Capacitance of Parallel Plate Capacitor filled with Dielectrics

The initial capacitance of the parallel plate capacitor filled with air is given by: $C_1 = \frac{\varepsilon_0 S}{d}$

From the given diagram, the region between the plates (separated by distance $d$ and having area $S$) is divided horizontally into two equal halves, each having a plate area of $\frac{S}{2}$:

1. Upper Half (Area = $\frac{S}{2}$): This region is divided into two slabs along the direction of the electric field, each having a thickness of $\frac{d}{2}$:

   • Top-Left Slab: Filled with dielectric $\varepsilon_1 = 2$, thickness $\frac{d}{2}$, and area $\frac{S}{2}$. $C_{\text{TL}} = \frac{\varepsilon_1 \varepsilon_0 (S/2)}{d/2} = \frac{2 \varepsilon_0 S}{d} = 2 C_1$
   • Top-Right Slab: Filled with dielectric $\varepsilon_2 = 4$, thickness $\frac{d}{2}$, and area $\frac{S}{2}$. $C_{\text{TR}} = \frac{\varepsilon_2 \varepsilon_0 (S/2)}{d/2} = \frac{4 \varepsilon_0 S}{d} = 4 C_1$

   Since these two slabs are arranged one after another along the direction of current flow, they are in series. The equivalent capacitance of the upper half ($C_{\text{upper}}$) is: $C_{\text{upper}} = \frac{C_{\text{TL}} \cdot C_{\text{TR}}}{C_{\text{TL}} + C_{\text{TR}}} = \frac{(2C_1)(4C_1)}{2C_1 + 4C_1} = \frac{8}{6} C_1 = \frac{4}{3} C_1$

2. Lower Half (Area = $\frac{S}{2}$): This region is filled continuously with dielectric $\varepsilon_1 = 2$ across the full plate separation $d$. $C_{\text{lower}} = \frac{\varepsilon_1 \varepsilon_0 (S/2)}{d} = \frac{2 \varepsilon_0 S}{2 d} = \frac{\varepsilon_0 S}{d} = C_1$

3. Total Capacitance ($C_2$): The upper and lower halves share the same potential difference across the plates, so they are connected in parallel: $C_2 = C_{\text{upper}} + C_{\text{lower}} = \frac{4}{3} C_1 + C_1 = \frac{7}{3} C_1$

Thus, the ratio $\frac{C_2}{C_1}$ is: $\frac{C_2}{C_1} = \frac{7}{3}$

Correct Answer: D