The
rated slew rate for the DH-220 is 30V/µS. If one assumes a maximum 100V step from -50V to 50V, that would translate to a 3.3333µs rise time in SPICE.
Let's assume for argument that the rise time is a faster 3µS for a maximum 100V step at 1KHz. Then how much power is dissipated in the snubber?
To find out I ran an LTSPICE simulation. It turns out the average power dissipation in the 10Ω snubber resistor is 470mW for a 100VPP 1KHz slew rate limited square wave. Therefore the average power dissipation in the series 0.1µF capacitor with an ESR=0.1Ω would only be 4.7mW. Not nearly enough to make it flame.
The peak power in the 10Ω resistor reaches 100W. Therefor the peak power in the capacitor would only be 1W. However, the entire power pulse V(PR1) only lasts 4µs, just a bit longer than the rise time. That works out to be less than a 0.8% duty cycle for the power waveform since it is not rectangular.
The first order thermal time-constant of something the size of the capacitor is at least on the order of tens of ms if not longer. Thus the transient heating is not going to be that great. Again not enough to hurt a properly working capacitor
If the capacitor had degraded with time in the damp salty marine environment as I suspect, then the power dissipation could be much greater as the ESR increases.
Thus the only way that capacitor fried so spectacularly was that is was defective, and even perhaps partially shorted. If the capacitor was working correctly even a full output 1 KHz square wave should not damage it.