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Does a.c. pass through a dielectric in a capacitor?

  1. Sep 24, 2007 #1
    Hi all

    My teacher says an a.c. passes through the dielectric of a capacitor. She says that the dielectric filters the a.c. passing through it.

    My book has no views on this. It just states that E lags behind the current I by a phase angle of 90 degrees. No comments on whether the a.c. is actually passing.

    On many sites, I read that the a.c. never passes through the dielectric; only the polarities of the plates are changed every half cycle. If a.c. does pass, it is called dielectric breakdown. However, the dielectric does not respond well with an a.c.; its resistance to the "flow of current" decreases with increase in the frequency of a.c.

    So what is the reality? Do electrons really pass through the dielectric? If yes, than please explain my teacher's explanation. If no, then what is I, the value of a.c. "flowing" in a capacitor at any time t.

    Thank you
    Mr V
  2. jcsd
  3. Sep 24, 2007 #2
    There may be some ambiguity in semantics here.

    It appears you understand the process.

    In typical well-designed circuits there is most certainly never any dielectric breakdown. (Oh, there might be some specialty applications where breakdown is intended for some purpose, but breakdown will almost certainly damage the dielectric).

    The ambiguity might be in the fact that "current" means "flow" which means "movement of charge". Charge is certainly moving when a dipole in a dielectric changes its orientation because of a change in an external field. There is no other movement than that, though.
  4. Sep 24, 2007 #3
    But just to make sure you do understand...

    A.C. certainly passes through the capacitor. Those moving dipoles really are allowing charge to move. If there is no breakdown, then the movement stops once the external field is counteracted by the field created by the dipoles alignment (same thing as saying once the capacitor is charged to the external voltage), or once all the dipoles are aligned and no more are available (in practice, saturating the dielectric is normally avoided).
  5. Sep 24, 2007 #4
    When you apply an electric potential on a capacitor (say from a battery, d.c), the capacitor charges and for a slight moment, there is effectively some current passing through the circuit until the capacitor is full charged. But the current never pass through the dielectric (or at least in usual cases as explained above). The charges are simply accumulated in the plates of the capacitor.
    if you reverse the polarity of the battery on the charged capacitor, there will be a current in the reverse direction, firt discharging the capacitor until 0, then charging it in the reverse polarity (this second process will take longer time than the first).
    For a a.c., it basically the same process of charging / discharging of a capacitor, hense the idea of having a current in the circuit might mislead to consider that the capacitor lets current through (which is false in some extent).

    Correct me if I am wrong.
  6. Sep 24, 2007 #5
    With AC, you should not be concerned with the actual linear movement of charge, but you should focus on how the changing electric field transfers energy.

    AC "passes through" a capacitor not by literal charge flow, but instead by an electric field on one plate acting on the charges of the other plate and transferring energy. By electrostatic induction.

    Imagine a large cylinder with a dividing plate that was free to move along the length of the cylinder while still dividing the cylinder, like a piston head. If you fill both sides of the cylinder with water, then pressure on one side can push the plate, and cause water on the other side to move without actually tranfering water between the two sides
  7. Sep 27, 2007 #6
    I understand that you are talking about displacement current. But then, displacement current is also present when a d.c. charges a capacitor. Then why is it said that a capacitor blocks d.c. but allows a.c. to pass through.

    What I am actually saying is that our teacher told us that a.c. "really" passes through the capacitor. That is what my confusion is. I am not talking about displacement current but actual passage of a.c. or d.c. which, according to me, is known as dielectric breakdown.

    Thanks for your replies nevertheless.

    Mr V
  8. Sep 27, 2007 #7
    Electronics is a LOT like plumbing, and there are quite a few good analogies. Voltage (EMF) is pressure. Amperes (current) is flow (in Coulombs per second instead of gallons per second).

    A resistor is like a restriction in a pipe (inducing turbulence and thus wasting energy in the form of heat).

    An inductor is like a long straight pipe that holds a lot of water. That water has a notable inertia and when you stop the flow by cutting off a valve, the momentum of that water applies a force to the valve. The force can be very high if you shut off the valve quickly. That's why, in real plumbing, there is often an intentional air pocket (a vertical pipe closed at the top) to help soften that bang when a faucet at the end of a long pipe is turned off quickly.

    A capacitor is like a rubber diaphragm blocking flow. The diaphragm can stretch a bit as the pressure differential between one side and the other changes, and thus there is current but only for a short time (until the tension on the diaphragm counteracts the pressure differential). If you continually reverse the pressure across the diaphragm, you repeatedly see water flowing in and out the pipes leading to each side of the diaphragm--remember that current through a device does not mean, "the same electron that entered, also exits". It means, "when an electron enters, an electron (not necessarily the same one) exits".

    The higher the frequency, the more current passes. The "frequency" of D.C. is certainly ambiguous until infinite time has passed because it takes that long to know whether that applied voltage is going to change or not! So over finite time you get some current. The longer the time (while D.C. is applied) the less the current.
    Last edited: Sep 27, 2007
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