I was reading about dielectrics and wondered why there was no formula for their parallel connection between two parallel plate capacitors. So I thought of deriving it myself.
Let $k_1$ and $k_2$ be the dielectric constants of two dielectric slabs of width $d$ each and cross section area $A_1$ and $A_2$ connected in parallel across a parallel plate capacitor.
Let $Q_1$ and $Q_2$ be the charges on the upper and lower surface of one of the parallel plate capacitors. (Note that $Q_1$ can be equal the $Q_2$ or else $E$ across the slabs would be different as $k_1$ and $k_2$ is different. This would lead to a different potential difference across them, which contradicts that they are in a parallel connection)?
$E1 = \dfrac{Q_1}{A_1k_1 \epsilon_{0}}$ and similar for $E_2$.
Equating $V=E_1d=E_2d$, I got $\dfrac{Q_1}{A_1K_1} = \dfrac{Q_2}{A_2K_2}$
Assuming $A_{eq}$ to be the area of the cross-section of the equivalent dielectric slab(with same width $d$) with which the system is with, and equivalent dielectric constant $k_{eq}$, I couldn't find an expression relating these quantities.
Could someone please help me obtain a relationship for this?