Geezer said: "In running round ANY closed loop in a static electric field the field force always integrates to zero". Hmm... not so sure about that in this case. Even if we do the math without any approximation, there should be some difference in this case. The fact that approximation is possible means that field is close to zero function (compared to field between the plates) outside the plates. No matter how strong the field is outside the plates it is significantly stronger between the plates (or so I was told

). Having all this in mind, I really cant integrate field force to zero around this loop.

But, even if this is totally bad setup, we're missing the point.

To find a way to ask a question, I'll leave just charged and disconnected capacitor.

The diagram:

[diagram=635_0]

Let's say that this system is zero friction (you can add friction later)

If a charged particle with initial velocity v

_{0} and kinetic energy E

_{k1}=m*v

_{0}^{2}/2(going just horizontally) and charge -Q

_{0} enters our capacitor like illustrated in diagram, the force of the capacitor electric field should add vertical component to that velocity Δv during its motion between the plates. So, when our charged particle exits capacitor, it's kinetic energy should be E

_{k2}=E

_{k1}+m*Δv

^{2}/2.

Since capacitor wasn't discharged by this (can't see where would charges +Q and -Q really go), it's energy should stay the same.

To summarize all this:

We had capacitor with it's energy E

_{c} which didn't change, we had a particle with its energy that did change (at cost of what?).

I hope that this really explains my confusion about energy of a field.

PS We have similar experiment repeating itself for millions of years, only this time, it's not an electric, but gravity field.

Moon is orbiting the Earth and it's gravity is crating ocean tides (a loooooooot of energy), and is probably the reason of Earth's core being molten and hot(even more energy), and yet, Moon's gravitational field is powerful enough to keep astronauts from flying out of it's orbit.