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. 1.8 pF
B. 45 pF
C. 40.5 pF
D. 20.25 pF

Answer

To solve the problem, we need to find the capacitance of a parallel plate capacitor filled with two different dielectrics. The steps are as follows:

### Step 1: Understand the initial conditions
The initial capacitance of the capacitor with air between the plates is given as:
\[ C_0 = 9 \, \text{pF} \]
The formula for capacitance \( C \) of a parallel plate capacitor 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: Identify the dielectric materials
The capacitor is filled with two dielectrics:
- Dielectric 1: \( k_1 = 3 \) with thickness \( \frac{d}{3} \)
- Dielectric 2: \( k_2 = 6 \) with thickness \( \frac{2d}{3} \)

### Step 3: Calculate the capacitance of each section
For the first dielectric (thickness \( \frac{d}{3} \)):
Using the formula for capacitance with a dielectric,
\[ C_1 = \frac{k_1 \varepsilon_0 A}{\frac{d}{3}} = 3 \cdot \frac{\varepsilon_0 A}{\frac{d}{3}} = 9 \cdot \frac{\varepsilon_0 A}{d} = 9 \, \text{pF} \]

For the second dielectric (thickness \( \frac{2d}{3} \)):
\[ C_2 = \frac{k_2 \varepsilon_0 A}{\frac{2d}{3}} = 6 \cdot \frac{\varepsilon_0 A}{\frac{2d}{3}} = 9 \cdot \frac{\varepsilon_0 A}{d} = 13.5 \, \text{pF} \]

### Step 4: Combine the capacitances
Since the two dielectrics are in series, the equivalent capacitance \( C_{eq} \) can be calculated using the formula for capacitors in series:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{9 \, \text{pF}} + \frac{1}{13.5 \, \text{pF}} \]

Calculating the right side:
\[ \frac{1}{C_{eq}} = \frac{1}{9} + \frac{1}{13.5} = \frac{3}{27} + \frac{2}{27} = \frac{5}{27} \]

Taking the reciprocal to find \( C_{eq} \):
\[ C_{eq} = \frac{27}{5} = 5.4 \, \text{pF} \]

### Step 5: Final Result
The equivalent capacitance of the capacitor filled with the two dielectrics is:
\[ C_{eq} = 5.4 \, \text{pF} \]