# How dielectric transmits AC

1. Mar 25, 2012

### Saw

I had always thought that a capacitor with a dielectric transmits AC precisely because it polarizes in response to the applied field.

To tell you the truth, I realize now that I had two different views on how this process works:

One: the AC is oscillating on both sides of the dielectric and the dipoles of the latter are turning round in harmony with those oscillations.

The other: an oscillation of the AC happens in a conductor in one side of the dielectric; the dipoles of the dielectric oscillate confronting that oscillation, so that the + charge faces the electrons when they approach and the – charge faces them when they fly away and this oscillation of the dipoles sort of “emits” a new AC through the conductor of the other side (something akin to the electron transitions in an atom emitting an EM wave).

As to why the transmission is better (reactance is lower) for higher frequencies, I was thinking that it is so because the lower frequency did not have enough energy to make the dipoles turn round.

“Driven by an AC supply, a capacitor will only accumulate a limited amount of charge before the potential difference changes polarity and the charge dissipates. The higher the frequency, the less charge will accumulate and the smaller the opposition to the current.”

it seems that it is the opposite, that is to say, the AC goes through the dielectric in spite of the polarization, because it is not affected by the latter to the extent that the frequency is high enough…

What is closer to the truth? Can anyone provide some explanation helping me to visualize the process?

2. Mar 25, 2012

### sophiecentaur

It may help to point out that this quote leaves out an essential part of the argument. The rate at which a Capacitor can charge is limited by the Resistance of the source. It is the RC time constant that limits the amount by which a capacitor can charge and discharge during an AC cycle. For a Zero source resistance, there is no difference in how much a capacitor will charge at any frequency.

3. Mar 25, 2012

### Antiphon

The current through the capacitor is I = C dV/dt. This equation holds whether there is resistance in the circuit or not. Its the reason higher frequencies of applied voltage pass more current through the capacitor- the time rate of change of voltage defines the current flow.

4. Mar 26, 2012

### sophiecentaur

You need to go back to the OP and to decide what it means, exactly. What does the expression "transmits AC" mean?
Does it refer to Power?
Perhaps it would be better to put a Capacitor into a specific circuit and discuss power delivered to a particular value of resistive load rather than to use a vague term like "transmits". By including other circuit elements (for matching) I could eliminate the effect of the C entirely.
The conventional analysis gives reliable and understandable results.

5. Mar 26, 2012

### Saw

It does not refer to power.

What did I mean? I understand that the term may be unconventional, if used in this context. I will try to explain myself.

In a DC circuit, a capacitor blocks the current. But it does not block an AC. That is, put very simply, what I meant by "transmission" of the current: not blocking it.

Then the question is the detail about how this happens at microscopic level.

I start by assuming what I do understand from the books:

- The dipoles in the capacitor rotate, oscillate following the oscillation of the AC.

- If the frequency of oscillation is zero, i.e. if we have a DC, the current is blocked. To the extent that the frequency increases, the AC is less impeded and more "transmitted" (not blocked).

Given this, I see two possibilities:

(i) The capacitor does not block the AC because its dipoles oscillate. The problem with low frequency is that there is not enough energy to make the dipoles oscillate.
(ii) The capacitor does not block the AC in spite of the oscillation of its dipoles. The problem with low frequency is that it gives time to dipoles to oscillate.

I had thought (i) was right but wikipedia favors (ii). I presume the right approach is (ii) but can you confirm?

6. Mar 26, 2012

### sophiecentaur

Why do you bring in the concept of dipoles to explain the basic operation of a Capacitor? A Vacuum capacitor works just as well - with no dielectric. The capacitance may be less but the principle is still the same and there are no "dipoles" in a vacuum.

It can always be a problem when someone approaches a topic with their own private terminology. The word "transmit" has strong implications of Power or Energy transmission. If you are just referring to the current that passes through a Capacitor as it charges then why not discuss it in conventional terms? The conventional words deal with what happens with no trouble and everyone knows what is being talked about. Any 'introduced term' should be defined strictly so that there can be no confusion.

In a capacitor with no dielectric, an imbalance of charge accumulates on the surfaces of the two plates until the field across it balances the supply voltage (the rate at which this happens will depend upon the source resistance and will be instant if there is no source resistance). If a voltage source is applied, the PD establishes itself instantly. The action of the dipoles in a dielectric is to allow more charge to flow until the balance is achieved (the capacitance is increased). The details of how these dipoles behave is mainly of interest when the AC frequency is high enough for losses to occur, when a resistive element reveals itself in the component.

But the reason for the 'High Pass' behaviour of a Capacitor is just that, when the applied voltage alternates, (and there is a finite source resistance) current has not finished flowing before the polarity changes. The faster the direction changes, the closer the voltage on the output terminal of the capacitor tracks the supply signal. (The formulae are all there for all to read if they want to)

7. Mar 26, 2012

### Antiphon

First understand how the capacitor works without a dielectric. Then come back and put the dielectric in.

8. Mar 26, 2012

### Saw

Well, you are right, I got distracted by the dipoles issue. I will think it over in view of your comments.

9. Mar 26, 2012

### sophiecentaur

Go for the easy one first. ;-)