In a `p-n` junction diode having depletion layer of thickness `10^(-6)m`, the potential across it is `0.1V`. The electric field produced is

Options

A. `10^(7)Vm^(-1)`
B. `10^(-6) Vm^(-1)`
C. `10^(5) Vm^(-1)`
D. `10^(-5) Vm^(-1)`

Answer

To find the electric field produced in a p-n junction diode with a given depletion layer thickness and potential, we can use the formula for electric field (E):

### Step-by-Step Solution:

1. **Identify the given values**:
- Thickness of the depletion layer, \( D = 10^{-6} \, \text{m} \)
- Potential across the depletion layer, \( V = 0.1 \, \text{V} \)

2. **Use the formula for electric field**:
The electric field \( E \) across a region is given by the formula:
\[ E = \frac{V}{D} \]
where \( V \) is the potential difference and \( D \) is the width of the depletion region.

3. **Substitute the known values into the formula**:
\[ E = \frac{0.1 \, \text{V}}{10^{-6} \, \text{m}} \]

4. **Calculate the electric field**:
\[ E = 0.1 \times 10^{6} \, \text{V/m} \]
\[ E = 10^{5} \, \text{V/m} \]

5. **Conclusion**:
The electric field produced in the p-n junction diode is \( 10^{5} \, \text{V/m} \).

### Final Answer:
The electric field produced is \( 10^{5} \, \text{V/m} \).

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