A simple capacitor mental model (not good for quantitative predictions but useful for "just imagining") includes gravity and water and a tank: with these provisos:
- The area of the tank base can be likened to the capacitance of the capacitor.
- The tank height is related to the maximum voltage allowed, if any, for the capacitor.
- The amount of water in the tank is related to the stored charge in a capacitor.
- It takes work to pump water into the tank and the higher the water level already in the tank, the more work it takes to pump more water into it.
- A tank filled higher can perform more work from the same released amount of water because of the higher pressure available.
Now let's look at one of the fundamental relationships of a capacitor, where the total charge on the capacitor has this relationship to the capacitance and voltage measured between its two terminals:
$$Q=C\: V$$
Returning to the water tank model, this is basically saying that the volume of water in the tank is equal to the area of the tank (capacitance) times the height of the water level (voltage across the two terminals) in the tank.
Which makes sense.
Let's look at another relationship of a capacitor:
$$U=\frac{1}{2}C\: V^2$$
This is pretty easy, as well. If you imagine that using a certain amount of water from a tank that has a particular water level, and therefore the ability to put a certain pressure behind that water as you use it, then it should be pretty obvious that you can do more work with a gallon of water when the tank is fuller than when it is not so full. The gallon bursts out harder with a fuller tank.
With a full tank, the pressure starts out harder working and gradually becomes weaker and weaker until at the end even with still a fair bit of water left there just isn't that much pressure behind it. If you imagine this measure of "hard work" as the y-axis, and time on the x-axis, you might draw a line that starts high at first and then slopes downward to the right until you run out of water and the line hits the x-axis at some time in the future. The total work done with the entire tank will just be the area in that triangle. If the starting pressure height is \$C\: V\$ (the total water in the tank) and the time is \$V\$ (the total height of water you have to use), then the total area here is that of a simple triangle's area, which is what the above equation of \$U\$ shows you.
A \$1\:\textrm{F}\$ capacitor is then a very large area tank. If you fill it up to the height of \$1\:\textrm{V}\$, then the total volume will be \$1\:\textrm{C}\$. But there's less "pressure" despite the large volume. So the work it can do is only \$\frac{1}{2}\:\textrm{J}\$ (optimistically assuming 100% efficiency.)
In comparison, a \$\frac{1}{2}\:\textrm{F}\$ capacitor is then a tank with half the area of base. You'd need to fill it twice as high, to a height of \$2\:\textrm{V}\$, so that the total volume will be same \$1\:\textrm{C}\$ as before. But there's more usable "pressure" overall now, despite the same volume as before. So the work it can do is twice as much, or \$1\:\textrm{J}\$!
That's because for some of the time, the water current (in this analogy) is under more pressure and therefore can do enough more work to make the difference in the end. Both have the same total water to work with. But one of them has a harder working stream for part of the time, so to speak.
None of that changes the fact that in both cases there is only so much water to work with. But some of it can do more work in one case, than in the other.
Also, notice that in both cases, the height (voltage) does decline. In the case of a garden hose hooked up to the bottom of a tank, you might find that it will spin a lawn sprinker while the tank height is above some line but not below that. Similarly, a circuit might work so long as the voltage on the capacitor is above some voltage, but not below that. It's not all that dissimilar, in this sense.
Does that help?