itself a rather arbitrary quantity tied to the choice of units, the free space impedance $\zeta _0= \sqrt{\frac{\mu_o}{\epsilon_0}}$ is a real physical quantity just like $c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ is. $\endgroup$
    – hyportnex
    Commented Sep 20, 2022 at 23:31
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    $\begingroup$ The free space impedance and $c$ are no more real than $\epsilon_0$. Just as $\epsilon_0$ disappears in Gaussian units, so also $c$ disappears in Planck units and the free space impedance disappears in Lorentz-Heaviside units. Real physical quantities, like the fine structure constant $\alpha$, don't disappear in different units. In US units Newton's 2nd law is $F=kma$ where $k=1/32.2 \mathrm{\ lbf \ lbm^{-1} \ ft^{-1} \ s^2}$. In this formula $k$ disappears if we use different units. Do we believe that $k$ is a real physical quantity, or just an artifact of the units? $\endgroup$
    – Dale
    Commented Sep 23, 2022 at 13:33

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