A parallel plate capacitor is formed by two plates each of area `30pi cm^2` separated by 1 mm. A material of dielectric strength `3.6 xx 10^7 Vm^(-1)` is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is `7xx10^(-6)` C, the value of dielectric constant of the material is :
[Use `1/(4pi epsilon_0 =9xx10^9 Nm^2 C^(-2)` ]

Options

A. 1.66
B. 1.75
C. 2.25
D. 2.33

Answer

To find the dielectric constant \( K \) of the material filling the parallel plate capacitor, we will follow these steps:

### Step 1: Identify the given values
- Area of the plates, \( A = 30\pi \, \text{cm}^2 = 30\pi \times 10^{-4} \, \text{m}^2 \)
- Separation between the plates, \( D = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Dielectric strength, \( E = 3.6 \times 10^7 \, \text{V/m} \)
- Maximum charge, \( Q = 7 \times 10^{-6} \, \text{C} \)
- Value of \( \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

### Step 2: Calculate the capacitance \( C \) of the capacitor
The capacitance of a parallel plate capacitor filled with a dielectric is given by:
\[ C = \frac{K \cdot \epsilon_0 \cdot A}{D} \]
Where \( \epsilon_0 = \frac{1}{4\pi \cdot 9 \times 10^9} \).

### Step 3: Relate charge \( Q \), capacitance \( C \), and voltage \( V \)
The charge stored in the capacitor is given by:
\[ Q = C \cdot V \]
The voltage \( V \) across the plates can be expressed in terms of the electric field \( E \) and the separation \( D \):
\[ V = E \cdot D \]

### Step 4: Substitute \( V \) into the charge equation
Substituting \( V \) into the charge equation gives:
\[ Q = C \cdot (E \cdot D) \]

### Step 5: Substitute \( C \) into the charge equation
Substituting the expression for \( C \) into the charge equation:
\[ Q = \left(\frac{K \cdot \epsilon_0 \cdot A}{D}\right) \cdot (E \cdot D) \]
This simplifies to:
\[ Q = K \cdot \epsilon_0 \cdot A \cdot E \]

### Step 6: Solve for \( K \)
Rearranging the equation to solve for \( K \):
\[ K = \frac{Q}{\epsilon_0 \cdot A \cdot E} \]

### Step 7: Substitute the values
Now we will substitute the known values into the equation:
1. Calculate \( \epsilon_0 \):
\[ \epsilon_0 = \frac{1}{4\pi \cdot 9 \times 10^9} \approx 8.85 \times 10^{-12} \, \text{F/m} \]
2. Substitute \( Q \), \( A \), \( E \), and \( \epsilon_0 \):
\[ K = \frac{7 \times 10^{-6}}{(8.85 \times 10^{-12}) \cdot (30\pi \times 10^{-4}) \cdot (3.6 \times 10^7)} \]

### Step 8: Calculate the values
1. Calculate \( A \):
\[ A = 30\pi \times 10^{-4} \approx 9.42 \times 10^{-3} \, \text{m}^2 \]
2. Substitute and calculate \( K \):
\[ K = \frac{7 \times 10^{-6}}{(8.85 \times 10^{-12}) \cdot (9.42 \times 10^{-3}) \cdot (3.6 \times 10^7)} \]

### Step 9: Final calculation
After performing the calculations, we find:
\[ K \approx 2.33 \]

### Conclusion
The value of the dielectric constant \( K \) of the material is approximately \( 2.33 \).

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