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Imagine a capacitor with capacitance $C$, plate distance $d$ and voltage $U$. Now we decrease the distance between the plates with connected power-source ($U = const.$). For the energy $W$ of the field inside the capacitor we get: $$W(d) = \frac{1}{2} C U^2 = \frac{\epsilon_0AU^2}{2d} \sim \frac{1}{d}.$$ This means the field energy increases, when the plates get closer to each other. Thus we're performing work when decreasing the distance between the plates, which implies there is a force $$F = -\frac{\mathrm{d}W}{\mathrm{d}d} = \frac{\epsilon_0AU^2}{2d^2}$$ seperating the plates. Now i wonder, what that force could be. The electrical force would rather cause an attraction between the plates, since they're oppositely charged.

I also considered the magnetic field that builds up between the plates of the capacitor due to the increasing electrical field, but i don't see why it should cause a seperating force.

I'd be really thankful if someone could either find a mistake in my thought process or explain where that force comes from.

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It seems to me that you are forgetting that when you move the plates of your capacitor closer, the capacitance C changes. This means that more charge will have to flow into the capacitor (to maintain constant voltage) and thus the power source is doing work.

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  • $\begingroup$ So that means the work that is done to increase the energy field is performed solely by the power source and the electric force still attracts the two plates? That makes a lot of sense, thank you! $\endgroup$ Commented Jul 19, 2017 at 16:55

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