Does the capacitance of a capacitor depend on the type of dielectric that is NOT between the capacitor? 
In the image above there are two infinite planes. The infinite plane at $z = a$ has a surface charge $\sigma$ while the infinite plane at $z = 0$ has a surface charge $-\sigma$. The dielectric that has a dielectric constant $\epsilon_1$ is present in the region $ 0 < z < a$, while the dielectric that has a dielectric constant $\epsilon_2$ is present in the regions $ a < z$ and $z < 0$. Using Gauss' law, the Electric field produced by the infinite plane at $z = a$ is:
$$
\vec{E}_1 = \frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z>a\quad \text{and} \quad \vec{E}_1 = -\frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z<a \tag{1}.
$$
While for the infinite plane at $z = 0$ the E-field is: 
$$
\vec{E}_2 = -\frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z>0 \quad  \text{and} \quad \vec{E}_2 = \frac{\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad z<0. \tag{2}
$$
Clearly the E-fields in regions $z > a$ and $z < 0$ are zero while the E-field in $ 0 < z < a$ is:
$$
\vec{E} = -\frac{2\sigma}{\epsilon_0(\epsilon_1+\epsilon_2)}\hat{z}\quad 0<z<a.\tag{3}
$$
If $\epsilon_1 = \epsilon_2$ then:
$$
\vec{E} = -\frac{\sigma}{\epsilon_0(\epsilon_1)}\hat{z}\quad 0<z<a.\tag{4}
$$
The potential difference between $z =a$ and $z = 0$ is: 
$$
V = \frac{\sigma}{\epsilon_0\epsilon_1}a.\tag{5}
$$
With equation (5) the capacitance is:
$$
C = \frac{Q}{V} = \frac{\sigma\cdot A}{\frac{\sigma}{\epsilon_0\epsilon_1}a} = \epsilon_1 \frac{A \epsilon_0}{a} = \epsilon_1 C_{\text{vacuum}} \tag{6}
$$
which is expected. However, if we say $\epsilon_1 \neq \epsilon_2$ then the capacitance becomes:
$$
C = \frac{\epsilon_1+\epsilon_2}{2} C_{\text{vacuum}} \tag{7}
$$
Somehow the external dielectric has changed the capacitance, if this is true why does this happen? If not where have I made the error?
Edit 1: I assumed that both the dielectrics were linear dielectrics. Furthermore, when calculating the flux I placed a square box that encapsulated a portion of the infinite plane. With this Gauss' law becomes:
$$
\iint_{s}\vec{D}\cdot\vec{dA} = \iint\vec{D_{above}}\cdot\vec{dA} + \iint\vec{D_{below}}\cdot\vec{dA} = (D_{top}+D_{below})A = A \sigma, \tag{8}
$$
where $A$ is the area of a face, also, The unit normal vector for $\vec{dA}$ always point in the same direction as the displacement field, that is why I added the contributions ($D_{top}+D_{below}$). In equation (8) only the contributions from the faces of the box whose normal is perpendicular to the plane are taken. Since I was looking at linear dielectrics I used: 
$$
\vec{D} = \epsilon_0\vec{E} + \vec{P} = \epsilon_0\epsilon_r\vec{E}. \tag{9}
$$
Using equations (8) and (9) I obtained equations (1) and (2), we get:
$$
D_{top}+D_{below} = \epsilon_0\epsilon_2 E+\epsilon_0\epsilon_1 E = \sigma. \tag{10}
$$
Edit 2:
It turns out that placing the same $E$ in equation (10) is wrong (physically not possible) and this was the source of my problems. To rectify the problem we must realize that the magnitude of $D$ is the same on the top and below. Using equation (8) we get:
$$
D = \frac{\sigma}{2}.\tag{11}
$$
Using equations (9) and (11) the E-fields for the infinite plane at $z = a$ is: 
$$
\vec{E} = \frac{\sigma}{2\epsilon_0\epsilon_2}\quad z>a\quad \text{and}\quad \vec{E} = -\frac{\sigma}{2\epsilon_0\epsilon_1}\quad 0<z<a. \tag{12}
$$
While for the infinite place at $z = 0$ the E-fields are:
$$
\vec{E} = -\frac{\sigma}{2\epsilon_0\epsilon_1}\quad 0<z<a\quad \text{and}\quad \vec{E} = \frac{\sigma}{2\epsilon_0\epsilon_2}\quad z<0.\tag{13}
$$
With this the fields in the region where dielectric 2 is present cancel and go to zero while the region where dielectric 1 is present the field is:
$$
\vec{E} = -\frac{\sigma}{\epsilon_0\epsilon_1} \rightarrow V = a\frac{\sigma}{\epsilon_0\epsilon_1}\rightarrow C = \epsilon_1 C_{\text{vacuum}}.
$$
 A: Draw a Gaussian surface around both plates and the net enclosed charge is zero. Consequently the net flux is zero and there is no electric field produced outside the capacitor. 
The external dielectrics are irrelevant.
Hope this helps 
A: No, it does not. The problem is in Eqs. (1) and (2), i.e. in your expressions for the electric fields. 
When you apply Gauss's law
in terms of $E$, you should consider all charges, also the ones
accumulated in the dielectrics. That is why in this case it is more convenient to apply Gauss's law in terms of the
electric displacement field, usually denoted as $D$.
Think also just a moment about it: Eqs. (1) and (2) cannot be right, since the electric field generated by the plane cannot have the same magnitude in two different dielectrics!
Applying Gauss's law in terms of $D$, you should be able to show that the electric field inside the capacitor, hence the capacitance, do not depend on the external dielectric.
I can add the derivation, if you want (I was once downvoted for writing too many details).
