The capacitance of a parallel plate capacitor with air as medium is `3 muF.` with the introduction of a dielectric medium between the plates, the capacitance becomes `15mu F.` The permittivity of the medium is

Options

A. `5C^(2)N^(-1)M^(-2)`
B. `15C^(2)N^(-1)m^(-2)`
C. `0.44xx10^(-10)C^(2)N^(-1)m^(-2)`
D. `8.854xx10^(-11)C^(2)N^(-1)m^(-2)`

Explanation

To find the permittivity of the dielectric medium introduced between the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the relationship between capacitance and dielectric constant
The capacitance \( C \) of a parallel plate capacitor is given by the formula:
\[ C = \frac{A \epsilon_0}{d} \]
where:
- \( A \) is the area of the plates,
- \( \epsilon_0 \) is the permittivity of free space (approximately \( 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( d \) is the separation between the plates.

When a dielectric material is introduced, the capacitance increases to:
\[ C' = \frac{A \epsilon}{d} = K \cdot C \]
where \( K \) is the dielectric constant of the material, and \( \epsilon = K \epsilon_0 \).

### Step 2: Set up the equations with given values
From the problem, we know:
- The initial capacitance \( C = 3 \, \mu F = 3 \times 10^{-6} \, F \)
- The capacitance with the dielectric \( C' = 15 \, \mu F = 15 \times 10^{-6} \, F \)

Using the relationship between the capacitances:
\[ K = \frac{C'}{C} = \frac{15 \times 10^{-6}}{3 \times 10^{-6}} = 5 \]

### Step 3: Calculate the permittivity of the dielectric medium
The permittivity \( \epsilon \) of the dielectric medium can be expressed as:
\[ \epsilon = K \epsilon_0 \]
Substituting the values:
\[ \epsilon = 5 \cdot (8.85 \times 10^{-12} \, F/m) \]

### Step 4: Perform the calculation
Calculating \( \epsilon \):
\[ \epsilon = 5 \cdot 8.85 \times 10^{-12} = 44.25 \times 10^{-12} \, F/m \]
or
\[ \epsilon = 0.4425 \times 10^{-10} \, F/m \]

### Final Result
The permittivity of the dielectric medium is:
\[ \epsilon \approx 4.425 \times 10^{-11} \, F/m \]