Why is $\vec E_{\text{Vacuum}}=\epsilon_{r}\cdot\vec E_{\text{Dielectric}}$?

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

gives

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

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 $$\vec E_{\text{Dielectric}}=\frac{\vec E_{\text{Applied}}}{\epsilon_{r}}?$$

• This is simply the definition of $\epsilon_r$ – By Symmetry Jun 21 '17 at 21:04
• @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. – BLAZE Jun 21 '17 at 21:37
• The definition is normally stated as $\vec{D} = \epsilon_0\epsilon_r\vec{E}$. Outside of the dialectic, however, the polarisation is $0$, so $\vec{D} = \epsilon_0 \vec{E}_{\mathrm{Applied}}$. – By Symmetry Jun 22 '17 at 4:31