What is an AC voltage rating for a capacitor?
Summary

Based on the guidelines in Selection of Capacitors for Pulse Applications
 the required voltage rating is liable to be surprising and annoying.


Capacitor voltage rating = DC volts + AC component / Kfactor.
Kfactor is dependant on frequency and &lt;= 1. 
Value  as per this chart (from above reference).
At 70 Khz K ~= 0.35 so AC voltage component is multiplied by a factor of 1/0.35 = 2.9 !
For polypropylene K ~~= 1.16 - 0.16 x log(f)
(Numerical values were correct. Formula has been corrected   ).
(log base 10) - 
for 10HZ &lt; f &lt; 1 MHz.
(empirically based on graph below)


eg
 at  1 MHz multiply any AC component x ~= 5
  at 100 KHz multiply any AC component x ~= 3
  at  10 KHz multiply any AC component x ~= 2  

For this specific example

 


Kf at 70 kHz ~= 0.35
Veffective = Vdc + (Vpeak-Vdc)/kf 
= 200 + (800-200)/0.35 
=~ 2000 Volt capacitor required !!!





This is more applicable for pulse applications or very high frequency AC (which your example is a case of), although it's worth noting that at 100 HZ the scaling factor is already down to 80% of the DC capacitance value.

The example graphs you have given are for Polypropylene film dielectric.
 Thye numerical values will vary with dielectric type.

The reason given is that the dielectric strength of the film decreases with increasing frequency.


The explanation behind the reason, which doesn't need to be known to apply the formulae, starts to get into deep magic and arcane physical properties but appears to relate to the increase of dissipation factor with frequency and the increasing likelihood of internal corona discharge with increasing thickness of material (or "effective thickness" with increasing frequency).

This interesting (or boring depending on ones interests) document
Mylar film - Product informationm from Dupont Teijin offers some insights for polyester / Mylar which can be expected to be generally applicable to other plastics. Figure 8 shows increasing dissipation factor with frequency (hence lowering of resistance to applied voltage and corona discharge) 


Application of the formula is easier than understanding the reason.  

(a) Solution for:
 +ve DC voltage with
  +ve going pulse
  or added AC such that Vmin >= 0V.

This applies to a capacitor with a (say +ve) DC offset and an added +ve going pulse OR DFC with an added AC waveform such that V is always > 0.
 For AC offset by a DC component such that the waveform still crosses 0 Volts see (b) below.  


Calculate a k multiplier value based on frequency.
From table K &lt;= 1.
This is a derating factor for the AC part of the waveform.
Calculate minimum voltage = Vmin  
Calculate Vpp = Vmax - Vmin.
Calculate effective voltage of the AC component

Vac effective = Vpp/k.
(Wghich will always be >= Vpp)
Add DC and AC values 

Veffective = Vdc_applies + Vac 
= Vdc_applied + Vpp/k.

QED.


(b) Solution for Vdc + Vac such that combined waveform still crosses 0v twice per cycle


Vmin = 0
Vpp = Vpeak [[= VAC_peak_to_peak/2 + Vdc]]
Obtain k from table as above.
Veffective = Vpp / k.


In your example case (a) applies.

Vdc = 200V
 You report that Vmax = 800V so 
 Vpp = (Vmax - 200) = (800-200) = 600v.
K calculation from referenced WIMA document.



K for 70 kHz =~= 0.35

Veffective = 200 + 600/0.35 = 1914v

2 kV capacitor required !!!!!!!!!!!!!!!!!!!!!!!!!