Let's set aside equations and think about this conceptually. To be clear about the question here, I'm assuming that you have put a fixed charge onto each plate (equal and opposite) and then isolated them (so the charges remain constant). Here are the key ideas that are significant for this:

* Assuming the separation between the plates is always small compared to their diameter/size, parallel plate capacitors give rise to a very simple electric field pattern: it is essentially zero everywhere outside, and it is uniform throughout the region between the plates with a strength that is independent of the separation. (Technically, it depends only on the area charge density: charge per unit area. All of this can be deduced from the electric field pattern of an infinite charged plane.)

* Capacitors store their energy within that electric field between the plates. All electric fields carry an energy density proportional to field strength squared, so the total energy stored in a uniform field is proportional to the volume occupied by that field.

So for your question, I can combine these ideas to recognize that pulling the plates apart will result in a larger volume between the plates, and therefore it must result in more stored energy (because the strength of the field in between stays the same as you pull). That energy has to come from somewhere, so I can deduce that I would have to put energy into the system while pulling: it would indeed be hard to pull apart. (The level of "hard" would depend on a lot of factors.)

(If you pull far enough so the separation isn't small compared to the plates' diameter anymore, you'll pretty quickly reach a case where you can approximate each plate as a point charge. Pulling those opposite charges apart clearly requires work, too, though the force gets smaller as they get farther apart.)

Mind you, you might instead keep the voltage constant (rather than the charges), perhaps by keeping both plates plugged in to a battery the whole time. In that case, the story changes:

* Voltage is (morally) equal to electric field times distance, so here if the separation between the plates doubles then the electric field must be cut in half.

* The plates' area is fixed, so the volume between them is proportional to their separation.

* The same fact from before about energy density being proportional to electric field strength squared applies, and total energy is still energy density times the volume between the plates.

So combining these ideas, we can see that the total energy stored between the plates will wind up decreasing as the plates are pulled apart, because the decreasing field winds up being squared when finding the total energy.

That seems very strange! Opposite charges attract, after all, so you'd still expect that you would need to do work to pull the plates apart. The subtlety here is that the battery's stored energy is changing in this process as well. (Let's assume a rechargeable battery for the moment.) As the plates separate, the charge on each plate goes down in proportion to the reduced electric field in between, so there's suddenly a lot of excess charge that needs to go somewhere. That means that the extra positive charges will be forced back into the battery's + side and the extra negative charges will be forced back into its - side. And that process stores additional energy. I haven't done the calculation, but I assume that this increase in energy will more than compensate for the decrease of energy in the capacitor itself (in exactly the right proportion to allow for the work of pulling the plates apart).

Thanks. This is starting to make sense. So I guess if you connect the two plates with a conductor after they were pulled apart, the increased energy would take the form of higher voltage?

That's right: the voltage in this case will be proportional to the separation between the plates, so if you discharged them across a light bulb it would (briefly!) glow much brighter if you pulled the plates apart first.

(Also what's the best way to learn the formulas so I can figure out these kind of questions for myself?) Buying an advanced Physics textbook seems too daunting.

As someone who regularly assigns the question you asked as homework (in an advanced physics course, mind you), let me put in a word of advice: don't think of what you're aiming for as "learn[ing] the formulas". In my experience, that mindset (which fits so very well in most high school science classes) is the single most common stumbling block for college-level physics students.

Equations and formulas exist purely in service to concepts. If you can't tell a story about "what's really going on" (qualitatively) without reference to the equations, then (in most cases) you probably shouldn't try to use the equations, either. I see student after student try to solve complicated problems via "equation hunting", where they just dig through their notes or the textbook looking for formulas that have the right variables in them, and then look for ways of combining them to find an answer. (Sometimes their thinking is a step more sophisticated than that, but it's a characteristic pattern.) Students start to become experts in physics once their mental model of the subject transforms from a jumbled pile of independent equations into a network of concepts with equations like little neurons binding them together.

Yes, that's an excellent point. So I guess the follow up question is how to learn the qualitative ideas behind electric fields?

I've got a lot of inventions I've thought of related to static charges, and I'm trying to basically figure out why they wouldn't work. Maybe I can email a couple over to you and you can point me to the right concepts? (email in my profile)

Alas, the trouble is that it's tremendously hard to learn (or teach!) the qualitative ideas without also learning the equations. (The only counterexample to that I've ever seen in physics is Feynman's amazing little book "QED", which isn't at all what you need here.) So I've got nothin' for ya. (And while I do enjoy conversations about this sort of thing, obviously, I can't possibly sustain one with my job these days: I'm already going to regret the time I've spent on this one. But it was fun.)

There's no substitute for actually working through the equations and math, but working with simulations can help drive concepts home. For example, PhET has a virtual capacitor lab (it's a Java applet).

http://phet.colorado.edu/en/simulation/capacitor-lab

If you have multivariable calculus under your belt, then don't be too afraid of jumping into something like Griffiths E&M.

The voltage wouldn't change, but the energy would decrease. The energy stored in a capacitor is a potential energy, which means that the energy depends on the orientation of the capacitor's plates. The decrease in energy is equal to the work required to pull the plates apart.

Intuitively it seems like the energy would increase because you're putting in work to pull them apart, no?

And you'd still have opposite charges on the two plates so there would always be some energy from that?