In p-n junction, having depletion layer of thickness `2 ×10^(-6)m` the potential difference across it is 0.2 V. The electric field is

Options

A. `10^6V/m`
B. `4×10^5V/m`
C. `10^5V/m`
D. `10^(-5)V/m`

Answer

To find the electric field (E) in a p-n junction with a given depletion layer thickness and potential difference, we can use the relationship between electric field, potential difference, and distance. The formula is given by:

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

Where:
- \( E \) is the electric field (in volts per meter, V/m),
- \( V \) is the potential difference (in volts, V),
- \( D \) is the thickness of the depletion layer (in meters, m).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Thickness of the depletion layer, \( D = 2 \times 10^{-6} \, m \)
- Potential difference, \( V = 0.2 \, V \)

2. **Substitute the Values into the Formula:**
\[ E = \frac{V}{D} = \frac{0.2 \, V}{2 \times 10^{-6} \, m} \]

3. **Calculate the Electric Field:**
- First, calculate the denominator:
\[ 2 \times 10^{-6} = 0.000002 \, m \]
- Now perform the division:
\[ E = \frac{0.2}{2 \times 10^{-6}} = \frac{0.2}{0.000002} = 100000 \, V/m \]

4. **Express the Result in Scientific Notation:**
\[ E = 1 \times 10^{5} \, V/m \]

### Final Answer:
The electric field across the depletion layer in the p-n junction is \( 1 \times 10^{5} \, V/m \).

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