In the biased p-njunctions shown in Fig. 41-15, there is an electric field $\overrightarrow{\mathbf{E}}$in each of the two depletion zones, associated with the potential difference that exists across that zone. (a) Is the electric field vector directed from left to right in the figure or from right to left? (b) Is the magnitude of the field greater for forward bias or for back bias?

There are two biased p-n junctions, where the electric field is present in the two depletion zones: forward bias and reverse bias.

When we apply a potential difference across the p-n junction, if we connect the positive terminal of the battery with the p-type semiconductor and the negative terminal with the n-type semiconductor, the p-n junction is said to be forward biased. In this case, the current can easily flow through the junction.

When the p-n junction is forward biased, the current can easily through the junction. Thus, the direction of the electric field is from left to right inside the junction.

When a reverse bias is applied across the junction, the electric is so developed such that it does not allow the flow of any majority charge carriers a small current is still observed due to the flow of minority charge carriers which move through the junction with drift velocity. Thus, the electric field is from left to right.