You seemed to be suggesting that you would have to increase the rate of change to maintain a constant current. Maybe I misunderstood, or it wasn't clear. But a constant rate of change will continue to produce a constant current indefinitely.

Poorly explained by me. What I meant was, increasing the rate of voltage change to increase the current **level** so as to "deplete/saturate" the plates quicker

But there is no such concept. This is not how capacitors work.

There is no such thing as charge saturation in a capacitor. The voltage can rise until the breakdown voltage of the insulation occurs. Until that time the linear equation is obeyed.

I think this is the crux of my confusion. Isn't the "charge" related to the capacitors capacitance.

No, the charge is related to the voltage established between the plates. The capacitance is a constant, not a capacity. It is not like a volume that can be filled.

The voltage continues to increase due to the externally applied voltage. If we were to continue to increase the applied voltage to a capacitor past the point were the current is constant (ie charged up to its capacitance) and then suddenly remove that applied voltage,

Again, this is a misunderstanding of how capacitors behave. You cannot apply a voltage to a capacitor

^{1}, the capacitor applies the voltage to you. It owns its own voltage and delivers it when measured. You change the voltage of a capacitor by applying current to it. The voltage then changes for as long as you apply current, proportional to the length of time you apply the current (strictly the integral of the current over time, but I simplify).

would the voltage across the now disconnected capacitor be the same voltage that was last applied or would it "drop" to a level that is a function of the charge stored in it? (Arghh, seeing I couldn't explain myself before, I'm sure I have done equally bad here )

As soon as you stop applying current to the capacitor the voltage stops changing and remains forever at whatever value it had reached. (In the real world capacitors leak, but ideal capacitors will hold their voltage forever.)

Oh, and another thing, is it fair to say that I=C*(dv/dt) only holds true once the capacitor is charged or is this some idiosyncratic spice thing

It is not meaningful to say "once the capacitor is charged". The charge on a capacitor is a number that can vary from zero to as much as you like (essentially as much as the capacitor can sustain before voltage breakdown occurs). The equation

*I* =

*C* d

*V*/d

*t* holds always. It is the basic defining equation of a capacitor.

(It is possible you may have trouble with this if you have not studied mathematics or physics to a level where you deal with differential equations. Rates of change can be a difficult concept to grasp until you can relate the mathematical notation to real world behavior. A physical analogy might be mass and acceleration. Newton's second law of motion

^{2} says that a body will experience a constant acceleration--a constant rate of change of speed--for as long as a constant force is applied. Think of the force as current and the speed as voltage. The speed goes up and up without limit for as long as the force is applied. There is no "maximum speed" and no "saturation" that occurs, at least until you reach relativistic velocities. The speed can be as much as you like, 100 mph, 1000 mph, 1,000,000 mph. It just goes up without limit for as long as you apply the constant force.)

[1] To understand the statement "you cannot apply voltage to a capacitor", think of how a car changes speed. A similar statement is true: "you cannot apply speed to a car". A car changes speed if it is given a push by the engine, but it always takes time for the new speed to come up on the speedometer. You can measure the speed, but you cannot "set" the speed. If you are driving along at a steady 30 mph and you suddenly press your right foot to the floor, the speed does not immediately change. The instant after you put your foot down the speed is still 30 mph just as it was before. But then the speed gradually ramps up with a curve until it tops out and the car is fully "charged".

Capacitors obey the same rules: you change the voltage on a capacitor by giving it a "push", by feeding current through it. You can't set the voltage directly, but you can move the voltage where you want by applying a current, a "force", to it.

[2] Newton's second law of motion is just like the capacitor formula. The capacitor formula says:

*I* = *C* d*V*/d*t*

"Current equals capacitance times the rate of change of voltage"

Newton's second law says:

*F* = *m* d*v*/d*t*

"Force equals mass times the rate of change of velocity"

In this equation the mass is the "capacitance" of the object. More massive objects accelerate more slowly with the same force, and when they are moving they hold more "charge" (momentum) than small objects.

Just as with a capacitor you have the charge equation, *Q* = *C**V*, so with a moving object you have the momentum equation:

*p* = *m**v*

where *p* is the momentum, *m* is the mass and *v* is the velocity.