Electric field inside a diode When a voltage is applied to a diode (forward or reversed bias) the depletion zone is changed due to charges change in this region. My question is in both case (forward or reversed bias), how the electric field that is responsible of moving the charges in the P and N region is established ? Is it the same mechanism of electric field establishment inside a conductor i.e surface charges density making the electric field? 
 A: I don't think any of the current answers address your question as I understand it. 
My understanding is that your question is effectively: What spatial distribution of charge gives rise to the electric field in a diode?
In the case of a uniform dielectric, the internal field is created by surface charges on the two surfaces,  such are usually very small in width. In a pn junction this is not the case. 
https://ecee.colorado.edu/~bart/book/book/chapter4/gif/fig4_3_1.gif
 Figure a of the above image shows the charge density profile of a pn junction.  Notice that you have both nonzero charge density in a sizeable area,  the depletion region. In the case of two dielectrics, instead of a pn junction,  you would get charge accumulating only along the surface of the interface. This non uniform charge density also means that the electric field is not simply a constant in the material, but rather varies along the depletion region, something that ia very different than for a simple resistor. 
This charge density comes the bare donors/acceptors that cannot move,  and are neutralized at the ends of the depletion region by the electrons and holes which accumulate in very small regions due to being free to move. 
A: You are talking about a p-n junction which forms a diode. Now a diode junction is very small compared to the rest of semiconductor. The diode junction can in a first approximation be taken as an insulator since it does not have any mobile charges compared to the rest of the region. When a voltage is applied at the ends the region outside the junction can be viewed as a conductor and hence in a fist approximation has no potential drop or electric field. Therefore all the applied voltage shifts to the p-n junction and an electric field is added or subtracted from original electric field arising from the built-in voltage. In case of forward voltage the electric field due to the applied voltages oppose that due to the built in voltage and opposite is for reverse voltage. Remember the built-in voltage is always from n side to p side.
A: The electric field inside the diode will be opposite to that of the electric field of the source. Thus will cause a barrier potential of 0.3 or 0.7 volts opposite to the source when its forward biased. That is why the the diode start conducting when forward biased when source is at more than 0.3 or 0.7 volts. Less than 0.3 or 0.7 volts it will conduct very low current or no current. When reversed biased diode will again built polarity opposite to the source which would be equal to the potential difference of the battery. Thus there is no conduction in reverse biased diode. 
A: The inner workings of a diode is a very complex topic, so I will start with some basic diode theory.
The current in a diode consists of two parts, diffusion current and drift current.
The diffusion current is due to differing electron or hole concentrations at a location in the diode. If electrons are concentrated at one location they will diffuse into the neighbouring area similar to how a drop of inc diffuses in a glass of water.
Then we have the drift current which is due to the electric field which exists the depletion region (I will come to the origin of this field later). If an electron diffuses into the depletion region it is swept by the field to the positive side. This is the drift current. Of course, it applies to both types of charge carriers, that is, both to electrons and holes. Electron-hole pairs formed in the depletion region (for example by thermal excitation or light absorption) will also be swept out of it and contribute to the drift current.
Let's consider a p-n junction. At the junction interface, we have   excess holes (holes in the conduction band) in the p-region, say to the left; and negatively charged electrons (electrons in the valence band) in the n-region, to the right. Consider the hypothetical situation when we suddenly bring a p-semiconductor together with an n-semiconductor. What will happen? Excess holes will now diffuse to the right because the hole concentration to the right is lower in the n-region than in the p-region. Similarly, excess electrons will diffuse to the left because of the lower electron concentration in the n-region. These holes and electrons occasionally collide and recombine which means that an electron "fall" into a hole and those charge carriers are "lost" because they are no longer in the conduction and valance bands respectively. At this point, they become immobile because charge can only be transported in the conduction band for holes and in the valence band for electrons. However, they have diffused from their original positions which were initially uncharged (they are built from atoms, which are neutral after all), but because they have left, the holes leave behind a negative charge while the electrons leave behind a positive charge and so the p-side becomes negatively charged and the n-side becomes positively charged. It is this charge that gives rise to the electric field inside the depletion region.
At equilibrium (0 V over the diode) the two currents, diffusion and drift, must cancel as no net current can pass through the diode. (If there would indeed be current at 0 V, we would be able to create energy from nothing from that diode, violating conservation of energy.) The question may now arise as to what happens when we short a p-n junction. We indeed have a field over the depletion region so we we might expect current to flow. However, what actually happens is that so called contact potentials arise at the semiconductor-metal interface cancelling the potential of the field in the depletion region precisely, so in the end, we do not get any current flow. (I'm not going to go into the details of exactly why these potentials arise.)
When we have zero bias, due to the electric field in the depletion region, charge carriers have a hard time passing through it, and only diffusing carriers that happen to have enough kinetic energy will go through (around 0.7 eV for silicon diodes), and hence there will be very little diffusion current.
Looking at an equation called the Shockley diode equation: (which is an approximation)
$I = I_S(e^{V/nV_t} - 1)$
The left term within the parentheses is due to diffusion current, while the right term is due to drift current. (Note that at 0 V, they cancel precisely.) Here, $I_S$ is called the scale current and is proportional to the junction interface area, which is constant for a given diode; $n$ is the grading coefficient, which is also constant and depends on the doping profile of the junction; and $V_t$ is the thermal voltage, which is constant for constant temperature. $I$ and $V$ are current through and voltage of the diode, respectively.
We see that increasing the voltage increases the diffusion current, but leave the drift current constant. This is somewhat unintuitive because normally drift current is proportional to voltage as in Ohm's law. However, because there are so few charge carriers in the depletion region, Ohm's law does not apply there and virtually every charge carrier ending up there is very quickly swept into one side of the junction by the field. The drift current is due to minority carriers (for example, electrons in the p-region) diffusing into the depletion region or electron-hole pairs forming in the depletion region. Because the amount of minority carriers in each side and electron-hole pair formation are quite independent of applied voltage, so is drift current.
So what happens when we forward bias the diode then? We are essentially counteracting the electric potential barrier created by the field in the depletion region, lowering the barrier and in turn lowering the required energy that a charge carrier needs to pass the barrier. Hence, current can pass in the forward direction in the diode. Because we have a lower effective field over the depletion region, less charge must be displaced and thus the depletion region must be smaller than for the equilibrium case.
When a diode is reverse-biased, the opposite happens; The electric potential barrier is essentially increased and even less diffusion current will go through. The drift current will remain constant, though, so we will see a small current (called leakage current) going in the reverse direction of the diode. For reverse bias, more charge is displaced and the depletion region is larger compared to the equilibrium case. For forward bias, drift current is usually considered negligible.
Even though this answer is lengthy, it is a major simplification of of the inner workign of a diode and many explanations have been left out on why certain things happens as they do. To get a better understanding of p-n junctions (and semiconductors in general) I can recommend the book "Solid State Electronic Devices" by Streetman and Banerjee.
A: In a diode, two processes are at play, first one is the "diffusion" and the second one is the "drift" , now even at room temperature electrons on the n-side try to go the p-side so as to stabilize themselves or to reduce their potential energy to a minimum, but when they do that they get stuck in the holes on the p-side, now those electrons were first in the conduction band that's why they were able to go haywire but now when they enter that hole ( a hole is essentially a vacancy in the valence band ) , they release their extra potential energy, and gets stabilized by entering in the valence band, now they are stuck and can't move due to the normal thermal energy that we get at the room temperature, due to this a "depletion region" is formed, we call it depletion because electrons are not able to go crazy in that region, same happens on the n- side, as electrons initially move from n-side to p-side, this exposes the +ve charge of the dopant atom and we call it a "hole" ( I don't like holes though ). Now this depletion layer has a voltage associated to it that tries to prevent more diffusion and tries to revert it back, so for example diffusion tries to kick electrons from n-side to p-side whereas drift tends to take them back to their initial configurations, when these two processes are balanced, we get stable equilibrium.
Now to distort that equilibrium, we can do two things.


*

*Forward Bias : We know that "diffusion" and "drift" are fighting with each other, but under normal conditions no one wins, but what if we make one of them win! Aha! Thats what we do when we bias a diode. In forward bias we attach +ve pole of battery on the p side and -ve pole of battery on the  n-side, as we do that -ve side of battery pushes electrons in the n-side and slides them towards the p-side, due to this that depletion layer starts vanishing, and now is the fun part when it vanishes electrons are no longer prevented to diffuse over to the p-side  and diffusion current increases heavily.electrons were trying to go all the way to p side ever since we made a pn junction but they were not able to do so due to that depletion layer, what we did in forward bias is reduced that depletion layer and then diffusion did its work.

*Reverse Bias : exactly opposite happens, in this case depletion layer increases and diffusion almost stops  but drift continues its work in the opposite direction, that's why we get a small current in the opposite direction.
