Just wishing to clear this up, so then let's suppose we have a 50v 1 farad cap to play with.

If i feed 12v into this cap, what happens, the voltage will rise to 12volts, but does it also fill the cap with 1farad's worth of energy? regardless of voltage supplied... eg, 3v@1f 12v@1f or does it work more like a battery, when it reaches 50v only then will it store 1farad's worth of energy?.. take a 12v lead acid battery, if you feed 12v, as soon as it hits 12v, it's not going to store any more energy but supply 14.5volts it will continue to consume until it hits 14.5v (as it gradually cuts off)

so, voltage input = voltage out at whatever it's rated? or will it only store say 1 farad providing 50 volts are fed if it's rated 50v.... 15v cap, 2farads, it could store 2farads but say at only 3volts? or would you need to supply 15volts to get 2 farads stored.....?

 

I'm no expert on the physics here (and there will be tiny differences, I'm sure, due to ESR and all that), but I believe a power source will fill ANY capacitance, becoming static, or fully charged, at the voltage of the source (we're talking DC here). So you will have X Farads of electrons stored in your capacitor(s) at the supply voltage. This is providing the source can deliver that quantity of electrons without dying, as I am sure a small battery would.

Think of the voltage as a pressure that is filling up the caps, and Farads as quantity of charge (electrons sent in under that pressure).

 

Farads are a measure of capacitance, not power or energy storage.

The storage of energy in capacitors is measured in Joules.
The formula is 1/2C V ^2 (half C times V Squared) with C in Farads.

Think of capacitance like a balloon, and voltage as pressure.

A bigger balloon stores more air, or the same balloon at higher pressure (voltage) also stores more air, up to the point it bursts (or the insulation breaks down).