200J bank to 2kv eh?
As Sceadwian said you're going to need transformers for that sort of voltage. I guess for stupidly low efficiency a boost converter with a single inductor, followed by many cockroft/walton multiplier stages will manage it. But with all the losses your C cell will be dead before it finishes charging.
Most batteries with low self-discharge have the highest capacity when discharged at a low current. That means low wattage for a long time. With your 100uf cap bank, as low as that seems, by the time you've boosted your little battery to a significant voltage, the amount of current delivered to the capacitor will be silly. Also note capacitors don't hold charge forever.... so if the uA you're supplying to it doesn't over come the self discharge, your caps voltage won't go very far :/
Some numbers, just for 'fun'. Single cell boost with 90% to 5v. Max output 100mA (battery supplies 550mW). thats a half a watt. Joules = watts*secconds. For 200, thats 400 seconds, or just under 7 minutes. To boost 5v to 2kV... thats a pretty large ratio transformer you're looking at. So with 50% efficiency (which is optimistic) it'll take twice as long = 800 seconds, or 13 and a half minutes. Working backwards (I'm cool, I've avoided any specific circuits here!) Q=CV=IT, I=CV/T
CV = 100u * 2000 = 0.2. 0.2/800 seconds = 0.25mA. - you better make sure you have no resistors in parallel with the cap bank. A megohm resistor would take most if not all that current at 2kV.
Theoretically yes. Practically, naaaa