What happens to a polarity sensitive capacitor when you feed a square wave through it?
I disagree with the answer given. I think the answer confuses a negative voltage (with respect to zero or ground) with potential difference between two points in a circuit (i.e. the plates of the capacitor).

The voltage ACROSS  a capacitor cannot be changed instantly, so the positive plate is still MORE POSITIVE than the negative plate at the transition, thus it always maintains the normal polarisation.



Case 1: Suppose the input was 0V to 5V (or 0 to +V).

On the negative edge of the input pulse the positive plate falls to 0V BUT the negative plate falls to -5V at the same time maintaining the correct polarisation direction (positive plate more positive than the negative). The negative plate then discharges through R1 back to zero.

On the positive edge the voltage across the capacitor is zero. The positive plate rises to +5V as does the negative plate. We still have zero volts across the capacitor. The negative plate then discharges through R1 to zero BUT the positive plate remains at +5 volts. The capacitor is polarised normally.

At no time is the negative plate more positive than the positive plate. 

Case 2: (As suggested in comments) Suppose the input was +5v to -5V (or +V , -V)

This simply cannot be the circuit as drawn above. There is a common (ground) connection clearly shown between the pulse signal and R1 with a positive end shown connected to the capacitor.