A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant $$\varepsilon_1$$ and $$\varepsilon_2$$, as shown in figures. The distance between the plates is d and area of each plate is A. If capacitance in first configuration and second configuration are $$C_1$$ and $$C_2$$ respectively, then $$\frac{C_1}{C_2}$$ is :

First configuration

Two dielectrics fill half the thickness each, so they are in series.

Each slab has thickness

$$\frac{d}{2}$$

and full area AAA.

Capacitances:

$$C_a=\frac{\varepsilon_1A}{d/2}$$

$$=\frac{2\varepsilon_1A}{d}$$

$$C_b=\frac{\varepsilon_2A}{d/2}=\frac{2\varepsilon_2A}{d}$$

Series combination:

$$\frac{1}{C_1}=\frac{1}{C_a}+\frac{1}{C_b}$$$$\frac{1}{C_1}=\frac{d}{2A}\left(\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}\right)$$

Thus

$$C_1=\frac{2A\varepsilon_1\varepsilon_2}{d(\varepsilon_1+\varepsilon_2)}$$

Second configuration

Each dielectric occupies half area

$$\frac{A}{2}$$

and full thickness ddd, so they are in parallel.

Capacitances:

$$C'_1=\frac{\varepsilon_1(A/2)}{d}$$

$$C'_2=\frac{\varepsilon_2(A/2)}{d}$$

Hence

$$C_2=C'_1+C'_2$$$$C_2=\frac{A}{2d}(\varepsilon_1+\varepsilon_2)$$

Now ratio:

$$\frac{C_1}{C_2}=\frac{\frac{2A\varepsilon_1\varepsilon_2}{d(\varepsilon_1+\varepsilon_2)}}{\frac{A}{2d}(\varepsilon_1+\varepsilon_2)}$$

Simplifying,

$$\frac{C_1}{C_2}=\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$$

So correct answer is

$$\frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1+\varepsilon_2)^2}$$