Why aren't the majority charge carriers attracted by terminals of a battery when the p n junction is reverse biased?

Question

It is clear that when the junction is reverse biased, the free electrons from the n side can not pass the depletion layer and fill the holes in the p side. But why will the free electrons not move toward the battery's positive terminal instead? at least to some extent, it should happen: free electrons from the n side go to the positive terminal of the battery, and the holes in the p side be filled by the electrons from the negative terminal of the battery while no electron passes the depletion layer.

Answer 1

why will the free electrons not move toward the battery's positive terminal instead? at least to some extent, it should happen: free electrons from the n side go to the positive terminal of the battery, and the holes in the p side be filled by the electrons from the negative terminal

That does happen, at least for a short while. The result is that the reverse biased N region becomes positively charged, and the reverse biased P region becomes negatively charged, and one has, in effect, a capacitor. Of course, the net migration of charges stops (except for leakage cuurent) when the voltage created by these charges matches the externally applied voltage, (or the voltage exceeds the breakdown voltage) as is the case in capacitors in general.

Depending upon your application, the capacitance of a reverse biased diode may be troublesome, irrelevant, or useful. It may be troublesome in a high frequency circuit, where it allows high frequency ac signals to pass where the designer would prefer they be blocked. It may be useful in the case of varactor diodes where the reverse bias controls the capacitance and by controlling the capacitance, controls the frequency of an oscillator, or the characteristics of a filter.