Concept:In reverse bias, the external voltage applies a positive potential to the N-side and a negative potential to the P-side, creating an applied field that points in the same direction as the built-in field.
Formula:$$E_{\text{net}} = E_{\text{built-in}} + E_{\text{applied}}$$
Solution:- The built-in electric field (\( E_{\text{built-in}} \)) points from the positive donor ions (N-side) to the negative acceptor ions (P-side).
- Reverse bias establishes an external electric field (\( E_{\text{applied}} \)) in the same direction (from N to P).
- Because both electric field vectors align, their magnitudes add directly: \( E_{\text{net}} = E_{\text{built-in}} + E_{\text{applied}} \).
Why other options are incorrect:- Option A: Field subtraction (\( E_{\text{built-in}} - E_{\text{applied}} \)) occurs under forward bias.
- Option B: The net field increases in reverse bias and is non-zero.
- Option C: Superposition of collinear electric fields involves vector addition, not division.
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