Pulsing 20V supply in 3 ns with 200 nF bypass capacitors would involve 1330 A current peak.
--> Wikipedia --> Capacitor --> Current–voltage relation:
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Taking the derivative of this and multiplying by C yields the derivative form:
for C independent of time, voltage and electric charge.
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In words: If you want to charge a capacitor ... you need to put in current.
* The bigger the capacitor --> the higher the current
* the less time (for the same charge) -> the higher the current
* the more voltage --> the higher the current
thus I = C * delta_V / delta_t
Now in your case:
C = 200nF
delta_V = "from 0V to 20V" = 20V
delta_t = "within 3ns" = 3ns
Now you get I = 200nF * 20V / 3ns ...
THIS simply is what a capacitor does. It´s what makes a capacitor a capacitor.
Thus it´s highly recommended to be understood an known for anyone who works with a capacitor. (not only circuit designer).
I recommend to keep basic capacitor working principle in mind ... at least one should know "where" to find the information /formula and "how to use it".
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Charging a capacitor basically similar to "filling water into a bucket".
Klaus