# Understanding the capacitor deeply

1. Jun 14, 2008

### finenil

hello every body,
I am trying to understand the capacitor behavior as deeply as i can ...I posted this thread in Classical Physics section but I later think it could be better if I post it here in EE.
consider a simple capacitor with two plates A and B

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1- why the charge on both plates are equal in magnitude ?
when we apply a voltage to a capacitor,charge will accumulate on side A.this charge will build an E field which will pass thru the insulator and exert force on the plate B.So charges on plate B move and accumulate together.Ok but how do we know that this E field caused by Charges on plate A will exactly attract the same amount of charge that was on plate A on plate B??
why these two has to be equal ? Conservation of Charge Law ?

2 - why is that the higher the frequency , the more is the current ? is the following intuition wrong ?
when we apply a ac voltage to plate A of a capacitor , charges density on plate A will change with voltage.
when charge density changes , E field caused by them in the Capacitor changes too.
This changing E filed will cause the plate B charge density changing .
Ok but I am thinking that when the E filed is changing very fast(high freq.) the plate B charges may not find enough time to respond to it and move to the new configuration.so the charge density(voltage) on plate B may lag the charge density(voltage) of plate A. Isnt this in contradiction to that at high freq. the capacitors impedance is lower ?

2. Jun 14, 2008

### Staff: Mentor

Yes, conservation of charge.

In a conductor, the electrons are the mobile charges, and when an electric potential is applied to a capacitor (starting with neutral conductors on both plates), the electrons leave one plate and travel to the other plate. The + charge is due to the proton charges in the absence of the electrons, and the - charge is due to the excess of electrons (which came from the other plate, which is now positive).

In part b, is one asking about reactance?

$$\frac{1}{\omega\,C}$$

3. Jun 14, 2008

### NoTime

2) A real capacitor has both inductive and resistive components as well as capacitance.
As frequency increases the inductive and resistive will become more noticeable.

4. Jun 14, 2008

### finenil

yes but first we are discussing the ideal case and second, this does not explain why the impedance drops when freq. arise.

electrons do not leave one plate do they!? I think that's the electron gun when electrons actually leave the plate.In capacitor their electrical field as I explained causes the other plate to became positive.

questions Still remained...

5. Jun 15, 2008

### Eidos

The current does not flow through the dielectric, it goes from one plate through the circuit to the other plate.

Its quite difficult to visualise electrons moving through a capacitor under ac conditions, especially when it comes to filters and what not. But for DC you're spot on, the charge accumulates at one plate and depletes at the other when we apply an external voltage to the capacitor.

In answer to your question, why the capacitors impedance is frequency dependent the way that it is, you need to have done the Fourier Transform.

The voltage current law for an ideal capacitor is:
$$i=C\frac{dv}{dt}$$
Where i is the current, v is the applied voltage and C is the capacitance of the capacitor. What this is telling us is that the current 'through' the capacitor is dependent on the rate of change of voltage across the capacitor.

Lets say the capacitor is charged to some voltage V1 and kept there. If we change the voltage to some new level, V2, very slowly, then the amount of charge leaving one plate and accumulating on the other, per unit time, is less than if we have to very quickly change the voltage. That should make intuitive sense. Since flow of charge per unit time is defined as the current, the above differential equation tells us in mathematics what we've just now written in words.

Now how does this relate this to frequency?

Its possible to represent a signal as a sum of cosine waves of different frequencies, amplitudes and phase. This is what Fourier transforms are all about. Now it turns out that a quickly changing signal, in this case the applied voltage across the capacitor, has high frequency cosine waves that make it up. Whereas a slow changing signal does not have those high frequencies present. Hence the dependence of impedance on frequency.

This is a hand wavy argument and I hope that Im not confusing you, I've left out some important stuff about the types of signals and so on. What is cool though, is that the signal need not be periodic, necessarily, to be made out of an infinite sum of cosine waves. So the fact that our voltage in the example changes and then stays at a value is no big deal.

Below we can see why the magnitude of the impedance is the way it is.
If we take the fourier transform of both sides we get.

$$I=C j\omega V$$.
Where $$j=\sqrt{-1}$$
We then rearrange so that we can get a 'resistance' like in Ohm's law V/I.

$$\frac{V}{I}=\frac{1}{j \omega C}$$
Now the modulus of the right hand side is, $$\frac{1}{\omega C}$$ which is what we were trying to find.

Last edited: Jun 15, 2008
6. Jun 15, 2008

### NoTime

In the Ideal case the +(hole) charge density will always equal the -(electron) charge density by definition.

For the Ideal case there is no free space transit of electrons, as in a CRT, by definition.
In the real world there is.

7. Jun 16, 2008

### Eidos

Even in a parallel plate capacitor?

8. Jun 16, 2008

### NoTime

In general insulators (or dielectrics) do not have infinite resistance.
Most of the time this is of little consequence since the capacitor leakage parameter is usually very small for most dielectrics.
Occasionally, you need to pay attention.

9. Jun 16, 2008

### edmondng

capacitance has inductance as well. its generally Ceff = C + L//C or some other combination. As frequency increases, impedance drops, and once it gets beyond a point where inductance is dominant over capacitance, impedance goes up.

10. Jun 16, 2008

### mheslep

That is: practical capacitors have inductance; not capacitance has inductance.

Last edited: Jun 16, 2008
11. Jun 16, 2008

### Integral

Staff Emeritus
Energy is transferred between the plates of a capacitor via the electric field. Excess charge introduced via a potential difference to one plate creates an electric field which redistributes the free charges on the other plate. In the DC case charge distributions are static. Once the plates are charged there is no further motion of the free charges. When the voltage on one plate changes additional charge is added or removed the charge distribution on the other plate must change to balance this new charge state.

So a changing voltage on one plate creates a changing charge this changing charge is transmitted via the E field to the other plate. This causes a movement of the charge on the other plate, this is equivalent to a change in voltage, thus a changing current passes through the capacitor.

12. Jun 17, 2008

### edmondng

yes, thanks for correcting me