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 capac…

capacitor capacitance electrolytic-capacitor supercapacitor

# Why conductor having charge(let say 1C) have tremendous potential but same conductor used in capacitor have low voltage?

Asked 8 days ago

Active 8 days ago

Viewed 81 times

1

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)

capacitorcapacitanceelectrolytic-capacitorsupercapacitor

ShareCite

EditFollowFlag

asked Jan 29 at 3:03

**71**77 bronze badges

- 1because they were charged with a MV power supply – jsotola Jan 29 at 3:17
- 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. – SamGibson**♦**11 hours ago

## 1 Answer

1

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.

ShareCite

EditFollowFlag

answered Jan 29 at 4:14

**316k**1313 gold badges275275 silver badges681681 bronze badges

- 1Since 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 Jan 29 at 4:28
- 1Where 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 Jan 29 at 5:25
- 3If 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 Jan 29 at 9:35
- can you please explain your point in terms of mathematical modeling of what potential and charge and electric field – Aniket Kumar yesterday
- 1All 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 = e
*er*V/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. – Spehro Pefhany yesterday - Michael Faraday himself doubted we would ever be able to achieve 1F capacitance. – Spehro Pefhany 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.

How height of water tank from the given analog is related to capacitor or conductor?

@tlfong01 , can you please clear your point in the light of my question

My apologies for missing your question, and therefore not replying. Let me refresh my memory and give you an answer perhaps over this weekend.

(4) I read your question a second time and found it a bit confusing. You seem to ask the voltage and charge of two kinds of capacitors, one big capacitance but low voltage, another small capacitance but very high charge. And you wish to know why. (5) There are a couple of ways to explain. Let me try to use the ** water tank** analog, and add one more factor to consider –

**. I am just brainstorming, so it is not at all organized, and may have many typos.**

*energy*(6) I also wish to discuss the science not technology. ie I won’t consider why aluminium or other substance is used, what I wish to consider is if (a) they can store more or less charges and how high voltage they can stand. As said earlier, I will also consider energy. (7) I will first consider the water tank, their parameters (volume vs capacitance, pressure vs voltage, and mechanical energy vs electrical energy.

(8) I will also consider the inductor, which is the mirror of capacitor. In short, two paradims Capacitance and Inductance at the same time, to give a complete picture, which helps understanding two sides more deeply. (9) We need to assume some prerequisiste or background to ensure we will be using the same level of knowledge and most importantly, concepts and terms. (10) Let me start with the capacitor, mirroring to inductor, then move on to the water tank, explain the concepts, then come back to capacitor, and finally to inductor, for the deeper and more solid understanding of the concepts.

(11) Let me first ask you a couple of questions, to make sure we are at the same level, using the same jargons. The first questions (12) I know the energy stored in a capacitor is 1/2 * C * V#2. Can you, or have you followed the steps in arriving this formula? (13) Do you know for the inductor, the corresponding energy is 1/2 * L * I#2. Note: # symbol stands for two stars sticking together.

(14) Now moving to the water tank, do you know what the is energy stored in the tank, in terms of classical mechanics, ie Newton’s Laws, with terms mass, velocity, acceleration, distance (height) etc. You don’t need to give any explanation, perhaps just say something like water tank energy E = 1/N * X * Y**2. and what are N, X, Y?

Ah, locking down bed time. See you tomorrow. Cheers.

The last message was posted 12 hours ago.

Categories: Uncategorized