Why is the barrier potential and electric field from n side to the p side?

Question

In the above image taken from my class notes you can see that I have been taught that when a semi-conductor is doped with both P-type impurity and N-type impurity, one on each side, the barrier potential that is generated is from the side of the n type impurity to that of the p type and hence the electric field is from n type to p type. But aren’t the excess electrons due to the n-type impurity attracted by the positively charged holes of the p-type impurity so shouldn’t the direction of the electric field be the other way around ? I know that electric field lines go from Higher potential to lower potential but isn’t this counter-intuitive here ?

electric fields electromagnetism electrostatics semiconductor physics

Cluse · Accepted Answer

[quote]  But aren't the excess electrons due to the n-type impurity attracted by the positively charged holes of the p-type impurity...[/quote]
Once the depletion region has formed, there exists no attraction between the excess electrons in the n-type region and holes in the p-type region. This is because of the barrier potential which is set up by the acceptor ions and donor ions.
So when the electrons do try to diffuse across the region towards the p-type, they come across the negative​ acceptor ions and are repelled. The same thing happens with the holes being repelled by the positively charged donor ions.
Thus we can say that a electrostatic potential exists over the barrier due to the immobile ions and it is from the n-type region (positive donor ion) to p-type (negative donor ion).
Hope this helps.

Why is the barrier potential and electric field from n side to the p side?

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