Animated Solution for Physics - Electrostatics: A parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C_1$ in air. When two dielectrics of different relative permittivities (${\epsilon }_1=2$ and ${\epsilon }_2=4$) are introduced between the two plates as shown in the figure, the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

Imagine you are looking at a complex architectural structure, but instead of concrete and steel, it's made of electric fields and dielectrics. At first glance, a capacitor filled with multiple different dielectric materials might look intimidating. However, the secret to solving these problems lies in the art of slicing. By breaking down the complex geometry into simpler, recognizable parts, we can conquer even the most daunting setups.

Let's carefully examine the physical setup of our capacitor. The total area of the plates is $S$, and the separation distance is $d$. The space between the plates is divided into three distinct regions:

1. Top Left Region: This part has a dielectric constant ${\epsilon }_2=4$. It occupies half the area ($S/2$) and spans half the distance ($d/2$). 2. Top Right Region: This part has a dielectric constant ${\epsilon }_1=2$. It also occupies half the area ($S/2$) and spans the remaining half of the distance ($d/2$). 3. Bottom Region: This part has a dielectric constant ${\epsilon }_1=2$. It occupies the other half of the area ($S/2$) but spans the entire separation distance ($d$).

Because the electric field lines pass sequentially through the top left and top right regions, these two act as capacitors connected in series. Conversely, the entire top combination and the bottom region share the same potential difference across the main plates, meaning they are connected in parallel.

We know the fundamental formula for the capacitance of a parallel plate capacitor is:

Let's apply this to each of our three sliced regions. We will express everything in terms of the original air capacitance, $C_1=\frac{{\epsilon }_{0}S}{d}$.

For the top left capacitor ($C_{top1}$):

For the top right capacitor ($C_{top2}$):

For the bottom capacitor ($C_{bot}$):

Now that we have our individual building blocks, it's time to assemble the equivalent circuit. First, we resolve the series combination of the top two capacitors. The equivalent capacitance for two capacitors in series is their product divided by their sum:

Finally, we combine this top equivalent capacitor with the bottom capacitor. Since they are in parallel, we simply add their capacitances together to find the total new capacitance, $C_2$:

The problem asks for the ratio of the new capacitance $C_2$ to the original capacitance $C_1$. From our final equation, it is clear that:

This elegant result shows how breaking a complex problem into fundamental atomic steps leads directly to the solution. The correct option is (d).