1. **Problem statement:** A parallel plate capacitor with plate area $S$ and separation $d$ has capacitance $C_1$ in air. Two dielectrics with relative permittivities $\varepsilon_1=2$ and $\varepsilon_2=4$ are introduced, each occupying half the plate area ($S/2$). We need to find the ratio $\frac{C_2}{C_1}$ where $C_2$ is the new capacitance. 2. **Key concepts:** - Capacitance of a parallel plate capacitor: $C=\frac{\varepsilon_0 \varepsilon_r S}{d}$ where $\varepsilon_0$ is permittivity of free space and $\varepsilon_r$ is relative permittivity. - When dielectrics are placed side by side (in parallel along the plate area), the total capacitance is the sum of capacitances of each section. 3. **Capacitance in air:** $$C_1 = \frac{\varepsilon_0 S}{d}$$ 4. **Capacitance with two dielectrics:** Each dielectric occupies area $S/2$ with thickness $d$. Capacitance of first dielectric slab: $$C_{2,1} = \frac{\varepsilon_0 \varepsilon_1 (S/2)}{d} = \frac{\varepsilon_0 \times 2 \times S/2}{d} = \frac{\varepsilon_0 S}{d}$$ Capacitance of second dielectric slab: $$C_{2,2} = \frac{\varepsilon_0 \varepsilon_2 (S/2)}{d} = \frac{\varepsilon_0 \times 4 \times S/2}{d} = \frac{2 \varepsilon_0 S}{d}$$ 5. **Total capacitance $C_2$:** Since the two capacitors are in parallel (side by side), capacitances add: $$C_2 = C_{2,1} + C_{2,2} = \frac{\varepsilon_0 S}{d} + \frac{2 \varepsilon_0 S}{d} = \frac{3 \varepsilon_0 S}{d}$$ 6. **Ratio $\frac{C_2}{C_1}$:** $$\frac{C_2}{C_1} = \frac{\frac{3 \varepsilon_0 S}{d}}{\frac{\varepsilon_0 S}{d}} = 3$$ 7. **Check options:** None of the options (6/5, 5/3, 7/5, 7/3) equals 3 exactly. Re-examine the problem: The figure description says the block height is $d/2$, but the plate separation is $d$. This suggests the dielectrics are stacked vertically, each with thickness $d/2$, not side by side. 8. **Revised model:** Dielectrics are stacked in series, each with thickness $d/2$ and permittivities $\varepsilon_1=2$, $\varepsilon_2=4$. 9. **Capacitance of each slab:** $$C_{2,1} = \frac{\varepsilon_0 \varepsilon_1 S}{d/2} = \frac{2 \varepsilon_0 S}{d/2} = \frac{4 \varepsilon_0 S}{d}$$ $$C_{2,2} = \frac{\varepsilon_0 \varepsilon_2 S}{d/2} = \frac{4 \varepsilon_0 S}{d/2} = \frac{8 \varepsilon_0 S}{d}$$ 10. **Total capacitance in series:** $$\frac{1}{C_2} = \frac{1}{C_{2,1}} + \frac{1}{C_{2,2}} = \frac{1}{\frac{4 \varepsilon_0 S}{d}} + \frac{1}{\frac{8 \varepsilon_0 S}{d}} = \frac{d}{4 \varepsilon_0 S} + \frac{d}{8 \varepsilon_0 S} = \frac{2d}{8 \varepsilon_0 S} + \frac{d}{8 \varepsilon_0 S} = \frac{3d}{8 \varepsilon_0 S}$$ Thus, $$C_2 = \frac{8 \varepsilon_0 S}{3d}$$ 11. **Ratio $\frac{C_2}{C_1}$:** Recall $C_1 = \frac{\varepsilon_0 S}{d}$ $$\frac{C_2}{C_1} = \frac{\frac{8 \varepsilon_0 S}{3d}}{\frac{\varepsilon_0 S}{d}} = \frac{8}{3}$$ 12. **Final answer:** $$\boxed{\frac{C_2}{C_1} = \frac{8}{3}}$$ Since $\frac{8}{3} \approx 2.67$, closest option is (d) $\frac{7}{3} \approx 2.33$, but none exactly matches. Possibly a typo in options or problem statement. **Key concepts involved:** - Capacitance formula with dielectrics - Capacitors in series and parallel - Relative permittivity effect on capacitance