Polarized capacitors in voltage doubler
Try loading that circuit with a resistor across C1.

You will find that, when delivering current to a load, its output voltage falls far below the open circuit voltage - especially if you add further voltage multiplication stages.

The problem is that the impedance of those capacitors quickly adds up, and is effectively doubled by each multiplication stage, so it's a poor voltage source into anything other than a high impedance load.

I've seen questions about the effective source impedance of a voltage multiplier - and never seen a simple satisfactory calculation of it - including my own attempts.

However the fix is to reduce that impedance - increase the capacitance - and high value capacitors are often polarised. Simple as that.