A parallel plate capacitor contanins a mica sheet (thickness ` 0. 5+xx10^(-3) m`) . And a sheet of fiber (thickness `0. 5 xx 10^(-3) m`) . The dielectric constant of mica is `8` and that of thye fiber is `2.5` Assuming that the fiber breaks down when subjected to an electric field of ` 6.4 xx 10^6 Vm^(-1)`. , find the maximum safe voltage that can be applied to the capacitor.

Explanation

To find the maximum safe voltage that can be applied to the capacitor containing a mica sheet and a fiber sheet, we will follow these steps:

### Step 1: Identify the parameters
- Thickness of mica sheet, \( d_1 = 0.5 \times 10^{-3} \, \text{m} \)
- Thickness of fiber sheet, \( d_2 = 0.5 \times 10^{-3} \, \text{m} \)
- Dielectric constant of mica, \( K_1 = 8 \)
- Dielectric constant of fiber, \( K_2 = 2.5 \)
- Breakdown electric field of fiber, \( E_{\text{breakdown}} = 6.4 \times 10^6 \, \text{V/m} \)

### Step 2: Calculate the total thickness of the capacitor
The total thickness \( L \) of the capacitor is the sum of the thicknesses of the mica and fiber sheets:
\[ L = d_1 + d_2 = 0.5 \times 10^{-3} + 0.5 \times 10^{-3} = 1.0 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the electric field in each dielectric
The electric field \( E \) in each dielectric can be expressed as:
\[ E_1 = \frac{V_1}{d_1} \quad \text{(for mica)} \]
\[ E_2 = \frac{V_2}{d_2} \quad \text{(for fiber)} \]
Where \( V_1 \) and \( V_2 \) are the voltages across the mica and fiber, respectively.

### Step 4: Relate the voltages and electric fields
Since the capacitors are in series, the total voltage \( V \) across the capacitor is:
\[ V = V_1 + V_2 \]
Using the electric fields:
\[ V_1 = E_1 \cdot d_1 \quad \text{and} \quad V_2 = E_2 \cdot d_2 \]
Thus,
\[ V = E_1 \cdot d_1 + E_2 \cdot d_2 \]

### Step 5: Express the electric fields in terms of the breakdown field
The maximum electric field that can be applied to the fiber is \( E_{\text{breakdown}} \). Therefore, we set:
\[ E_2 = E_{\text{breakdown}} = 6.4 \times 10^6 \, \text{V/m} \]

### Step 6: Calculate the voltage across the fiber
Using the relationship for the fiber:
\[ V_2 = E_{\text{breakdown}} \cdot d_2 = 6.4 \times 10^6 \cdot 0.5 \times 10^{-3} \]
Calculating \( V_2 \):
\[ V_2 = 6.4 \times 10^6 \cdot 0.5 \times 10^{-3} = 3200 \, \text{V} \]

### Step 7: Calculate the electric field in the mica
Using the ratio of the electric fields (considering the dielectric constants):
\[ \frac{E_1}{E_2} = \frac{K_2}{K_1} = \frac{2.5}{8} \]
Thus,
\[ E_1 = E_2 \cdot \frac{K_2}{K_1} = 6.4 \times 10^6 \cdot \frac{2.5}{8} \]
Calculating \( E_1 \):
\[ E_1 = 6.4 \times 10^6 \cdot 0.3125 = 2000 \, \text{V/m} \]

### Step 8: Calculate the voltage across the mica
Now, using the electric field in the mica:
\[ V_1 = E_1 \cdot d_1 = 2000 \cdot 0.5 \times 10^{-3} = 1000 \, \text{V} \]

### Step 9: Calculate the total voltage
Now, we can find the total voltage \( V \):
\[ V = V_1 + V_2 = 1000 + 3200 = 4200 \, \text{V} \]

### Step 10: Conclusion
The maximum safe voltage that can be applied to the capacitor is:
\[ \boxed{4200 \, \text{V}} \]