Basically, I'm asking why the electric field in a vacuum (or the applied electric field) is related to the electric field in a dielectric by the relative permittivity $\epsilon_{r}$.

For context I'll provide the following question with a known solution:

To estimate the effective separation, $\vec d$, in an induced atomic dipole we assume that only electrons in the outer shell of the atom are displaced. Sulphur atoms have $4$ electrons in their outer shells. Sulphur has $3.8 \times 10^{28}$ atoms per meter cubed and a relative permittivity $\epsilon_{r} = 4.0$.

Estimate the value of $d$ when an external field of $1\mathrm{kV}\mathrm{m}^{−1}$ is applied to a block of sulphur.

The solution to this is (with more details added):

$$\vec P=\chi_e\epsilon_0\vec E_{\text{Dielectric}}=4N\cdot q \vec d$$ Now since $$\fbox{$\color{red}{\vec E_{\text{Dielectric}}=\frac{\vec E_{\text{Applied}}}{\epsilon_{r}}}$}$$ Therefore $$4N\cdot q \vec d=\frac{(\epsilon_r -1)}{\epsilon_r}\epsilon_0\,\vec E_{\text{Applied}}\tag{1}$$ Since $\epsilon_r =1+\chi_e$

Solving equation $(1)$ for $\vec d$ and substituting $$\epsilon_{r} = 4.0$$ $$\vec E_{\text{Applied}}= 10^3 \mathrm{V}\mathrm{m}^{-1}$$ $$q=\text{elementary charge}=e^{-}=1.6\times 10^{-19}\mathrm{C}$$ $$\mathrm{N}=\text{Number density of sulphur}=1.38\times 10^{28}\,\mathrm{m}^{-3}$$


$$\vec d =\frac{4-1}{4}\epsilon_0\times 10^3\times\frac{1}{4\times 3.8\times 10^{28}\times 1.6\times 10^{-19}}\approx 2.7\times 10^{-19}\mathrm{m}$$

I understand everything about this solution apart from the fact why the formula boxed in red holds.

The author simply stated this fact without justification. I would like to know why the electric field for the Dielectric is the applied Electric field reduced by a factor of $\epsilon_r$.

Why not $$\vec E_{\text{Dielectric}}=\frac{\vec E_{\text{Applied}}}{3\epsilon_{r}}$$ or why is it even divided by $\epsilon_r$ in the first place?

This may seem a futile and obvious question to some of you but I have just started learning about electromagnetic fields in matter (instead of just in vacuums) so this is far from trivial to me.

Could anyone please provide me some insight/intuition as to why $$\vec E_{\text{Dielectric}}=\frac{\vec E_{\text{Applied}}}{\epsilon_{r}}?$$

  • $\begingroup$ This is simply the definition of $\epsilon_r$ $\endgroup$ – By Symmetry Jun 21 '17 at 21:04
  • 1
    $\begingroup$ @By Symmetry $\epsilon_r$ is simply the ratio of the applied field to the field due to dielectric? Seems a little strange to me. Is there anyway you could provide a link and/or answer that explains this in a little more detail? Many thanks. $\endgroup$ – BLAZE Jun 21 '17 at 21:37
  • 1
    $\begingroup$ The definition is normally stated as $\vec{D} = \epsilon_0\epsilon_r\vec{E}$. Outside of the dialectic, however, the polarisation is $0$, so $\vec{D} = \epsilon_0 \vec{E}_{\mathrm{Applied}}$. $\endgroup$ – By Symmetry Jun 22 '17 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.