A parallel plate capacitor, dielectric constants

Question:
There is a parallel plate capacitor of area $L^2$ and separation distance of $d$. A half of the space, $d/2$, is filled with a material of dielectric constant, $\kappa_1$. The other half is filled with the other material of constant, $\kappa_2$. What is the capacitance of this capacitor if the free space capacitance is $C_0$?

Answer:
The capacitance of a parallel plate capacitor is
\[ C = \frac{\epsilon_0\kappa A}{d} \]
where $A=L^2$. When a capacitor does not have any dielectrics (free space between plates), it will be
\[ C_0 = \frac{\epsilon_0 A}{d} \]
We can consider the particular capacitor is like series connection of two different capacitors. Namely,
\[ \frac{1}{C_{\mathrm{Total}}} = \frac{1}{C_1}+\frac{1}{C_2} \\
\rightarrow C_{\mathrm{Total}} = \frac{C_1C_2}{C_1+C_2} \]
From the condition, we have each capacitance of $C_1$ and $C_2$.
\[ C_{1,2} =  \frac{\epsilon_0\kappa_{1,2} A}{\frac{1}{2}d} = \frac{2\epsilon_0\kappa_{1,2} A}{d} \]
Then, we can rewrite it in terms of $C_0$.
\[ C_{1,2} = 2\kappa_{1,2}C_0 \]

Plug in the above.
\[ C_{\mathrm{Total}} = \frac{2\kappa_1C_0\times 2\kappa_2 C_0}{2\kappa_1C_0+2\kappa_2 C_0}    \\
      = \frac{4\kappa_1\kappa_2 C^2_0}{2C_0(\kappa_1+\kappa_2)}   \\
      =  \frac{2\kappa_1\kappa_2 C_0}{\kappa_1+\kappa_2}  \]

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