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

Answer

To solve the problem, we need to find the capacitance of a parallel plate capacitor that has been modified by filling the space between the plates with two different dielectrics. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The initial capacitance of the parallel plate capacitor with air between the plates is given as:
\[ C_0 = 9 \text{ pF} \]

### Step 2: Identify the Dielectrics
The space between the plates is filled with two dielectrics:
- Dielectric 1: Dielectric constant \( k_1 = 3 \) and thickness \( \frac{d}{3} \)
- Dielectric 2: Dielectric constant \( k_2 = 6 \) and thickness \( \frac{2d}{3} \)

### Step 3: Calculate the Capacitance of Each Section
The capacitance of a capacitor filled with a dielectric can be calculated using the formula:
\[ C = k \cdot \frac{A \epsilon_0}{d} \]
Where:
- \( k \) is the dielectric constant
- \( A \) is the area of the plates
- \( \epsilon_0 \) is the permittivity of free space
- \( d \) is the separation between the plates

#### Capacitance of Dielectric 1 (\( C_1 \)):
For the first dielectric:
- Thickness = \( \frac{d}{3} \)
- Dielectric constant = \( k_1 = 3 \)

Using the formula:
\[ C_1 = k_1 \cdot \frac{A \epsilon_0}{\frac{d}{3}} = 3 \cdot \frac{A \epsilon_0}{\frac{d}{3}} = 9 \cdot \frac{A \epsilon_0}{d} \]

Since \( \frac{A \epsilon_0}{d} = C_0 = 9 \text{ pF} \):
\[ C_1 = 9 \cdot 9 \text{ pF} = 81 \text{ pF} \]

#### Capacitance of Dielectric 2 (\( C_2 \)):
For the second dielectric:
- Thickness = \( \frac{2d}{3} \)
- Dielectric constant = \( k_2 = 6 \)

Using the formula:
\[ C_2 = k_2 \cdot \frac{A \epsilon_0}{\frac{2d}{3}} = 6 \cdot \frac{A \epsilon_0}{\frac{2d}{3}} = 9 \cdot \frac{A \epsilon_0}{d} \]

Again, since \( \frac{A \epsilon_0}{d} = C_0 = 9 \text{ pF} \):
\[ C_2 = 9 \cdot 9 \text{ pF} = 81 \text{ pF} \]

### Step 4: Combine the Capacitances
The two capacitors \( C_1 \) and \( C_2 \) are in series. The formula for the equivalent capacitance \( C_{eq} \) of capacitors in series is:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]

Substituting \( C_1 \) and \( C_2 \):
\[ \frac{1}{C_{eq}} = \frac{1}{81 \text{ pF}} + \frac{1}{81 \text{ pF}} = \frac{2}{81 \text{ pF}} \]

### Step 5: Calculate the Equivalent Capacitance
Now, we can find \( C_{eq} \):
\[ C_{eq} = \frac{81 \text{ pF}}{2} = 40.5 \text{ pF} \]

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

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