A parallel plate capacitor with air between the plates has capacitance of `9pF`. The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant `k_1=3` and thickness `d/3` while the other one has dielectric constant `k_2=6` and thickness `(2d)/(3)`. Capacitance of the capacitor is now

Options

A. (a) `1.8pF`
B. (b) `45pF`
C. (c) `40.5pF`
D. (d) `20.25pF`

Explanation

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

### Step 1: Understand the Initial Conditions
The initial capacitance \( C_0 \) of the capacitor with air between the plates is given as:
\[ C_0 = 9 \, \text{pF} \]
The formula for capacitance is:
\[ C = \frac{\varepsilon_0 A}{d} \]
where \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

### Step 2: Determine the Configuration of Dielectrics
The space between the plates is filled with two dielectrics:
- Dielectric 1:
- Dielectric constant \( k_1 = 3 \)
- Thickness \( d_1 = \frac{d}{3} \)

- Dielectric 2:
- Dielectric constant \( k_2 = 6 \)
- Thickness \( d_2 = \frac{2d}{3} \)

### Step 3: Calculate the Capacitance of Each Section
The total capacitance of the capacitor can be found by treating each dielectric as a separate capacitor in series.

1. **Capacitance of Dielectric 1**:
\[ C_1 = \frac{k_1 \varepsilon_0 A}{d_1} = \frac{3 \varepsilon_0 A}{\frac{d}{3}} = \frac{9 \varepsilon_0 A}{d} \]

2. **Capacitance of Dielectric 2**:
\[ C_2 = \frac{k_2 \varepsilon_0 A}{d_2} = \frac{6 \varepsilon_0 A}{\frac{2d}{3}} = \frac{9 \varepsilon_0 A}{d} \]

### Step 4: Combine the Capacitances in Series
The equivalent capacitance \( C_{eq} \) of two capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C_{eq}} = \frac{1}{\frac{9 \varepsilon_0 A}{d}} + \frac{1}{\frac{9 \varepsilon_0 A}{d}} = \frac{2}{\frac{9 \varepsilon_0 A}{d}} = \frac{2d}{9 \varepsilon_0 A} \]
Thus, the equivalent capacitance is:
\[ C_{eq} = \frac{9 \varepsilon_0 A}{2d} \]

### Step 5: Substitute the Initial Capacitance
We know from the initial conditions:
\[ C_0 = \frac{\varepsilon_0 A}{d} = 9 \, \text{pF} \]
Substituting this into the equation for \( C_{eq} \):
\[ C_{eq} = \frac{9}{2} C_0 = \frac{9}{2} \times 9 \, \text{pF} = 40.5 \, \text{pF} \]

### Final Answer
The capacitance of the capacitor after filling the dielectrics is:
\[ C_{eq} = 40.5 \, \text{pF} \]

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