$D$ is the electric displacement field or commonly the flux density and $E$ is the field intensity. There is a fundamental difference between them which will be understood to certain extent as you go through the following answer. Consider a point charge of $Q$ coulombs. This means that the number of flux lines emitted by the charge is $Q$ coulombs. ![enter image description here][1].

Let the hypothetical sphere shown in figure has a radius $r$. Then $D$
is given by 
\begin{equation}
D = \frac{Q}{4\pi r^2}.
\end{equation}
That is, $D$ is the number of flux lines passing per area. So, to get an intuitive grasp, interpret $Q$ as a number (number of flux lines) and $D$ as a number density (number of flux lines per area). Now, what about $E?$ $E$, which is the electric field intensity, is actually a force ($E$ is defined as force per coulomb) per flux line, that is the force carried by each flux line. So, the relation $D = \varepsilon E$ connects the number density of flux lines, D,  with a force per flux line term, $E$. Now, the permittivity $\varepsilon$ is defined as the ability to pass lines of electric flux through it. This is a qualitative way of saying. Quantitatively, it can be seen as the ratio $\frac{D}{E}$, that is, $\varepsilon$ is the number of electric flux lines (unit is coulomb, as mentioned earlier) passing through unit area for unit force/flux (which is unit field intensity). That is, say $\varepsilon = 5$ (this value of $\varepsilon$ is hypothetical and considered only for the sake of explanation) means, there are 5 flux lines in a unit area considered normal to an electric field with each flux line carrying $1 N$ force.
  [1]: https://i.sstatic.net/HXjRA.jpg