A parallel plate capacitor with air between the plates has a capacitance of 8 pF. Calculate the capacitace if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant `(epsilon_(r)=6)`

Options

A. 72pF
B. 81pF
C. 84pF
D. 96pF

Explanation

To solve the problem, we will use the formula for the capacitance of a parallel plate capacitor, which is given by:

\[ C = \frac{\epsilon_r \epsilon_0 A}{d} \]

where:
- \( C \) is the capacitance,
- \( \epsilon_r \) is the dielectric constant,
- \( \epsilon_0 \) is the permittivity of free space (approximately \( 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( A \) is the area of one of the plates,
- \( d \) is the distance between the plates.

### Step 1: Identify the initial conditions
The initial capacitance \( C_1 \) is given as 8 pF (picoFarads), which is:

\[ C_1 = 8 \times 10^{-12} \, \text{F} \]

The dielectric constant for air is:

\[ \epsilon_{r, \text{air}} = 1 \]

### Step 2: Write the initial capacitance formula
Using the formula for capacitance, we can express the initial capacitance as:

\[ C_1 = \frac{\epsilon_{r, \text{air}} \cdot \epsilon_0 \cdot A}{d} \]

Substituting the values:

\[ 8 \times 10^{-12} = \frac{1 \cdot \epsilon_0 \cdot A}{d} \]

### Step 3: Determine the new conditions
Now, we reduce the distance \( d \) by half:

\[ d' = \frac{d}{2} \]

We also fill the space between the plates with a substance that has a dielectric constant:

\[ \epsilon_{r, \text{new}} = 6 \]

### Step 4: Write the new capacitance formula
The new capacitance \( C_2 \) can be expressed as:

\[ C_2 = \frac{\epsilon_{r, \text{new}} \cdot \epsilon_0 \cdot A}{d'} \]

Substituting \( d' \):

\[ C_2 = \frac{6 \cdot \epsilon_0 \cdot A}{\frac{d}{2}} \]

This simplifies to:

\[ C_2 = \frac{6 \cdot \epsilon_0 \cdot A \cdot 2}{d} = \frac{12 \cdot \epsilon_0 \cdot A}{d} \]

### Step 5: Relate the new capacitance to the initial capacitance
From the initial capacitance formula, we know:

\[ C_1 = \frac{\epsilon_0 \cdot A}{d} \]

Thus, we can express \( C_2 \) in terms of \( C_1 \):

\[ C_2 = 12 \cdot C_1 \]

### Step 6: Calculate the new capacitance
Substituting the value of \( C_1 \):

\[ C_2 = 12 \cdot (8 \times 10^{-12}) = 96 \times 10^{-12} \, \text{F} = 96 \, \text{pF} \]

### Final Answer
The new capacitance \( C_2 \) is:

\[ C_2 = 96 \, \text{pF} \]

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