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.
* 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.)