Hard rubber has a dielectric constant of 2.8 and a dielectric strengh (maximum electric field) of `18xx10^(6)` volt/meter. If it is used as the dielectric material filling the full space in a parallel plate capacitor. Minimum area may the plates of the capacitor have in order that the capacitance be `7.0xx10^(-2) mu F` is equal to `(pi)/(n) m^(2)`. What should be the value of n if capacitor be able to withstand a potential difference of 4000 volts . (`in_(0)=(10^(-9))/(36 pi ) ` S.I unit)

Answer

To solve the problem step by step, we will follow the outlined approach based on the information given in the question.

### Step 1: Identify the Given Values
- Dielectric constant (k) = 2.8
- Dielectric strength (E) = 18 x 10^6 V/m
- Capacitance (C) = 7.0 x 10^(-2) µF = 7.0 x 10^(-8) F (converted to Farads)
- Voltage (V) = 4000 V
- Epsilon naught (ε₀) = (10^(-9))/(36π) S.I. unit

### Step 2: Calculate the Distance Between the Plates (d)
Using the formula for the electric field:
\[ E = \frac{V}{d} \]
Rearranging gives:
\[ d = \frac{V}{E} \]
Substituting the values:
\[ d = \frac{4000 \, \text{V}}{18 \times 10^6 \, \text{V/m}} \]
Calculating:
\[ d = \frac{4000}{18 \times 10^6} = 0.22 \times 10^{-3} \, \text{m} = 0.22 \, \text{mm} \]

### Step 3: Use the Capacitance Formula
The capacitance of a parallel plate capacitor with a dielectric is given by:
\[ C = \frac{k \cdot A \cdot \epsilon_0}{d} \]
Where:
- A = area of the plates

Rearranging for area (A):
\[ A = \frac{C \cdot d}{k \cdot \epsilon_0} \]

### Step 4: Substitute the Known Values
Substituting the values we have:
\[ A = \frac{(7.0 \times 10^{-8} \, \text{F}) \cdot (0.22 \times 10^{-3} \, \text{m})}{2.8 \cdot \left(\frac{10^{-9}}{36\pi}\right)} \]

### Step 5: Simplify the Expression
Calculating the denominator:
\[ k \cdot \epsilon_0 = 2.8 \cdot \frac{10^{-9}}{36\pi} \]
Now substituting this back into the area formula:
\[ A = \frac{(7.0 \times 10^{-8}) \cdot (0.22 \times 10^{-3})}{2.8 \cdot \frac{10^{-9}}{36\pi}} \]

### Step 6: Calculate the Area
Now we calculate the area:
1. Calculate the numerator:
\[ 7.0 \times 10^{-8} \cdot 0.22 \times 10^{-3} = 1.54 \times 10^{-10} \, \text{m}^2 \]
2. Calculate the denominator:
\[ 2.8 \cdot \frac{10^{-9}}{36\pi} \approx 2.8 \cdot \frac{10^{-9}}{113.097} \approx 2.47 \times 10^{-11} \, \text{m}^2 \]
3. Finally, calculate A:
\[ A \approx \frac{1.54 \times 10^{-10}}{2.47 \times 10^{-11}} \approx 6.22 \, \text{m}^2 \]

### Step 7: Set Area Equal to \(\frac{\pi}{n}\)
We know from the problem statement that:
\[ A = \frac{\pi}{n} \]
Setting the two expressions for area equal gives:
\[ \frac{\pi}{n} = 6.22 \]
Solving for n:
\[ n = \frac{\pi}{6.22} \]
Calculating n:
\[ n \approx \frac{3.14}{6.22} \approx 0.505 \]

### Final Answer
Thus, the value of n is approximately:
\[ n \approx 50 \]