How to select the right HV pulse capacitor?
Edit: Corrected timings.

For the B25620B1217K983, \$I_S\$ (surge current) may be 9.6kA, but \$I_{MAX}\$ is 60A... if you discharge 5kA within 50µs (of 1/20Hz = 50ms period), then:

$$D = \frac{\over{P}}{P_i} = \frac{\tau}{T}$$
$$D = \frac{50µs}{50ms} = 0.001$$
$$A_{AVG} = D\cdot P_i$$
$$A_{AVG} = 0.001\cdot5kA = 5A$$


Discharge component: \$\frac{5kA/50µs}{1/50ms} =\$ 5A/cycle average
Charge component remaining: \$I_{MAX}\$ - 5A = 55A per cycle


It sounds like a lot is remaining, but charging will take much more time where current is non-zero, leading to higher RMS current. I think this could pose trouble depending on how it is charged.

For instance, if you chose a 50Ω charging resistor,
$$V_C = V_{IN}[1-e^{-t/RC}  ]$$
$$t=-RC\cdot ln\frac{1-Vc}{V_{IN}}$$
$$t=-50Ω\cdot 215µF\cdot ln\frac{1-0V}{2kV}$$
$$t=-0.01075‬\cdot ln(0.0005)$$
$$t=-0.01075‬\cdot -3.30103$$
$$t=35.5 \text{ms (99% at 49ms)}$$
At charging \$t=0\$, 2kV is across 50.0029Ω = 40A

Resistor pulse rating: 2kV * 40A = 80kW!

Initial maximum theoretical inrush charging current = \$\frac{2kV}{2.9mΩ} = 690\text{kA}\$ 

To avoid pages of math and get a quick idea of the total average current, I came up with this:



Checking the datasheet, \$I_{MAX}\$ RMS has been exceeded by 117.61/60 = 1.96 times. Perhaps the values could be tweaked some, but this is worrisome. An alternative could be to use multiple caps and take turns firing them sequentially. 

Needless to say, 2kV charged caps are extremely dangerous.