Assembling and Dis-assembling a Charged Capacitor
Then what will happen to the electrical field (energy) which was stored in the capacitor before un-rolling its electrodes?


Unrolling and separating the plates of the capacitor will require work, because the electric field between them creates an attractive force. Therefore, you will actually increase the energy stored in the field. The capacitance goes down as you increase the separation, but the voltage between the plates goes way up.

The relevant equations are

$$C = \epsilon \frac{A}{d}$$

$$V = \frac{Q}{C}$$

$$E = \frac{1}{2}C V^2$$

If area \$A\$ and permittivity \$\epsilon\$ are held constant, then capacitance \$C\$ is inversely proportional to the distance between the plates \$d\$.

If the charge \$Q\$ is also constant, then the voltage \$V\$ is inversely proportional to \$C\$, which makes it directly proportional to \$d\$.

Finally, the total energy \$E\$ turns out to be proportional to \$d\$ because the rising \$V^2\$ term overrides the falling \$C\$ term. This increase in energy is the physical work you have to put into increasing the separation.


  If I now bring back together the two electrode-papers again and roll-them-up into its original shape will I again get back the original energy which was there in it when the capacitor was charged originally?


Yes. The energy will drop to its original level, assuming that you haven't allowed any of the charge to leak away in the meantime.



Note that I'm glossing over a lot of physical details associated with the construction of electrolytic capacitors specifically. The paper separator is NOT the dielectric in this case &mdash; the dielectric is actually a thin layer of aluminum oxide on one of the foil plates.

Therefore, if you keep the foils immersed in a sufficient quantity of electrolyte as you separate them, there will be no change in capacitance or energy.

However, if you try to separate them in air, you will in effect be creating two capacitors in series, one with the original oxide dielectric, and another one with air as the dielectric. The latter will have very low capacitance, and the voltage will appear across it.