Concept:The electric field across a depletion layer is given by the potential difference divided by the layer thickness: \( E_{\text{avg}} = V_0 / W \). Peak electric field in an abrupt junction is \( E_{\text{max}} = 2V_0 / W \).
Formula:$$E_{\text{avg}} = \frac{V_0}{W} = \frac{0.70\text{ V}}{1.0 \times 10^{-6}\text{ m}} = 7.0 \times 10^5\text{ V/m}$$
Solution:- Given: \( V_0 = 0.70\text{ V} \) and \( W = 1.0\ \mu\text{m} = 1.0 \times 10^{-6}\text{ m} \).
- Average electric field: \( E_{\text{avg}} = \frac{0.70}{10^{-6}} = 7.0 \times 10^5\text{ V/m} \approx 10^6\text{ V/m} \).
- This strong electric field rapidly sweeps minority carriers across the junction and prevents majority carrier diffusion.
Why other options are incorrect:- Option A: 0.70 V/m incorrectly divides by 1 meter instead of 1 micrometer.
- Option C: 70 V/m assumes \( W = 1\text{ cm} \).
- Option D: \( 7.0 \times 10^{-6}\text{ V/m} \) multiplies by \( W \) instead of dividing.
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