Hi! I am not sure if what I have once learned about capacitance is right, can you give it a look : So two capacitors say one 1nF, the other 10microF and they are both charged to the same voltage. However, the capacitor with the more capacitance takes more time to charge than the smaller one. So what happens is that they both take charge, however the one with the more capacitance have larger plates(I know that this is not the only way to increase the capacitance, but it's only for the sake of simplicity) and when charge arrives at it's plate, it gets dispersed. The more the charges and the closer to one another the stronger the overall electric field would be, and so if they are further away from each other the electric field would be weaker. So the bigger the plate the more dispersed the charges would be and it would take more charge to build up the same voltage as the capacitor with lower capacity.
Is that right? Please just use "Yes" or "no".
My second question is if I have a capacitor with say 3volts, in theory I can get a very large instantaneous current out of it, is that true? Because theoretically if you have a battery and it has no internal resistance nor it can't be physically destroyed, you could get 100 amps out of a 1.5v battery, right? It's just that in real world 100 amps could burn down a house, what's left for a tiny power cell..

 

Well I don't know what you are talking about in your first question, but the answer is yes to question two in theory.
SG

 

Hi! I am not sure if what I have once learned about capacitance is right, can you give it a look : So two capacitors say one 1nF, the other 10microF and they are both charged to the same voltage. However, the capacitor with the more capacitance takes more time to charge than the smaller one. So what happens is that they both take charge, however the one with the more capacitance have larger plates(I know that this is not the only way to increase the capacitance, but it's only for the sake of simplicity) and when charge arrives at it's plate, it gets dispersed. The more the charges and the closer to one another the stronger the overall electric field would be, and so if they are further away from each other the electric field would be weaker. So the bigger the plate the more dispersed the charges would be and it would take more charge to build up the same voltage as the capacitor with lower capacity.
Is that right? Please just use "Yes" or "no".

No.

If you would like something other than a simple Yes or No, read on.

You almost got it. It takes more charge to build up the same voltage on the capacitor with the LARGER capacity.

As you say, there are more ways to achieve higher capacitance than just spreading out the charge over a larger area, but within the simple model you are working with, that is fine.

My second question is if I have a capacitor with say 3volts, in theory I can get a very large instantaneous current out of it, is that true?

Yes.

In theory you could get an infinite impulse of current lasting an infinitesimally short amount of time. In practice this never happens, but you can get very large short-duration currents out of capacitors. It's one of the things they are useful for.

Because theoretically if you have a battery and it has no internal resistance nor it can't be physically destroyed, you could get 100 amps out of a 1.5v battery, right? It's just that in real world 100 amps could burn down a house, what's left for a tiny power cell..

Yes.

And in practice you can get batteries with low voltage that can deliver huge currents. When I was working for NIST our 1000 A supply used a 2 V submarine battery. It could deliver 1000 A continuously for some time.

When you start your car you are typically pulling a couple hundred amps out of that 12 V battery, depending on the size of the engine and other things. A large, high-compression engine on a cold day can pull well over a thousand amps.

 

NO!
Capacitor current is determined by (a) size of wires and soldering inside the bank (b) the power loss factor inside the insulator between plates tg(fi) what is an order 0,05 for bad materials like PCB and 0,0001 for most brilliant materials. The thermal loss then is Q(therm)=tan(fi)*N(circul in cap). And (c) the Foucault effect what is whirl produced losses in the cap metallized layer.
Normally, under the 5 Amps the main loss are (b), however in very high freq like GHz probably (c) may have an over, at least some cases. The (a) mostly aren`t a problem.
For example: DIY cap for 100 MHz 1kW oscillator (Clapp circuit). It must be about 100 pF and 3 kV. What gives an power consumption from 24V PS about 1000W/24V=20A. Those 20A will cause the voltage multiplication in the resonance tank Q=3000/24=120x. Thus, the tank circulating energy will be 1 kW * 120=120 kVAR. If that capacitor is made from ordinary glass-textollite (FR-4) material, then heat loss will be 120 000*0,05=6kW (!!!) - more than device "eats". Or other words, it will not functioning at all, and will make a short-circuit.
The same circuit using a high quality Vishay capacitor (tan(fi)=0,0001; price 3800-5600 USD per piece !!!!) will dissipate the 0,0001*120 000=12W or thermal energy what is just nothing.
About Foucault - if 100 MHz takes a swirl deapth about 6 micrometers, but pcb material has 35 micron thick copper, the effect will be notable but not detrimental. Just fresh therm0-camera measurements shown only pair centigrades difference because this effect. So, it is much smaller than (b).