I recently learned about the dielectric that is used between the plates of a capacitor. If $E_0$ is electric field between the plates of capacitor in free space and $E_i$ is electric field due to induced charge in a dielectric after it is inserted inside the capacitor, (Let the dielectric constant be $K$) It is known that electric field becomes $\dfrac{E_0}{K}$. That is,

$$\frac{E_0}{K} = E_0 - E_i$$

Now from where I studied it, they use $\sigma_i$ as charge density of induced charge. And write $E_i=\dfrac{ \sigma_i}{\epsilon_0}$

$\epsilon_0$ being the permittivity of free space. What I don't get is why use the permittivity of free space instead of permittivity of dielectric material? After all, the electric field inside the dielectric is being calculated.

So I went to MIT 802 Walter Lewin's lectures and this is what I got.


He also uses $\epsilon_0$

Why is it that we not use the permittivity of dielectric medium?


2 Answers 2


You do this one of the two ways, you don't both take into count the induced electric field of the dielectric and use the permittivity of the dielectric. Think of it in this way, writing out the permittivity of the dielectric or writing a dielectric constant is an easy, express way to define the counter-effect the dielectric has on the external electric field passing through it. So in the 'base' calculation, you subtract the two electric fields, every other way to write it follows that calculation. Therefore using the dielectric permittivity in that case would be doing the same thing twice.

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    $\begingroup$ I understand you but the measurement of field due to dielectric must include some factor of the dielectric? To be different from electric field of free space? $\endgroup$
    – Rew
    Jun 29, 2020 at 17:53
  • $\begingroup$ It includes the induced charge, that changes from material to material, depending on their molecular structure. All dielectrics will get polarized in different magnitudes, even if the outside effect on them is the same. The dielectric constant of the material is deduced by measuring how much induced charge it creates in a certain situation, I believe Prof. Lewin also continues the lecture with that calculation. $\endgroup$ Jun 29, 2020 at 18:02
  • $\begingroup$ but doesn't it make more sense to use $\epsilon$ of the material of dielectric because it tells us how much the dielectric will permit its own field to interfere with the induced field in the dielectric? It is like an independent problem of E fields passing through a material. Why would your answer change if I ask you what is the E field through a material with permittivity $\delta$ produced by opposite charges lying on it's surfaces which are very large (so the field is uniform) It will only be logical to answer $\frac{Q}{A\delta}$ $\endgroup$
    – Rew
    Jun 30, 2020 at 2:49
  • $\begingroup$ @Rew no, what I'm saying is, its own field IS what changes the $\epsilon_0$ to the $\epsilon$ :=D $\endgroup$ Jul 1, 2020 at 0:07

So I think I found the answer and I'll just post it here for future users who face the same problem. See the thing is the $K$ I've been treating all together as a new constant which is absolute for any given dielectric is NOT ANYTHING NEW! but it is the same thing as $\epsilon_r$ the relative permittivity! That is where the $\epsilon$ of the medium comes into play. $K$ is not absolute but relative to the permittivity of free space. Hell $$K= \frac{\epsilon_m}{\epsilon_0}$$

The electric field decreases as $E_m = \frac{E_0}{\epsilon_r}$

This is where $K$ comes from.

$E_m = \frac{E_0}{K}$

The gist is, you can write induced E field as follows

$$E_i= \frac{\sigma_i}{\epsilon_0} = \frac{\sigma_i \times K}{\epsilon_{medium}}$$

this link here has been useful.


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