A capacitor of capacitance 15 nF having dielectirc slab of `epsilon_(r)=2.5` dielectric strength 30 MV/m and potential difference = 30 volt. Calculate the area of plate

Options

A. `6.7xx10^(-4)m^(2)`
B. `4.2xx10^(-4)m^(2)`
C. `8.0xx10^(-4)m^(2)`
D. `9.85xx10^(-4)m^(2)`

Explanation

To solve the problem step by step, we will follow the necessary formulas and calculations.

### Step 1: Understand the given data
We have the following data:
- Capacitance (C) = 15 nF = \(15 \times 10^{-9}\) F
- Relative permittivity (\( \epsilon_r \)) = 2.5
- Dielectric strength (E) = 30 MV/m = \(30 \times 10^6\) V/m
- Potential difference (V) = 30 V

### Step 2: Calculate the distance (d) using dielectric strength
The dielectric strength (E) is defined as the potential difference (V) divided by the distance (d):
\[ E = \frac{V}{d} \]
Rearranging this gives:
\[ d = \frac{V}{E} \]
Substituting the values:
\[ d = \frac{30 \, \text{V}}{30 \times 10^6 \, \text{V/m}} = \frac{30}{30 \times 10^6} = 10^{-6} \, \text{m} \]

### Step 3: Use the formula for capacitance
The capacitance of a capacitor with a dielectric is given by:
\[ C = \frac{\epsilon_0 \cdot \epsilon_r \cdot A}{d} \]
Where:
- \( \epsilon_0 \) (permittivity of free space) = \(8.85 \times 10^{-12} \, \text{F/m}\)
- \( A \) is the area of the plates.

### Step 4: Rearranging to find the area (A)
Rearranging the capacitance formula to solve for area \( A \):
\[ A = \frac{C \cdot d}{\epsilon_0 \cdot \epsilon_r} \]
Substituting the known values:
\[ A = \frac{(15 \times 10^{-9} \, \text{F}) \cdot (10^{-6} \, \text{m})}{(8.85 \times 10^{-12} \, \text{F/m}) \cdot (2.5)} \]

### Step 5: Perform the calculation
Calculating the denominator:
\[ \epsilon_0 \cdot \epsilon_r = 8.85 \times 10^{-12} \cdot 2.5 = 2.2125 \times 10^{-11} \, \text{F/m} \]
Now substituting back into the area formula:
\[ A = \frac{(15 \times 10^{-9}) \cdot (10^{-6})}{2.2125 \times 10^{-11}} = \frac{15 \times 10^{-15}}{2.2125 \times 10^{-11}} = \frac{15}{2.2125} \times 10^{-4} \approx 6.77 \times 10^{-4} \, \text{m}^2 \]

### Final Result
The area of the plates \( A \approx 6.77 \times 10^{-4} \, \text{m}^2 \).