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, we need to find the value of \( n \) such that the area of the capacitor plates can be expressed as \( \frac{\pi}{n} \, m^2 \). We will follow these steps:

### Step 1: Calculate the distance between the plates (D)

The dielectric strength gives us the maximum electric field \( E \) that the dielectric can withstand. The relationship between the potential difference \( V \), electric field \( E \), and distance \( D \) is given by:

\[ V = E \cdot D \]

Rearranging this gives:

\[ D = \frac{V}{E} \]

Substituting the values:

\[ V = 4000 \, V, \quad E = 18 \times 10^6 \, \text{V/m} \]

\[ D = \frac{4000}{18 \times 10^6} = \frac{4000}{18000000} \approx 0.22 \times 10^{-3} \, m \]

### Step 2: Use the capacitance formula

The capacitance \( C \) of a parallel plate capacitor filled with a dielectric is given by:

\[ C = \frac{K \cdot \epsilon_0 \cdot A}{D} \]

Where:
- \( K = 2.8 \) (dielectric constant)
- \( \epsilon_0 = \frac{10^{-9}}{36 \pi} \, \text{F/m} \) (permittivity of free space)
- \( A \) is the area of the plates
- \( D \) is the distance calculated in Step 1

Rearranging for \( A \):

\[ A = \frac{C \cdot D}{K \cdot \epsilon_0} \]

### Step 3: Substitute known values

We know \( C = 7.0 \times 10^{-2} \, \mu F = 7.0 \times 10^{-8} \, F \) and \( D \approx 0.22 \times 10^{-3} \, m \):

\[ A = \frac{7.0 \times 10^{-8} \cdot 0.22 \times 10^{-3}}{2.8 \cdot \frac{10^{-9}}{36 \pi}} \]

### Step 4: Simplify the equation

Calculating the denominator:

\[ K \cdot \epsilon_0 = 2.8 \cdot \frac{10^{-9}}{36 \pi} = \frac{2.8 \times 10^{-9}}{36 \pi} \]

Now substituting back into the equation for \( A \):

\[ A = \frac{7.0 \times 10^{-8} \cdot 0.22 \times 10^{-3} \cdot 36 \pi}{2.8 \times 10^{-9}} \]

### Step 5: Calculate \( A \)

Calculating the numerator:

\[ 7.0 \times 10^{-8} \cdot 0.22 \times 10^{-3} \cdot 36 \pi \approx 7.0 \times 0.22 \times 36 \times \pi \times 10^{-11} \]

Calculating the value gives:

\[ \approx 0.0005 \, m^2 \]

Now, substituting back into \( A \):

\[ A \approx \frac{0.0005}{2.8 \times 10^{-9}} \approx 50 \, m^2 \]

### Step 6: Express \( A \) in terms of \( \frac{\pi}{n} \)

We have \( A = \frac{\pi}{n} \):

\[ 50 = \frac{\pi}{n} \]

Thus,

\[ n = \frac{\pi}{50} \]

### Final Value of \( n \)

To find \( n \):

\[ n = 50 \]

### Conclusion

The value of \( n \) is \( 50 \).

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