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

A thin half ring of radius $35 \text{ cm}$ is uniformly charged with a total charge of $Q$ coulomb. If the magnitude of the electric field at centre of the half ring is $100 \text{ V/m}$, then the value of $Q$ is ______ $\text{nC}$.
($\epsilon_o = 8.85 \times 10^{-12} \text{ C}^2/\text{Nm}^2$ and $\pi = 3.14$)

Answer 1

To find the electric field intensity (E) in the depletion region of a p–n junction, we use the formula for electric field: E = V/d, where V is the potential difference across the junction and d is the width of the depletion region.

Given: V = 0.6 V and d = 6×10⁻⁶ m.

Step 1: Substitute the values into the formula:

E = 0.6 / (6×10⁻⁶) = (0.6 / 6)×10⁶ N/C

Step 2: Simplify the expression:

E = 0.1×10⁶ N/C = 1.0×10⁵ N/C

Conclusion: The electric field intensity in the depletion region is 1.0×10⁵ N/C. This value falls within the given range of 1,1.

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.

Correct Answer: 1

Solution and Explanation

Top Questions on Electric Field

Questions Asked in JEE Main exam

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.

Correct Answer: 1

Solution and Explanation

Top Questions on Electric Field

Questions Asked in JEE Main exam

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.