Does anyone know why the permittivity of free space is defined as: $\epsilon_0 = \frac{1}{\mu_0 c^2}$ ?

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    $\begingroup$ Ummm.. because the right FudgeFactor(tm) makes the answers match observed behavior? $\endgroup$ – Carl Witthoft Dec 24 '14 at 21:27
  • $\begingroup$ I want an actual answer lol, that's why I came here. I expect to get answers from people more knowledge-able than me. $\endgroup$ – NavyColors_Blue Dec 24 '14 at 21:29
  • $\begingroup$ Coming from electrostatics vacuum permitivity is needed to get the ratio between electrostatic charge and mechanical force right, which basically indicates that we are using the wrong units for either charge or force. $\endgroup$ – CuriousOne Dec 24 '14 at 21:29
  • $\begingroup$ If you want an actual answer, then ask an actual question. You haven't even indicated whether you have learned what $A$ and $B$ and $E$ fields are. $\endgroup$ – Carl Witthoft Dec 24 '14 at 21:30
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    $\begingroup$ @CarlWitthoft: How could one have indicated if one learnt of those concepts? Where from could one know that one should indicate such things? An answer to a question can't consists in a arguing that the question isn't actual. For whom asks, the question is actual. Which sin did he/she committed by asking? $\endgroup$ – Sofia Dec 24 '14 at 21:39

Ultimately, $\epsilon_0$ is there to make the numbers right in SI units. Both $c$ and $\mu_0$ are defined quantities (the latter from the definition of the Ampere in terms of the force between current carrying wires) and this leads to a requirement for, and definition of, $\epsilon$ through the relationship you quote.

This relationship arises from a manipulation of Maxwell's equations that result in a wave equation with a wave speed given by $1/\sqrt{\epsilon_0 \mu_0}$.

However, there is nothing special about the SI system - others are often used, especially the natural one where $c=1$.


I subscribe to Rob Jeffries explanation, I just want to show the things more clearly: Two of the Maxwell equations say that

$ (1) \ \nabla $ x $E = -\frac {∂B}{∂t}$

Another equation is that in void, where there are no currents and no charges,

$ (2) \ \nabla $ x $B = \epsilon _0 \mu _0 \frac {∂E}{∂t},$

where $\epsilon _0 \mu _0$ are the permittivity and permeability of the vacuum.

Now, taking the derivative of (2) in time we get

$ \nabla $ x $\frac {∂B}{∂t} = \epsilon _0 \mu _0 \frac {∂^2E}{∂t^2},$

and substituting (1),

$ (3) \ \nabla $ x $\nabla $ x $E = -\epsilon _0 \mu _0 \frac {∂^2E}{∂t^2},$

If you take for simplicity $E$ along the axis $x$ and dependent only on $z$ you get

$ (4) \ \epsilon _0 \mu _0 \frac {∂^2E}{∂t^2} = \frac {∂^2E}{∂z^2}. $

Denoting $\sqrt (\epsilon _0 \mu _0 ) = v^{-2}$, (4) is exactly a wave-equation with a solution $E = Ae^{ik(z - vt)} + Be^{-ik(z + vt)} $.

But, we know that the velocity of the propagating e.m. field is the light velocity s.t. $v = c$.


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