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I was solving a question in an Electromagnetics book and need some help: two parallel plates were connected to a 100V battery. The battery was removed and then a slab of dielectric material was placed between the plates touching the upper one but leaving 0.25 mm of free space in the bottom. It is required to find electric field intensity and some other quantities after inserting the slab.

The idea of the question is know what remains constant before and after inserting the slab. It makes sense to assume that the charge would remain the same since it has no way to go after the battery is removed. However, after I solved the question with this assumption, I found that the total energy stored in the capacitor after inserting the slab is less than the total energy before inserting it. I thought the energy would also be conserved but apparently it is not. Can anybody give a physical explanation why the total energy of the system is less? Where did this energy go?

More details:
Plates area: 2x10^2 m2
Distance between the plates: 1.25 mm
Thickness of dielectric slab: 1 mm
Relative permittivity of the dielectric slab: 5


Solution: before inserting the slab:
Total capacitance = 1.42e-10 F
Voltage = 100 v
Energy (W) = .5cv^2 = 7.08e-7 Joules
surface charge density = electric flux density = 1.42e-8 c/m2

After inserting the slab:
in free space region: C=7.08e-10 F, E = 80,000 v/m, v=20V, W = 1.42e-7 Joule
in dielectric slab: C=8.85e-10 F, E=16,000 V/m, V=16 V, W=1.13e-7 Joule
Total W = 2.55e-7 Joules compared to 7.08e-7 before inserting the slab. Where did the 4.53e-7 Joule go?

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As the slab is brought to the plates, it will experience an attractive force that will tend to pull it into the gap. If you do not restrict it, the slab will accelerate.

In an ideal environment, the energy loss in the capacitor will be balanced by the increase in kinetic energy of the slab. Over time, this energy will be lost to friction or drag.

Once the slab is at rest inside the plates, work must be done to remove it against the electric field. This work will raise the energy stored in the capacitor.

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