Question

Two identical sized capacitors $C_1$ and $C_2$ are connected with a battery as shown in figure. Capacitor plates are square plates. A dielectric slab of dielectric constant $K=2$, are filled in half the region of the two capacitors are shown:
$C\to$ capacity, $q\to$ charge stored, $U\to$ energy stored.
Match the following two columns

Answer 1

Capacitance of air filled parallel plate capacitor is $C_0=\frac{A{ϵ}_0}{d}$ where $A=$ area of plate and $d=$ separation between plates.
Here $C_1$ contains two capacitor of area $A/2$ and separation $d$. They are in parallel.
Thus, $C_1=\frac{\left(A/2\right)K{ϵ}_0}{d}+\frac{\left(A/2\right){ϵ}_0}{d}=\frac{\left(A/2\right)2{ϵ}_0}{d}+\frac{\left(A/2\right){ϵ}_0}{d}=C_0+C_0/2=\frac{3}{2}{C}_0$
As $C_2$ is filled half by dielectric slab so $C_2=\frac{A{ϵ}_0}{d-d/2+\frac{d/2}{K}}=\frac{A{ϵ}_0}{d/2+\frac{d/2}{2}}=\frac{4}{3}\frac{A{ϵ}_0}{d}=\frac{4}{3}{C}_0$
$\therefore \frac{C_1}{C_2}=\frac{\left(3/2\right)C_0}{\left(4/3\right)C_0}=\frac{9}{8}$ thus $A\to 3$
As $C_1$ and $C_2$ are in series so charges on both are same.
$\therefore \frac{q_1}{q_2}=1$ thus $B\to 1$
Energy $U=\frac{q^2}{2C}$, As q is same for both so $U\propto \frac{1}{C}$
$\therefore \frac{U_1}{U_2}=\frac{C_2}{C_1}=\frac{8}{9}$
Thus $C\to 2$