Extra energy from capacitor problem

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When a parallel plate capacitor is charged and then disconnected from the battery, doubling the plate separation results in the electric field remaining constant while the voltage and energy stored in the capacitor both double. The extra energy comes from the work done to separate the plates, which requires energy input due to the attractive force between them. This work translates into increased energy stored in the electric field, not in the plates themselves. Consequently, the total energy stored in the capacitor increases when the plates are pulled apart. The discussion concludes that the work done on the plates effectively increases the energy stored in the electric field.
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a parallel plate capacitor is charged by a battery and then disconnected from the battery. the separation between the plates is doubled...

the E-field stays the same since charge is constant
the voltage doubles since V=Ed
the energy doubles since U=0.5QV

I was wondering, where does this extra energy come from?

does total energy increase?
 
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Hint: There will be a force of attraction between the two plates...
 
the force should stay the same since it's F=EQ

how does that play into the increase in energy?
 
It takes work to increase the distance between the plates. You have to do work on the plates to get them apart further.
 
so the work done on the plates translate to energy stored in the plates? so the total energy does increase.
 
indigojoker said:
so the work done on the plates translate to energy stored in the plates? so the total energy does increase.
Energy stored in the electric field increases, the energy isn't actually stored in the plates. And yes, the total energy stored increases.
 
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