It's well-known that if a diode is forward biased at a sufficient voltage, there is a voltage drop across the diode, i.e. it dissipates power.
However, I was reading Wikipedia's article on the depletion region of the diode (aka PN junction), and it occurred to me the electric field, E, as labeled below, is set up by immobilized ions and causes the voltage, V, as indicated to be positive:
I've also drawn the symbol of the diode in the above.
This result is definitely what I would expect, because if a battery (and resistor) is connected to the diode in the usual, forward-biased way, there should be voltage drop across the diode and the voltage V (as labeled) should be negative.
Where did I go wrong?