A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance `C_0` . Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area `1/3` A, dielectric constant 2K and another with area `2/3` A and dielectric constant K. If the capacitance of this new capacitor is` C` then `C//C_0` is :

Options

A. `1`
B. `4/3`
C. `2/3`
D. `1/3`

Answer

To solve the problem, we will calculate the capacitance of the new configuration of the parallel plate capacitor after replacing one-third of the dielectric material.

### Step-by-Step Solution:

1. **Understand the Original Capacitor**:
The original capacitor has:
- Plate area \( A \)
- Dielectric constant \( K \)
- Separation between plates \( d \)

The capacitance \( C_0 \) of the original capacitor is given by the formula:
\[ C_0 = \frac{\varepsilon_0 A K}{d} \]

2. **Identify the New Configuration**:
One-third of the dielectric material is replaced with another material that has a dielectric constant of \( 2K \). This results in two capacitors in series:
- Capacitor 1 (C1): Area \( \frac{1}{3}A \) and dielectric constant \( 2K \)
- Capacitor 2 (C2): Area \( \frac{2}{3}A \) and dielectric constant \( K \)

3. **Calculate the Capacitance of Each Capacitor**:
- For Capacitor 1 (C1):
\[ C_1 = \frac{\varepsilon_0 A_1 K_1}{d} = \frac{\varepsilon_0 \left(\frac{1}{3}A\right) (2K)}{d} = \frac{2\varepsilon_0 A}{3d} K \]

- For Capacitor 2 (C2):
\[ C_2 = \frac{\varepsilon_0 A_2 K_2}{d} = \frac{\varepsilon_0 \left(\frac{2}{3}A\right) K}{d} = \frac{2\varepsilon_0 A}{3d} K \]

4. **Combine the Capacitances**:
Since the two capacitors are in series, the total capacitance \( C \) is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C} = \frac{3d}{2\varepsilon_0 A K} + \frac{3d}{2\varepsilon_0 A K} = \frac{3d}{2\varepsilon_0 A K} + \frac{3d}{2\varepsilon_0 A K} = \frac{6d}{2\varepsilon_0 A K} = \frac{3d}{\varepsilon_0 A K} \]
Thus,
\[ C = \frac{\varepsilon_0 A K}{3d} \]

5. **Express \( C \) in terms of \( C_0 \)**:
From the expression for \( C_0 \):
\[ C_0 = \frac{\varepsilon_0 A K}{d} \]
Therefore, we can express \( C \) as:
\[ C = \frac{C_0}{3} \]

6. **Calculate the Ratio \( \frac{C}{C_0} \)**:
Now, we find the ratio:
\[ \frac{C}{C_0} = \frac{\frac{C_0}{3}}{C_0} = \frac{1}{3} \]

### Final Answer:
The ratio \( \frac{C}{C_0} \) is \( \frac{4}{3} \).