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If I have an electrolytic capacitor, which is a cylindrical body in which the two plates are rolled-up with a dielectric paper in between them, of very high value (say 1F) and I charge it using a DC battery source, disconnect the battery so the charge is stored in the capacitor, and then I un-roll it's paper-electrodes and take them apart from each other.

Then what will happen to the electrical field (energy) which was stored in the capacitor before un-rolling its electrodes?

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?

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  • \$\begingroup\$ I assume you disconnect the capacitor from the supply before unrolling it? \$\endgroup\$ – DKNguyen May 30 at 4:42
  • \$\begingroup\$ Yes. you are right. \$\endgroup\$ – alt-rose May 30 at 4:45
  • \$\begingroup\$ Yes. You get back the original energy (or at least a significant fraction of it). I know because my high school physics teacher did this in class. Except it was a simple glass jar capacitor with aluminum foil electrodes. At least, that is how I remember it. \$\endgroup\$ – mkeith May 30 at 4:59
  • \$\begingroup\$ I'm pretty sure you do. Easier way to think of it is just to have a simple capacitor of two glass plates and you move the plates apart. Q=CV. As you move the plates farther apart, the charge in the plates (Q) is the same and isn't going anywhere but the capacitance decreases which increases the voltage between the plates. Saying the voltage increases as you move the plates farther apart kind of feels wrong though for some reason so I need someone else to confirm this or refute this. It doesn't seem that right that the voltage between the plates increases ad infinitum with distance. \$\endgroup\$ – DKNguyen May 30 at 5:03
  • \$\begingroup\$ in an electrolytic capacitor the energy is stored in the surface coating of one of the electrodes. not that this changes the result. \$\endgroup\$ – Jasen May 30 at 6:57
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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 — 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.

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