A parallel plate capacitor has the space between its plates filled by two slabs of thickness `(d)/(2)` each and dielectric constant `K_(1)` and `K_2`. d is the plate separation of the capacitor. The capacitance of the capacitor is

Options

A. `(2epsilon_(0)A)/(d)((K_(1)+K_(2))/(K_(1)K_(2)))`
B. `(2epsilon_(0)A)/(d)(K_(1)+K_(2))`
C. `(2epsilon_(0)A)/(d) ((K_(1)K_(2))/(K_(1)+K_(2)))`
D. `(2epsilon_(0)d)/(A) ((K_(1)+K_(2))/(K_(1)K_(2)))`

Explanation

To solve the problem of finding the capacitance of a parallel plate capacitor with two dielectric slabs, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
We have a parallel plate capacitor with plate separation \( d \). The space between the plates is filled with two dielectric slabs, each of thickness \( \frac{d}{2} \) and dielectric constants \( K_1 \) and \( K_2 \).

2. **Capacitance of Each Dielectric Section**:
The capacitance \( C \) of a parallel plate capacitor filled with a dielectric is given by:
\[ C = \frac{K \epsilon_0 A}{d} \]
where \( K \) is the dielectric constant, \( \epsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

3. **Calculate Capacitance for Each Section**:
- For the first dielectric slab with dielectric constant \( K_1 \) and thickness \( \frac{d}{2} \):
\[ C_1 = \frac{K_1 \epsilon_0 A}{\frac{d}{2}} = \frac{2 K_1 \epsilon_0 A}{d} \]
- For the second dielectric slab with dielectric constant \( K_2 \) and thickness \( \frac{d}{2} \):
\[ C_2 = \frac{K_2 \epsilon_0 A}{\frac{d}{2}} = \frac{2 K_2 \epsilon_0 A}{d} \]

4. **Determine the Connection Type**:
Since the two dielectrics are arranged in series, we need to find the equivalent capacitance \( C_{eq} \) of the two capacitors \( C_1 \) and \( C_2 \).

5. **Use the Formula for Series Capacitance**:
The formula for the equivalent capacitance of capacitors in series is:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]

6. **Substitute \( C_1 \) and \( C_2 \)**:
\[ \frac{1}{C_{eq}} = \frac{1}{\frac{2 K_1 \epsilon_0 A}{d}} + \frac{1}{\frac{2 K_2 \epsilon_0 A}{d}} \]
This simplifies to:
\[ \frac{1}{C_{eq}} = \frac{d}{2 K_1 \epsilon_0 A} + \frac{d}{2 K_2 \epsilon_0 A} \]
\[ \frac{1}{C_{eq}} = \frac{d}{2 \epsilon_0 A} \left( \frac{1}{K_1} + \frac{1}{K_2} \right) \]

7. **Combine the Terms**:
\[ \frac{1}{C_{eq}} = \frac{d}{2 \epsilon_0 A} \cdot \frac{K_1 + K_2}{K_1 K_2} \]

8. **Invert to Find \( C_{eq} \)**:
\[ C_{eq} = \frac{2 \epsilon_0 A K_1 K_2}{d (K_1 + K_2)} \]

### Final Result:
The capacitance of the capacitor is:
\[ C_{eq} = \frac{2 K_1 K_2 \epsilon_0 A}{d (K_1 + K_2)} \]