Dielectrics and conductors in capacitors

I am studying composite parallel plate capacitors and came across two problems ( and ) where two different splitting techniques yield slightly different results. I want to understand which method is conceptually/physically correct for each setup, and why.

Problem 1: Container with a Metallic Partition

After $t = 5\text{ s}$, calculating the filled volumes gives:

• Left side: Fully filled ($h = 10\text{ m}, k=3$).
• Right side: Half-filled ($h_{\text{liquid}} = 5\text{ m}, k=2$; $h_{\text{air}} = 5\text{ m}, k=1$).

There is a thin metallic partition separating the left and right sides. I attempted to solve this using two different approaches:

• Method 1 (Vertical Partitioning): Assuming the metallic partition acts as an equipotential surface, isolating the left side ($30\varepsilon_0$) in series with the parallel combination of the right side ($10\varepsilon_0 + 5\varepsilon_0 = 15\varepsilon_0$). $C_{AB} = \frac{30\varepsilon_0 \cdot 15\varepsilon_0}{30\varepsilon_0 + 15\varepsilon_0} = 10\varepsilon_0 = 8.85 \times 10^{-11}\text{ F}$
• Method 2 (Horizontal Slicing): Slicing the capacitor horizontally at $h = 5\text{ m}$ into top and bottom parallel branches.
  • Bottom branch (series): $C_{\text{bot}} = \frac{15\varepsilon_0 \cdot 10\varepsilon_0}{15 + 10} = 6\varepsilon_0$
  • Top branch (series): $C_{\text{top}} = \frac{15\varepsilon_0 \cdot 5\varepsilon_0}{15 + 5} = 3.75\varepsilon_0$
  • Total: $C_{AB} = 6\varepsilon_0 + 3.75\varepsilon_0 = 9.75\varepsilon_0 \approx 8.63 \times 10^{-11}\text{ F}$

Both values appear in the options (Option A: $8.85 \times 10^{-11}\text{ F}$, Option D: $8.63 \times 10^{-11}\text{ F}$), and they are numerically extremely close ($2.5\%$ difference).

Problem 2: Pure Dielectric Blocks Without Partition

When applying both methods to a capacitor filled with dielectric blocks but no intermediate metal plate:

• Method 1 (Vertical Split / Imaginary Conductor): Placing the left region in series with the parallel combination of the two right regions. $\frac{C_2}{C_1} = \frac{4 \cdot 6}{4 + 6} = \frac{12}{5} = 2.400$
• Method 2 (Horizontal Split): Slicing horizontally into top and bottom parallel branches. $\frac{C_2}{C_1} = \frac{4}{3} + 1 = \frac{7}{3} \approx 2.333$

Here again, the results differ by only $\approx 2.86\%$.

My Questions:

1. Which method is conceptually correct for which physical situation? Does the presence of a physical conducting partition strictly demand Method 1 while making Method 2 invalid, and vice versa for pure dielectric setups?
2. Why do these two distinct methods yield numerically almost identical fractions/values ($\approx 2.5\% - 2.9\%$ relative difference) across both geometries?