# Electron distribution on capacitors

1. Aug 8, 2007

### gareth

Say we have a paralell plate capacitor which is charged using a DC source.

In my understanding there is a build up of excess positve and negative charge on opposing plates of the capacitor (depending on which plates are connected to the positive and negative terminals of the DC source).

My question is as follows; do we know where this excess charge actually is?

Does the charge distribute istself evenly over the surface of the metal which is in contact with the dielectric? and if so, what kind of distribution are we talking about?

Is it a sudden step? i.e. all the atoms on the surface acquire 1 extra electron and when the potential difference of the battery is equals the charge of these electrons it stops? so the higher the DC voltage the "thicker" the layer of excess electrons is on the plate?

Thanks for help, links are welcome aswell!

2. Aug 8, 2007

### Staff: Mentor

Atoms in matter tend to be neutral, i.e. the electrons (negative charge) balance the positive charge in the nucleus. The mobile charges in nature are electrons, which by convention are assigned a negative charge. The positive charge is due to a lack of electrons, and so the local electric field is due to the positive charge of the nuclei.

In charging a capacitor, electrons move from one plate to other, so one plate has less electrons and the other has more. One plate has charge +Q, while the other plate has charge -Q.

The electrons are in the conduction band and reside at the surface of a conductor. Generally the charge distribution is uniform over the surface since the charges (electrons) repel each other.

The movement of the electrons is relatively quick, almost like a step. The charge transfer rate slows as the charge builds up, and the rate, as a function of time, depends on the capacitance and internal resistance of the battery.

The voltage on the capacitor builds with charge and opposes the potential of the battery.

3. Aug 9, 2007

### gareth

I suspected that the excess charges reside on the surface of the capacitor plates.

I know you can estimate the amount of charge on each of the capacitor plates (Q=CV); but can you estimate the excess charge per unit volume on the surface of the plates?

Put another way, how "thick" is the surface layer.

We know the surface area of the plates, but to estimate the extra charge per volume we need to know the depth of the volume we are considering.

I'm assuming this depends on the DC voltage.

I have a feeling this may be a complex question, please do not hesistate in qiving a complex answer.

Many thanks
Gareth

4. Aug 22, 2007

### chuckschuldiner

In classical electrostatics, the charge resides only at the metal-dielectric interface as a delta-function. There is no charge distribution inside the metal for if this were the case , electric fields would exist inside the metal.

However, in real metal electrodes the free charges do form a layer of finite thickness and electric fields can penetrate inside the metal. This thickness is related to the Thomas-Fermi screening length of the metal. This thickness is usually around a few angstroms and is negligible for a capacitor of macroscopic dimensions. I am not sure but i think that the finite thickness of the charge distribution stems from the wavelike nature of the electrons (charge carriers).

If one employs a Drude model to model the metal electrode then the charge distribution will be of the form Q exp(x/l) where, Q is the charge induced, x is the spatial coordinate and l is the screening length. Typical screening lengths for most metals are of the order of 1 Angstrom. For more details, search for screening length in capacitors on google scholar.

5. Aug 22, 2007

### gareth

Thank you.

That is a HUGE help chuckschuldiner, just what I was looking for!

Now to pester you some more;

If I wanted to create a surface layer of charge on a metal, what is the best way to go about it?

Here are the options I am considering.......

1) I know if I electrostatically chrage a piece of metal which is not in contact with another conducting material, the charges will redistribute themselve in such a way as to have a zero net electric field inside the metal and hence no current will flow. These excess charges reside on the surface.

2) If I use a kind of capacitor arrangement with a vacuum as the dielectric, the charges wil also reside on the surface of the metal, the amount of charge proportional to Q=C*V where C=A*e/d (or something like that).

3) Biasing the metal. If I attach one end of DC source to ground and the other (say the positive) to the metal, will this cause an excess of electrons on the metal surface?

I know options 1 and 2 will work, but I'm not sure which will be most effective.

I'm not sure about option 3, since this could be considered as a capacitor with ground as one of it's plates and hence there would be no excess charge.

What option would you suggest for inducing the MOST charge on the surface of the metal?

Many Thanks
Gareth

6. Aug 22, 2007

### chuckschuldiner

To induce the most charge on the plates of a capacitor, you should have a high-permittivity material as the dielectric . Since, for a given applied voltage, Q is directly proportional to capacitance, and capacitance is in turn proportional to the dielectric constant, capacitors which use high-dielectric constant materials are ideal to realise high surface charge densities. Ferroelectrics like Barium Strontium Titanate are good examples of high-permittivity materials.

7. Aug 23, 2007

### gareth

Thanks again chuckschuldiner.

So to achieve high surface charge densities we need a dielectric with a high dielectric constant.

Say if we had a a capcitor with a low vacuum as the dielectric, and we drilled a hole in ONE of the plates (the one with excess positive charge), how would this effect the charge distribution on the OTHER plate?

Would there be a depletion of negative charge on the opposite area on the negative plate?

How would I incorporate A (the surface area) in this case? Is A the combined surface area? or do I have to subtract the area of the hole from A?

Gareth

8. Aug 27, 2007

### Gokul43201

Staff Emeritus
Subtracting the hole area will work only in the limit where the hole diameter is very large compared to the separation between plates. Outside that limiting regime, you must actually model how the field lines will run between the plates (i.e., solve the Poisson equation with given boundary conditions).

9. Aug 28, 2007

### gareth

thanks for the help Gokul43201!

This brings me to another question.

The excess charge on a plate of a capacitor is brought about by the electrical potential (say a DC source). But what if one of the plates is connected to the postive terminal and the other terminal (negative) is connected to ground. Do you still get an excess charge build up on one of plates? or does there need to be an electric field between the two plates for this to happen?

Thanks again!

Gareth

10. Aug 28, 2007

### Gokul43201

Staff Emeritus
Where ground is established in the circuit does not matter - the charge distribution will remain the same. So long as the other plate is connected to the negative terminal (or ground), you'll have the same distribution. If the circuit is left open, with the second plate completely disconnected, you will not really have any charge build-up on either of the plates.