CombinationsDielectrics in parallel

Question

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?

Answer 1

from the above situation as displayed in the diagram we notice that both the dielectrics have same potential drop through them. Let this be $ V $. So now we can say that if the fields in dielectric $ 1 $ and $ 2 $ are $ E $ and $ E' $ respectively then we have $ Ed =V =E'd $. Write E as $ \frac {Q}{Akε ₀} $ and a similar relation for dielectric two. So far we are good. Now we can split the capacitor into two parts connected in parallel with each other such that potential difference across both is same and equal to $ V $ . Capacitance of a parallel plate capacitor is $ \frac {KAε ₀}{d} $ hence we have$ C ₁= \frac {K₁A ₁ ε ₀}{d}$ and a similiar expression for $ C₂ $ using $ C= C₁+C₂ $ and setting $ C = \frac {K'(A ₁ + A₂) ε ₀}{d} $ and comparing the two we can solve for$ K' $. Hope it helped!