If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6 × 10-6 m, will be × 105 N / C.

Question

If the potential barrier across a p-n junction is 0.6 V. Then the electric field intensity, in the depletion region having the width of 6 × 10-6 m, will be × 105 N / C. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/cb9e8f6971aa764237fc189d6963f860-.png" /> E =( V / d )=( text Potential barrier Across Junction / text width of Depletion layer ) =(0.6 V /6 × 10-6 m )=1 × 105 V / m =1 × 105 N / C

Q. If the potential barrier across a p-n junction is $0.6 V$ . Then the electric field intensity, in the depletion region having the width of $6 \times 1 0^{- 6} m$ , will be ____ $\times 1 0^{5} N / C$ .

$E = \frac{V}{d} = \frac{Potential barrier Across Junction}{width of Depletion layer}$
$= \frac{0.6 V}{6 \times 1 0 ^{- 6} m} = 1 \times 1 0^{5} V / m$
$= 1 \times 1 0^{5} N / C$

Q. If the potential barrier across a p-n junction is 0.6V. Then the electric field intensity, in the depletion region having the width of 6×10−6m, will be ____ ×105N/C.

E=dV​= width of Depletion layer Potential barrier Across Junction ​ =6×10−6m0.6V​=1×105V/m =1×105N/C

Q. If the potential barrier across a p-n junction is $0.6 V$ . Then the electric field intensity, in the depletion region having the width of $6 \times 1 0^{- 6} m$ , will be ____ $\times 1 0^{5} N / C$ .

$E = \frac{V}{d} = \frac{Potential barrier Across Junction}{width of Depletion layer}$
$= \frac{0.6 V}{6 \times 1 0 ^{- 6} m} = 1 \times 1 0^{5} V / m$
$= 1 \times 1 0^{5} N / C$