Question

The potential barrier between the p−n junction is 0.8 V. If the width of the depletion region is 8×10⁻⁷ m then the intensity of electric field in this region is ____×10⁶ V/m

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Answer 1

The potential barrier in a p-n junction is 0.8 V and the width of the depletion region is 8×10⁻⁷ m. The intensity of the electric field in this region is 1 × 10⁶ V/m.

Step-by-step explanation:

The potential barrier in a p-n junction is the voltage difference that prevents the flow of current in the absence of an external power source. In this case, the potential barrier is 0.8 V. The intensity of the electric field in the depletion region is determined by the width of the region and the potential barrier.

The relationship between electric field, potential difference, and distance is given by the equation:

Electric field (E) = Potential difference (V) / Distance (d)

Given that the width of the depletion region is 8×10⁻⁷ m, we can calculate the electric field as follows:

E = 0.8 V / (8×10⁻⁷ m) = 1 × 10⁶ V/m

Therefore, the intensity of the electric field in the depletion region is 1 × 10⁶ V/m.