Question

The depletion region width of a p-n junction is \( 4 \times 10^{-6} \, \text{m} \) and the potential barrier is \( 0.8 \, \text{V} \). The electric field intensity in the depletion region is:

• A. \( 1 \times 10^5 \, \text{V/m} \)
• B. \( 2 \times 10^5 \, \text{V/m} \)
• C. \( 4 \times 10^5 \, \text{V/m} \)
• D. \( 8 \times 10^5 \, \text{V/m} \)

Hint:

Always double-check the power of ten in units. Writing \( 0.8 / 4 \) as \( 0.2 \) and then shifting the decimal to get \( 2 \times 10^5 \) prevents simple calculation errors.

Answer 1

The Correct Option is B

Step 1: Picture the junction.
The depletion layer acts like a tiny parallel-plate region where the barrier voltage is spread across a small width.
Step 2: Connect field, voltage and width.
For a roughly uniform field, voltage equals field times distance: $V = E\,d$, so $E = \dfrac{V}{d}$.
Step 3: List the data cleanly.
Barrier $V = 0.8\ \text{V}$ and width $d = 4\times10^{-6}\ \text{m}$.
Step 4: Substitute.
\[ E = \frac{0.8}{4\times10^{-6}} \]
Step 5: Simplify the powers of ten.
\[ E = \frac{0.8}{4}\times10^{6} = 0.2\times10^{6}\ \text{V/m} \]
Step 6: Write in standard form.
\[ E = 2\times10^{5}\ \text{V/m} \]
A neat sanity check: $4\ \mu\text{m}$ holding $0.8\ \text{V}$ naturally gives field of order $10^5\ \text{V/m}$, which is typical inside a junction. \[ \boxed{E = 2\times10^{5}\ \text{V/m}} \]