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This might look very fundamental but I am a bit stuck in this. How does the current pass(AC)between the plates when there is an insulator or dielectric between the plates.

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How does the current pass (AC) between the plates when there is an insulator or dielectric between the plates.

It does and it doesn't. This illustration may help a little.

enter image description here

Figure 1. A water pressure tank of the type used to even out water pressure supply. Image source: Pressure tank comparison on YouTube.

They're all doing the same job but I'll use the diaphragm pressure tank for illustration purposes. In this type of pressure tank a diaphragm separates the air from the water. This is analogous to the insulator in your capacitor. We'll say that the air represents the positive plat and the water represents the negative plate. When the water pressure rises water is pushed into the tank and when it falls the water is pushed out of the tank.

This is very similar to the smoothing capacitor in a DC power supply.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 2. The capacitor charges up when the voltage is high and discharges (into the load) when the rectifier voltage is low.

Now, how does the water go in and out of the pressure tank when there is no way through the diaphragm? It's because the pressure changes and energy is stored in the tank. In the same way charge can flow in and out of the capacitor because the voltage changes and energy is stored in the electric field of the capacitor.

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  • \$\begingroup\$ Thanks for the reply!!! \$\endgroup\$
    – sinecosine
    Apr 7, 2019 at 16:41
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It is more philosophical question. One way of thinking this is that current does not pass between plates, but current still needs to go into and out of capacitor terminals to charge it. It's called displacement current. But for all practical purposes, as current does go in and out of capacitor terminals, it may be just be modeled simply by saying that AC current passes through capacitor.

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Quoting "QED" by Richard Feynman: "Maxwell's theory of electricity and magnetism had to be changed to be in accord with the new principles of quantum mechanics that had been developed." Anyone who wants to understand the concept of "displacement current" needs to read Feynman's easy to understand book, "QED". Again quoting:"So now I present to you the three basic actions, from which all the phenomena of light (including RF) and electrons arise. _Action#1: A photon goes from place to place. _Action#2: An electron goes from place to place. _Action#3: An electron emits or absorbs a photon."

Here is how RF energy gets from one plate of a capacitor to the other plate: As the RF energy source rises in potential, it supplies a photon to electron#1 on one capacitor plate thus raising that electron's energy level higher than electron#2 on the other capacitor plate. As the RF energy source falls in potential, electron#1 gives up its photon which transfers its energy directly through the dielectric of the capacitor and is absorbed by electron#2 on the other capacitor plate. Since electrons are basically identical, the effect is the same as if the two electrons had traded places on the capacitor plates. When electron#2 absorbs the photon emitted by electron#1, it becomes identical to electron#1 before electron#1 gave up its photon. Likewise, electron#1 becomes identical to electron#2 before electron#2 absorbed the photon.

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  • \$\begingroup\$ thanks for your time for explaining this...so is it Quantum entanglement happening between two electrons? \$\endgroup\$
    – sinecosine
    Apr 7, 2019 at 16:37
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    \$\begingroup\$ No, nothing to do with entanglement. That simple transfer of photon energy from one electron to another is what allows the very common speed-of-light transfer of energy using the slow-moving free electrons in conductors. \$\endgroup\$ Apr 7, 2019 at 16:54
  • \$\begingroup\$ @ Cecil - W5DXP Yet we can very slowly change the charge on one plate of a capacitor. Does RF apply? Also seems that black paper would absorb the photons, and the displacement current would be ZERO. What am I missing here? \$\endgroup\$ Apr 8, 2019 at 7:17
  • \$\begingroup\$ Low frequency changes just means photons of a very low frequency. RF waves, consisting of photons, go through black paper like it wasn't there. RF is a much lower frequency than visible light. OTOH, gamma rays are a much higher frequency than light and go through almost anything. Fields and waves (photonic energy) have different characteristics depending upon frequency. [img]leadertechinc.com/blog/wp-content/uploads/2016/08/… \$\endgroup\$ Apr 8, 2019 at 12:59
  • \$\begingroup\$ The electromagnetic spectrum extends from extremely long wavelengths to extremely short wavelengths. Different wavelengths have different characteristics. Black paper is transparent to RF wavelengths. Please Google "electromagnetic spectrum". I tried to post a URL but it didn't work. \$\endgroup\$ Apr 8, 2019 at 13:17
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Until a capacitor is fully charged up, there is current flowing in one plate and out the other plate. That is the current we describe as "flowing".

More exactly:

(1) PLUS charges need to be cancelled by MINUS charges.

(2) more electrons on one plate need to be balanced by less electrons on the other plate.

(3) whenever electrons are added to one plate the same # of electrons have to leave the other plate.

Capacitors across sinusoids (ac power, audio, RF), such as 4*sin(60 * 2pi * Time) have the current being the derivative of the voltage, as 4* (60*2pi)*cos(60*2piT), where the (60*2pi) is 377. Note the 90 degree phaseshift, with voltage being SIN and the current being COSINE

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  • \$\begingroup\$ Thanks for the reply!!! \$\endgroup\$
    – sinecosine
    Apr 7, 2019 at 16:41

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