Q: laser --> charge capacitor? I'm not an EE, but I do have a question that I hope isn't totally trivial. Please tell me if I have the calcs below correct:   
Current research on the surface photoelectric effect shows that if one were to shine a 2MW/cm^2  KrF laser on a polished gold plate for 20ns , it produces a 2.5A/cm^2 current density.  
Now consider that  gold plate to be one plate of a plate capacitor whose area was 1 sq m (for easy computations), and whose 1mm thick dielectric had a constant of 8,  the capacitance would be 70.8nF.
So, 1 illumination would produce those 2.5A spread out over 1 sq meter, and would be a current density of  2.5X10^-4A/sq m. 
If there were 10 illuminations in 1 second into the 70.8 nF capacitor, it would charge to 35.3 kV   (assuming no leakage or breakdown of the dielectric).
(i.e., 2.5X10^-3A/70.8nF = 35.3kV).
Do I have this right?

A: You don't seem to factor the pulse duration (20 ns) into your calculation of the charge on the capacitor.
Even assuming the photoelectric effect is linear (so it still works at 10-4 of he reported illumination intensity) each pulse could be expected to produce 2.5 × 10-4 A/m2 for 20 × 10-9 s, or 5 × 10-12 coulumbs.
10 such pulses could be expected to charge 70.8 nF to q/C = 50 pC / 70.8 nF = 0.7 mV.
If you expect the current to continue after the input energy has ceased, you'll have to explain (or provide references to an explanation) how or why.
