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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?

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  • $\begingroup$ Welcome to Physics SE. I did not downvote. Your thoughts led to the definition of a permittivity equals 1. $\endgroup$ – Stefan Bischof Oct 5 '17 at 10:47
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    $\begingroup$ @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 '17 at 11:30
  • $\begingroup$ Keep in mind that empty space isn't empty space. It's full of quantum fluctuations. $\endgroup$ – Brad S Oct 5 '17 at 18:13
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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.

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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, $\epsilon=\epsilon_0$.

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