Discharge and charge rates of a capacitor - Comparing Energy Movements

If that is correct, then let me also ask something else: The voltage is irrelevant to the measured energy expenditure right? It doesn't matter if the first capacitor starts at 100 volts and discharges to 0 volts while the second capacitor starts at 0 and charges to 10 volts. All that matters for computing the energy movement is the actual charge/amps. Right?

This part is not right. The voltage is very relevant to the energy.

The energy stored in a capacitor is:

$$E = \frac{1}{2}CV^2$$

The voltage on a capacitor is determined by the charge and the capacitance:

$$V = \frac{Q}{C}$$

If you have a smaller capacitor, a given amount of charge will create a higher voltage, and it will take significantly more energy to put it there.

The problem (or rather, what may make this confusing) is that your thought experiment is not actually that trivial to turn into a real experiment. The simple capacitor experiments involve hooking up the cap to a near constant voltage source, which will give you the well-known charging graphs (charging current starts out high, then decreases).

Anyways, let's say you do have a constant current source to charge your caps with. If you charge C1 at 1A for 1s, you put 1C of charge on it. If you charge C2 at 2A for 1s, you put 2C of charge on it.

If you then disconnect the current source, the voltage between the two leads will be determined by the capacitance: If C1 ends up at 100V, that means it has a capacitance of 1C/100V = 10mF. If C2 ends up at 10V, it has a capacitance of 2C/10V = 200mF.

(For an explanation of the energy issues involved, see Dave Tweed's answer.)

One ampere is one coulomb per second, yes. Therefore, one coulomb is one amp-second.

The official definition of the farad is the amount of capacitance on which a charge of one coulomb will change its voltage by 1 volt.

We know that:

\$I_C = C \dfrac{dV}{dt} \$

or

\$ dV = \dfrac{I_c \cdot dt}{C}\$

Therefore:

\$ V = \dfrac{I_c \cdot t}{C} = \dfrac{Q}{C}\$ (omit the deltas for now)

\$ C = \dfrac{Q}{V}\$

Charge is charge regardless of the voltage. 1 coulomb is 1 coulomb. The capacitor voltages will vary depending on their sizes (the larger the cap, the lower the voltage a coulomb will produce).