Why is peak inverse voltage 2Vp in a half wave rectifier with a capacitve filter?

Question

See: https://www.quora.com/Why-is-peak-inverse-voltage-2Vp-in-a-half-way-rectifier-with-a-capacitor-filter

It makes sense to me that the PIV occurs at the bottom of the negative half cycle for the rectifier without the filter. What doesn't make sense to me is why the PIV for the rectifier with the filter occurs at the top of the positive half cycle. It seems like the reverse voltage is equal to the capacitor voltage minus the supply voltage, and at the top of the positive half cycle that would be zero. Am I missing something here?

Edit:

This image from the above link is what's confusing me. The above answers state that PIV is 2Vp because the voltage from the capacitor and the maximum negative voltage from the source are equal in magnitude and across the diode at the same time. However, this diagram shows the capacitor voltage is only equal in magnitude to Vp at the same peaks as the waveform without the capacitor (the top of the positive half cycle).

Answer 1

I believe that the second explanation "is" trying to say that the higher PIV (2Vp) is seen during the negative cycle. Perhaps the explanation could have been more clear if it were written as:

"Here, the capacitor is connected across the load. During the positive half cycle of input voltage, the capacitor charges to Vp. At the next negative half cycle of input voltage (before the capacitor can discharge), the diode becomes reverse biased, the diode then sees Vp at the cathode (capacitor side) and -Vp at the anode side."