Why conductor having charge(let say 1C) have tremendous potential but same conductor used in capacitor have low voltage?
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]]
: https://i.stack.imgur.com/IMvcV.gif![enter image description here](https://i.stack.imgur.com/MWhVz.jpg)capacitorcapacitanceelectrolytic-capacitorsupercapacitorShareCiteEditFollowFlagedited 20 hours agoasked yesterdayAniket Kumar4366 bronze badges
- 1because they were charged with a MV power supply – jsotola yesterday
- (1) Well, I would suggest to consider the following analog of two water tanks: one with big (cross section, floor) area but not tall, The other with very small area, but very very very tall (high). (2) Now the capacitance of a capacitor is like the volume of a water tank, the bigger the volume, the more water (electrons, charges) you can hold. (3) The pressure (at the bottom) of the tank is proportional to the height of the tank, the higher the water level at the tank, the higher pressure (at the bottom) of the tank. (4) You might like to ask me further questions before I move on. – tlfong01 2 mins ago Edit
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.ShareCiteEditFollowFlaganswered yesterdaySpehro Pefhany316k1313 gold badges275275 silver badges679679 bronze badges
- 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? – Aniket Kumar yesterday
- 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 – Aniket Kumar yesterday
- 1If 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. – Spehro Pefhany 20 hours ago