Charge removal of a capacitor

Hi,

I just came to know about basics about capacitors. I Understand that the displacement of charges cause the capacitor to store the electrical charge as electrostatic charge.

So what happens if i charged the capacitor and leave it without giving discharge path.

I mean how long it can have it's electrostatic energy?

and is there any equation governing this phenomena?

Thanks

In an ideal case without ANY discharge paths(i.e. no leakage, no nothing), the charge can be held forever.

But then, of course there will be some discharge paths...!

Normally its the insulation between the plates that leak a very small amount of current. Even when this is very good, the air its self conducts electricity especially when its moist. A good high voltage capacitor might hold a lethal charge for several days.
Frank

and is it different form corono discharge.. or similar to corono discharge...

please explain.

Corona discharge will exist only at very high voltages (» max. ratings for normal capacitors), especially at pointed corners or cusps, where high electric fields exist which are able to ionize the surrounding fluid or gas (air).

In commercially available capacitors (not valid for Leyden jars or bottles ;-) ) the a.m. leakage currents through the dielectric layer prevent the formation of sufficiently high voltage necessary for corona discharge.

Thanks for the replies. So all capacitors work in this principle. But how can a capacitor do coupling and decoupling?

In IC point of view, we are talking about coupling capacitors and decoupling capacitors.

What differentiates these two phenomena?

Thanks

As you will know, a capacitor (essentially) keeps (or blocks) its DC voltage, whereas it lets pass - more or less, i.e. frequency dependent - ac voltages.

Coupling: exactly the above! The intention is to let ac frequencies pass, blocking DC.
Decoupling: same as above! The intention is to short ac frequencies to GND, again blocking DC.

erikl said:

As you will know, a capacitor (essentially) keeps (or blocks) its DC voltage, whereas it lets pass - more or less, i.e. frequency dependent - ac voltages.

Coupling: exactly the above! The intention is to let ac frequencies pass, blocking DC.
Decoupling: same as above! The intention is to short ac frequencies to GND, again blocking DC.

Click to expand...

Theoretically the charge on the capacitor is retained forever.
In practice: There is a leakage path between the plates of the capacitor. The first and visible path is the path between the terminals. The air is a good conductor especially the moisture in the air.:razz:
Assume that the terminals are totally sealed, then the path(s) between the plates are involved.
In this second case the type of the capacitor is important.
The ceramic capacitors discharge easily. Then follows electrolytic capacitors. Then capacitors with polypropilen based (For instance reactive power compensation capacitors RPCC)
Once I saw a RPCC which retained the charge more than six years.

(By the way, I am manufacturing the RPCC in my factory)

To assist a grasp of this subject, I have a Youtube video which is an animated simulation of capacitor behavior.
With various AC and DC waveforms.
Visibly portrays capacitors charging and discharging. Electrons (or rather current bundles) flow through wires.

www.youtube.com/watch?v=eIWEU4pObJw

Thanks for the answers. Now We are talking about various capacitance effects
for example between transmission lines , between different metals in an IC.

So how exactly these capacitances are related to the parallel plate capacitance?

because in different metals we have different currents.. and different characterisitics .. but still we are speaking about capacitances. So how far it is true to tell these have capacitances.?

Thanks

In transmission lines we mention about the capacitances between the phases L1, L2, L3 and the capacitances between the phases and the ground. The value of these capacitances is again given by the approximate formula C=e x A/d Here we can assume A/d constant all the way thru the line but e changes due to the weather condition. There is no direct way of measuring the capacitances of the lines. But depending on the previous experiences the engineering department of the power company assumes a near correct value. Also by measuring the reactive capacitive power again a near correct value can be found. Once an estimation is done, the air core inductors are introduced (in parallel) to the system to compansate the capacitive loading.

The capacitance formula C=e x A/d is also correct for semiconductors. Here e is strictly defined. It is semiconductor. The area A and the distance between the "plates" are also known (at least for a certain part of the circuit) What is important is here is the frequency of the signal. The current thru the capacitor increases as the frequency increases. So there will be some unwanted coupling between the "let's say" gates.

This problem was encountered by Intel. I think they've solved it by introducing a vacuum layer between the conductors.
( Since the dielectric constant of a material, e=kxe0, where k a number greater than 1 and e0 is the dielectric constant of the vacuum, INTEL tried to make the dielectric constant of his IC's as small as possible. )