If K₁ > K₂ which condition for equivalent capacitance C_eq for the three configurations is correct?

The correct option is: C₁ > C₃ > C₂

To find the correct relationship between the equivalent capacitances of the three configurations, let us define the base parameters of the capacitor. Let the total plate area be $A$ and the total distance between the parallel plates be $d$. We define the capacitance of a vacuum-filled capacitor of area $A$ and thickness $d$ as:

$C_0=\frac{{\epsilon }_{0}A}{d}$

Each configuration is split horizontally into two layers of equal thickness $\frac{d}{2}$, and vertically into sections of area $\frac{A}{2}$. Let us calculate the equivalent capacitance for each configuration step-by-step.

1. Configuration C₁:
From the image, the top layer of $C_1$ is a single dielectric material of constant $K_1$ with thickness $\frac{d}{2}$ and area $A$. Its capacitance is:
$C_{\text{top}}=\frac{K_1{\epsilon }_{0}A}{d/2}=2K_1{C}_0$
The bottom layer consists of two halves connected in parallel, each having area $\frac{A}{2}$ and thickness $\frac{d}{2}$. The left half has dielectric $K_1$ and the right half has dielectric $K_2$:
$C_{\text{bottom}}=\frac{K_1{\epsilon }_0(A/2)}{d/2}+\frac{K_2{\epsilon }_0(A/2)}{d/2}=(K_1+K_2)C_0$
Since the top and bottom layers are connected in series, the reciprocal of the equivalent capacitance $C_1$ is:
$\frac{1}{C_1}=\frac{1}{2K_1{C}_0}+\frac{1}{(K_1+K_2)C_0}$

2. Configuration C₂:
In the second configuration, the top layer has a dielectric constant $K_2$ with thickness $\frac{d}{2}$ and area $A$:
$C_{\text{top}}=2K_2{C}_0$
The bottom layer is identical to that of configuration $C_1$:
$C_{\text{bottom}}=(K_1+K_2)C_0$
These layers are in series, so the reciprocal of the equivalent capacitance $C_2$ is:
$\frac{1}{C_2}=\frac{1}{2K_2{C}_0}+\frac{1}{(K_1+K_2)C_0}$

3. Configuration C₃:
In the third configuration, the dielectric boundaries are continuous from top to bottom, dividing the capacitor into two parallel-connected columns of area $\frac{A}{2}$:
- The left column has dielectric $K_1$ on top and $K_2$ at the bottom, in series:
$\frac{1}{C_{\text{left}}}=\frac{1}{K_1{C}_0}+\frac{1}{K_2{C}_0}=\frac{K_1+K_2}{K_1{K}_2{C}_0}\Rightarrow C_{\text{left}}=\frac{K_1{K}_2}{K_1+K_2}{C}_0$
- The right column has dielectric $K_2$ on top and $K_1$ at the bottom, in series, which is symmetrical to the left column:
$C_{\text{right}}=\frac{K_1{K}_2}{K_1+K_2}{C}_0$
Adding these parallel columns together gives the equivalent capacitance $C_3$:
$C_3=C_{\text{left}}+C_{\text{right}}=\frac{2K_1{K}_2}{K_1+K_2}{C}_0$
Taking the reciprocal to compare easily with the other configurations:
$\frac{1}{C_3}=\frac{1}{2K_1{C}_0}+\frac{1}{2K_2{C}_0}$

4. Direct Comparison under the condition K₁ > K₂:

First, comparing $C_1$ and $C_2$:
Since $K_1>K_2$, we have $\frac{1}{2K_1}<\frac{1}{2K_2}$.
Adding $\frac{1}{(K_1+K_2)C_0}$ to both sides yields:
$\frac{1}{C_1}<\frac{1}{C_2}\Rightarrow C_1>C_2$

Second, comparing $C_1$ and $C_3$:
Subtracting $\frac{1}{C_1}$ from $\frac{1}{C_3}$:
$\frac{1}{C_3}-\frac{1}{C_1}=\left(\frac{1}{2K_1{C}_0}+\frac{1}{2K_2{C}_0}\right)-\left(\frac{1}{2K_1{C}_0}+\frac{1}{(K_1+K_2)C_0}\right)=\frac{1}{C_0}\left(\frac{1}{2K_2}-\frac{1}{K_1+K_2}\right)$
Since $K_1>K_2$, we know that $K_1+K_2>2K_2$, which implies:
$\frac{1}{K_1+K_2}<\frac{1}{2K_2}\Rightarrow \frac{1}{2K_2}-\frac{1}{K_1+K_2}>0$
Thus, $\frac{1}{C_3}>\frac{1}{C_1}\Rightarrow C_1>C_3$

Third, comparing $C_2$ and $C_3$:
Subtracting $\frac{1}{C_3}$ from $\frac{1}{C_2}$:
$\frac{1}{C_2}-\frac{1}{C_3}=\left(\frac{1}{2K_2{C}_0}+\frac{1}{(K_1+K_2)C_0}\right)-\left(\frac{1}{2K_1{C}_0}+\frac{1}{2K_2{C}_0}\right)=\frac{1}{C_0}\left(\frac{1}{K_1+K_2}-\frac{1}{2K_1}\right)$
Since $K_1>K_2$, we know that $2K_1>K_1+K_2$, which implies:
$\frac{1}{K_1+K_2}>\frac{1}{2K_1}\Rightarrow \frac{1}{K_1+K_2}-\frac{1}{2K_1}>0$
Thus, $\frac{1}{C_2}>\frac{1}{C_3}\Rightarrow C_3>C_2$

Combining the inequalities, we get:
$C_1>C_3>C_2$