Q172 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
$\text { In a p-n junction, the depletion layer of thickness } 1 \mu \mathrm{m} \text { has } 0.05 \mathrm{~V} \text { potential across it. The electric field in } N C^{-1} \text { is }$
- A$5 \times 10^{2}$
- B$5 \times 10^{-8}$
- C$5 \times 10^{5}$
- Correct answerD$5 \times 10^{4}$
Solution
The electric field $E$ in a depletion layer is related to the potential difference $V$ and the thickness $d$ by the formula $E = \dfrac{V}{d}$.
Given values are $V = 0.05 \text{ V}$ and $d = 1 \mu\text{m} = 1 \times 10^{-6} \text{ m}$.
Substituting these values into the formula:
$E = \dfrac{0.05}{1 \times 10^{-6}}$
$E = 0.05 \times 10^{6} \text{ V/m}$
$E = 5 \times 10^{-2} \times 10^{6} \text{ V/m}$
$E = 5 \times 10^{4} \text{ V/m}$ (or $\text{N/C}$).
Answer: $5 \times 10^{4}$
Given values are $V = 0.05 \text{ V}$ and $d = 1 \mu\text{m} = 1 \times 10^{-6} \text{ m}$.
Substituting these values into the formula:
$E = \dfrac{0.05}{1 \times 10^{-6}}$
$E = 0.05 \times 10^{6} \text{ V/m}$
$E = 5 \times 10^{-2} \times 10^{6} \text{ V/m}$
$E = 5 \times 10^{4} \text{ V/m}$ (or $\text{N/C}$).
Answer: $5 \times 10^{4}$
Hence, the correct answer is option: 'D'.