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.

Solution

Physics sir · 3 steps

Treat the depletion region like a parallel-plate gap: field is just voltage divided by width.

1. Uniform field idea. Take the field across the depletion region as uniform. Then the barrier potential and the width are linked by V = Ed.

   E = V/d

2. Put in numbers. E = (0.6 V)/(6 × 10⁻⁶ m) = 0.1 × 10⁶ V m⁻¹ = 1 × 10⁵ V m⁻¹. One V m⁻¹ is the same as one N C⁻¹.
3. So. E = 1 × 10⁵ N C⁻¹, so the blank is 1.

The common slip: 10. Writing 0.6 / (6 × 10⁻⁶) as 0.1 × 10⁻⁶ instead of 0.1 × 10⁶ puts the power of ten on the wrong side. Reading the result as 10 × 10⁵ gives 10, not 1.

The idea to keep. For a barrier potential across a depletion width, use E = V/d. Watch the powers of ten, and remember V m⁻¹ and N C⁻¹ are the same unit.

Answer from the NTA final answer key. Solution by PrepWiser.