In a `p-n` junction diode having depletion layer of thickness `10^(-6)m`, the potential across it is `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, we can use the relationship between the electric field (E), the potential difference (V), and the thickness of the depletion layer (L). The formula we will use is:

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

### Step-by-Step Solution:

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

2. **Substitute the values into the formula:**
\[ E = \frac{V}{L} = \frac{1 \, \text{V}}{10^{-6} \, \text{m}} \]

3. **Calculate the electric field:**
\[ E = \frac{1}{10^{-6}} = 10^{6} \, \text{V/m} \]

4. **Conclusion:**
The electric field produced in the p-n junction diode is \( 10^{6} \, \text{V/m} \).

### Final Answer:
The electric field produced is \( 10^{6} \, \text{V/m} \).

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