1
\$\begingroup\$

When we talk about electrolytic capacitors or the supercapacitor(EDLC) then we know that, we can store large amount of charges in it. Since in electrolytic capacitors Aluminum plate acts as cathode and electrolytes acts as anode and due to very less width of dielectric (Aluminum oxide) it's capacitance increase and this results in more charge storing capacities (for safe value let say 1C charge as, such electrolytic capacitors are easily available) .....and they work at low voltage of around 2v to 50v.....but if the charges are stored on one of the aluminum plate (which acts as cathode of capacitor) then any conductor plate having such high amount of charge on it will have potential in millions of volts but these capacitor have very low relative voltage. So how is this possible? In the second picture (screen shot of results from wikipedia), why it is happening that capacitor stores more charges at same potential. What is the physics behind this phenomenon and how to explain in terms of calculations. [![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/IMvcV.gif![enter image description here](https://i.stack.imgur.com/MWhVz.jpg)

\$\endgroup\$
2
  • 1
    \$\begingroup\$ because they were charged with a MV power supply \$\endgroup\$
    – jsotola
    Jan 29 at 3:17
  • \$\begingroup\$ Comments are not for extended discussion. The conversation between Aniket Kumar and tlfong01 has been moved to chat. Please keep the discussion there. If a clear answer can be produced from the discussion then please post it as an answer. \$\endgroup\$
    – SamGibson
    Feb 6 at 14:10

1 Answer 1

1
\$\begingroup\$

The capacitance is very large (by design, this is what we typically want when significant energy storage is the goal) so the voltage is smaller for a given charge.

V=q/C so for very large C (compared to, say, a similar size parallel-plate capacitor with air dielectric) the voltage is relatively low.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ Since there will be more charge accumulation on the aluminum conductor so by this the potential of conductor must be very very high. But what we find that the potential of capacitors is very low. How to explain this? \$\endgroup\$ Jan 29 at 4:28
  • 1
    \$\begingroup\$ Where I got stuck is, let's say the shape of capacitor is sphere(just for the ease of calculations as we know here V=kq/R). And we know supercapacitors store lot of charge so if the charge is high then it's potential should be abruptly higher in terms of millions of volts, and as the opposite plates are charges with opposite charges in capacitor then net potential difference will be sum of mod of potential as Delta V=kq/R-(-kq/R).....this is what calculations says right....... But what we see in real is that potential difference between the capacitor is very low. So why is this happening \$\endgroup\$ Jan 29 at 5:25
  • 3
    \$\begingroup\$ If you take an example of an electrolytic capacitor- the internal electric field is actually very high, maybe 20MV/m. It only adds up to a few volts or tens of volts because the dielectric layer is only tens of nanometers thick. The dielectric constant of aluminum oxide is fairly high compared to air. And the effective area of the plates is huge because of etching. \$\endgroup\$ Jan 29 at 9:35
  • \$\begingroup\$ can you please explain your point in terms of mathematical modeling of what potential and charge and electric field \$\endgroup\$ Feb 5 at 4:16
  • 1
    \$\begingroup\$ All these can be simplified to the parallel-plate capacitor. In the case of a double layer capacitor there are effectively two in series, so consider just half. At 1V and a d of 0.3nm the field is 3 MV/m. If er = 8, the charge per unit area = eerV/d = 0.24 C/m^2. Activated carbon has a huge effective surface area so the coulombs stored at low voltage can be quite large in a small package. It seems counter-intuitive because both the effective area is huge for such a small package and the effective dielectric thickness is extremely thin. \$\endgroup\$ Feb 5 at 4:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.