Yes BennyF the voltage Vc (t) builds up in an initially discharged capacitor charging from a constant voltage Vb source as: Vc(t) = Vb (1-e^-t/rc)
IF the time constant RC is large compared to the time constant LC. I.e. then the inductance L can be ignored.
Typically, especially when charging from a battery, the time constant RC is several seconds. In your planned application two things are different:
(1) You do not have a constant voltage source.
(2) You must have RC very small compared to one second.
I.e. You want to use many (say N) individual capacitors with high voltage (>20kV) rating and at least C= 10 or more micro Farads in series. Thus the effective capacitance of the string, Ce will be C/N. For example, when N is one million Ce is a million times smaller than C (Ce is very small).
Likewise you will have R very small to avoid having all the power dissipated in the wires interconnecting these physically large individual capacitors. Thus your RC time constant should be less than 0.001 seconds* as the lightning bolt is over in about 0.03seconds.
SUMMARY:
With these conditions (many wires, or straps if you follow my advice, interconnecting you individual capacitors, which are physically large) the LC time constant cannot be ignored. In fact unless you pay a extra for low inductance capacitors, it very like controls the rate at which the current can build up in your capacitor string. I.e the simple large RC time constant curves you have copied and posted do not apply.
PS I am still waiting for you to reply to my post that explains why you are fundamentally wrong in most of your ideas. Read it at:
http://www.sciforums.com/showpost.php?p=2512112&postcount=153
* In fact your Ce being so small and your use of straps for low resistive loss as the interconnections between your many individual capacitors will probably make your RC time constant only a few micro seconds. Thus the LC time constant will probably about 100 times larger and dominate. - I.e. it will limit how fast you can put charge into your capacitors.
As I have already told you, most of the voltage will be across the inductance and the lightning will just continue its air arc to ground, by-passing your circuit. One does not want or need to see your circuit and given how ignorant you are about all this, it would only be good for laughs.