A parallel-plate capacitor with plate area $$A$$ has separation $$d$$ between the plates. Two dielectric slabs of dielectric constant $$K_1$$ and $$K_2$$ of same area $$\frac{A}{2}$$ and thickness $$\frac{d}{2}$$ are inserted in the space between the plates. The capacitance of the capacitor will be given by:

We need to find the total capacitance of a parallel-plate capacitor when it is partially filled with two different dielectric slabs of dielectric constants $$K_1$$ and $$K_2$$.

1. Analyze the Geometry of the Capacitor Slabs

Let's break the setup down into individual capacitor components by analyzing how the space is filled:

• Left Region (Unfilled): This region has an area of $$\frac{A}{2}$$ and a full separation distance of $$d$$. It forms a single air capacitor ($$C_3$$).
• Right Region (Filled): This region has a total area of $$\frac{A}{2}$$. It contains two dielectric slabs ($$K_1$$ and $$K_2$$), each having a thickness of $$\frac{d}{2}$$, stacked on top of each other. These two slabs form a series combination of two distinct capacitors ($$C_1$$ and $$C_2$$).

2. Calculate the Capacitance of Each Part

The standard formula for the capacitance of a parallel-plate capacitor with area $$A'$$, plate separation $$d'$$, and filled with a dielectric constant $$K$$ is given by:

$$C = \frac{K \varepsilon_0 A'}{d'}$$

Capacitors in the Right Stacked Region ($C_1$ and $C_2$):

Both capacitors share a reduced plate area of $$A' = \frac{A}{2}$$ and a halved thickness of $$d' = \frac{d}{2}$$.

• For the upper dielectric ($$K_1$$):

  $$C_1 = \frac{K_1 \varepsilon_0 \left(\frac{A}{2}\right)}{\frac{d}{2}} = \frac{K_1 \varepsilon_0 A}{d}$$

• For the lower dielectric ($$K_2$$):

  $$C_2 = \frac{K_2 \varepsilon_0 \left(\frac{A}{2}\right)}{\frac{d}{2}} = \frac{K_2 \varepsilon_0 A}{d}$$

Capacitor in the Left Air Region ($C_3$):

This section contains air ($$K = 1$$), has an area of $$A' = \frac{A}{2}$$, and spans the full distance $$d' = d$$.

$$C_3 = \frac{1 \cdot \varepsilon_0 \left(\frac{A}{2}\right)}{d} = \frac{\varepsilon_0 A}{2d}$$

3. Combine the Individual Capacitances

Step A: Find the equivalent capacitance ($$C_{\text{series}}$$) of the stacked region

Since $$C_1$$ and $$C_2$$ are connected back-to-back in series across the vertical voltage drop, their equivalent capacitance is calculated as:

$$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{d}{K_1 \varepsilon_0 A} + \frac{d}{K_2 \varepsilon_0 A}$$

$$\frac{1}{C_{\text{series}}} = \frac{d}{\varepsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) = \frac{d}{\varepsilon_0 A} \left( \frac{K_1 + K_2}{K_1 K_2} \right)$$

$$C_{\text{series}} = \frac{\varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$$

Step B: Find the total equivalent capacitance ($$C_{\text{total}}$$)

The empty air branch ($$C_3$$) and the dielectric-filled branch ($$C_{\text{series}}$$) sit side-by-side, sharing the same top and bottom plates. Therefore, they are arranged in a parallel configuration:

$$C_{\text{total}} = C_3 + C_{\text{series}}$$

$$C_{\text{total}} = \frac{\varepsilon_0 A}{2d} + \frac{\varepsilon_0 A}{d} \left( \frac{K_1 K_2}{K_1 + K_2} \right)$$

Factoring out the common term $$\frac{\varepsilon_0 A}{d}$$ gives the final expressions:

$$C_{\text{total}} = \frac{\varepsilon_0 A}{d} \left( \frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2} \right)$$

Final Answer: Option A $$\left[ \frac{\varepsilon_0 A}{d} \left( \frac{1}{2} + \frac{K_1 K_2}{K_1 + K_2} \right) \right]$$