Step-by-Step Solution

1. Initial Capacitance (C1)

The initial capacitor is a parallel plate capacitor with plate area S, plate separation d, and air as the dielectric medium. The permittivity of air is taken as $\varepsilon_0$.

The capacitance $C_1$ is given by the formula: $C_1 = \frac{\varepsilon_0 S}{d}$

2. Final Capacitance (C2) with Dielectrics

The problem states that two dielectrics with relative permittivities $\varepsilon_1 = 2$ and $\varepsilon_2 = 4$ are introduced as shown in a figure. Since the figure is not provided, we must deduce the configuration. A common configuration for such problems, which leads to one of the given options, is as follows:

• The capacitor is conceptually split into two parallel sections, each with an area of S/2.
• The first section (let's call it the left half) is filled entirely with the first dielectric ($\varepsilon_1 = 2$).
• The second section (the right half) is further split into two layers, each with thickness d/2. One layer is filled with dielectric $\varepsilon_1 = 2$ and the other with $\varepsilon_2 = 4$.

Let's analyze this composite capacitor:

a) Capacitance of the left half (C_A) This part is a single capacitor with area S/2, separation d, and dielectric $\varepsilon_1$. $C_A = \frac{\varepsilon_1 \varepsilon_0 (S/2)}{d} = \frac{2 \varepsilon_0 S}{2d} = \frac{\varepsilon_0 S}{d}$

b) Capacitance of the right half This half consists of two capacitors in series. Let's call them $C_B$ and $C_C$.

• Capacitor $C_B$ has area S/2, thickness d/2, and dielectric $\varepsilon_1 = 2$. $C_B = \frac{\varepsilon_1 \varepsilon_0 (S/2)}{d/2} = \frac{2 \varepsilon_0 S}{d}$
• Capacitor $C_C$ has area S/2, thickness d/2, and dielectric $\varepsilon_2 = 4$. $C_C = \frac{\varepsilon_2 \varepsilon_0 (S/2)}{d/2} = \frac{4 \varepsilon_0 S}{d}$

Since $C_B$ and $C_C$ are in series, their equivalent capacitance $C_{BC}$ is: $\frac{1}{C_{BC}} = \frac{1}{C_B} + \frac{1}{C_C} = \frac{d}{2 \varepsilon_0 S} + \frac{d}{4 \varepsilon_0 S} = \frac{2d + d}{4 \varepsilon_0 S} = \frac{3d}{4 \varepsilon_0 S}$ $C_{BC} = \frac{4 \varepsilon_0 S}{3d}$

c) Total Capacitance (C2) The left half ($C_A$) and the right half ($C_{BC}$) are in parallel. Therefore, the total capacitance $C_2$ is the sum of their individual capacitances: $C_2 = C_A + C_{BC}$ $C_2 = \frac{\varepsilon_0 S}{d} + \frac{4 \varepsilon_0 S}{3d} = \left(1 + \frac{4}{3}\right) \frac{\varepsilon_0 S}{d} = \frac{7}{3} \frac{\varepsilon_0 S}{d}$

3. The Ratio C2/C1 Now, we can find the ratio of $C_2$ to $C_1$: $C_1 = \frac{\varepsilon_0 S}{d}$ $C_2 = \frac{7}{3} \frac{\varepsilon_0 S}{d}$

Dividing the expression for $C_2$ by the expression for $C_1$: $\frac{C_2}{C_1} = \frac{\frac{7}{3} \frac{\varepsilon_0 S}{d}}{\frac{\varepsilon_0 S}{d}} = \frac{7}{3}$

4. Conclusion The calculated ratio is 7/3, which corresponds to option D. Even though the question is marked as multiple correct (MCQM), only this specific configuration leads to one of the options provided.