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What does "charge" on capacitor really mean?

Discussion in 'Electronic Basics' started by Email Invalid, Apr 19, 2004.

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  1. I am trying to understand what exactly Q=CV means for capacitors in


    C1 C2
    ----| |----| |----
    a b c

    Let's say that at time t=0, ideal voltage sources are connected to
    nodes a,b and c such that Va = 2V, Vb = 0V, Vc = 0V.
    Then the "charge on C1" would be Q1=C1 * 2.

    However, I thought that when we say that a capacitor stores charge
    Q=CV, what it really means is that the more positive plate has +Q and
    the more negative plate has -Q charge, right?

    It is my understanding that the "left" side of C1 would have charge
    +Q1 while the "right" side would have charge -Q1. But since VC2 = 0V,
    then C2 stores no charge and so the "left" side of C1 should have 0
    charge and the "right" side should have 0 charge.

    First of all, this seems confusing because how could node b have -Q1
    charge and 0 charge at the same time??? If the "right" plate of C1
    has -Q1 charge, how could the "left" plate of C2 have 0 charge
    simultaneously? Wouldn't node b have a total net charge of -Q1 so that
    the "left" plate of C2 would also have charge -Q1. But that doesn't
    make sense either since there is no voltage across C2 so C2 should
    have no charge on it whatsoever. I am confused!

    Okay, now let's say at t=t1, we remove the 0V voltage source on node b
    so that it now "floats" but keep Va=2V and Vc=0V Then what is the
    voltage and charge on node b?

    What if we removed all the voltage sources, what then?

    I guess I don't really understand Q=CV with a solid understanding.
  2. Yes. That is the charge transferred through C1.
    Right. Charge leaves one side and enters the other side. So the
    inside surface of one plate has extra electrons on it (the -Q) and
    one plate has a deficit of electrons on its inside surface (+Q)
    The nodes don't have the charge on them. It is localized on the
    surfaces of the capacitor plates. The internal electric field pulls
    those charges (+ and -) toward each other. From the outside, it just
    looks like a voltage, like a battery.
    0 volts.
    Then you have to pick one of the three nodes to measure the other two
    voltages with respect to. But the difference in voltage between any
    two nodes remains unchanged. This is exactly like storing static
    electricity on spots on the surface of a balloon. It just sits there.
  3. Bob Myers

    Bob Myers Guest

    I think your first problem may be with the term "voltage."
    The first question that comes to mind once you've said the
    above is "with respect to WHAT?" A "voltage" is never just
    an isolated value at a single node, but rather is the
    DIFFERENCE in electric potential between that node and a
    given reference. When we speak of "node voltages" as in
    the above, it would be with respect to a "ground" or "common"
    point not necessarily shown here. In your description, it
    sounds like you're describing what in effect are two isolated
    cases - a 2V source placed across C1, and nothing at all
    across C2 (effectively, C2 is shorted if nodes B and C
    are not only at zero potential difference, but actually tied to the
    same point).
    That's your misunderstanding of voltage coming to play again.
    A point at "zero volts" simply has no voltage difference with
    respect to some OTHER point we've decided to call our
    reference. Both points might be at a thousand or even a million
    volts with respect to yet another third point, but have zero volts
    with respect to one another. This says nothing at all about the
    "charge" on them.
    Charge isn't voltage. What the Q=CV equation is also telling
    you is that the voltage between any two points (at least in a
    static situation - remember, "static" electricty is what we're
    talking about when charge is "just sitting there" - depends on the
    voltage difference between two points and the capacitance
    between them.

    Consider this. Suppose I have two objects. I somehow force
    one unit (however many that is) of electrons to move from one
    of them to the other. The first winds up deficient in electrons
    (and so has a "positive charge") by one unit's worth, while the
    second has the same amount of excess (and so has a "negative
    charge," since electrons are said to be negative). What's the
    voltage between them? Answer - you don't know. You would
    have to know the physical configuration of the two object, and
    the distance between them (which together tell you the "capacitance"
    of the system) in order to figure that out. The voltage between
    these objects will be considerably different if they are, say, 1 mm
    spheres held 3 meters apart than if they were 1-meter-square
    flat plates held 1 mm apart.

    Does this help?

    Bob M.
  4. Let me try another example:

    a C1 b C2 c
    + + +
    Va Vb Vc
    - - -
    | | |

    Capacitor C1 and C2 are connected in series and initially three
    voltage sources Va, Vb and Vc are connected between ground and nodes
    a, b, and c respectively. All voltages are with respect to this common

    1. Initially at t=0, Va=2V, Vb=0V and Vc=0V.
    Then based on my understanding we have the following:

    Left plate of C1 has +Q1 charge (where Q1=C1 * (Va-Vb))
    Right plate of C1 has -Q1 charge (where Q1=C1 * (Va-Vb))
    Left plate of C2 has 0 charge
    Right plate of C2 has 0 charge

    Node b itself does not have a charge per se as it is localized to the
    plates on C1 and C2. But the net charge on node b would have to be -Q1
    (-Q1 + 0 = -Q1) right???

    2. Now at t=t1, we remove the voltage source Vb entirely so that node
    b "floats".

    It seems that if I did that, nothing has changed (in terms of node
    voltages and charge distribution), right???

    3. Okay now at t=t2 (with Vb already removed entirely), let's say I
    step voltage source Vc from 0V to -2V. What happens to the circuit in
    terms of nodal voltages and charge distribution, especially the charge
    distribution and voltage on node b??? What if instead I step voltage
    source Vc from 0V to +2V???

    If someone could explain what happens to the circuit in (3), then that
    might help greatly in my understanding of charge and capacitance.
  5. Yes. If these were air spaced capacitors and you set this up just as
    described, and then unhooked the batteries and physically separated
    the plates of the capacitors far from each other, then the right plate
    of C1, the left plate of C2 and the connecting wire would share the
    charge -Q1, which would redistribute itself in such a way to have the
    lowest energy (charge spread as far from similar charge as possible).
    Right. It is simpler to having a good battery cell in series with a
    dead one, with nothing connected between them.
    That -2 volt swing is divided between the two seriesed capacitors. If
    they are equal in capacity, each would take half of that swing (one
    volt added to each, with the negative side of that volt to the right.
    So Va is still 2 because the voltage source forces that, Vb is -1 volt
    (because of the -1 volt swing that is C1's share of the change) and Vc
    would be at -2, because a voltage source forces that.
    Remember that when there is a floating node such as your disconnected
    Vb, the capacitors divide voltage swings (change in voltage that
    represents shift of charge through the capacitors) applied to the
    ends, without regard to the voltage on the floating node.
  6. Ideal voltage sources are, basically, infinite sources of charge. What
    they do is to add or subtract however much charge it takes to make
    the node they are connected to assume a particular voltage relative
    to its ground.

    If you connect up a capacitor across a voltage source, then it will
    add or subtract enough charge so that the capacitor has a voltage
    across it equal to the supply voltage.

    For a capacitor that starts with both terminals at the voltage supply's
    ground, that means it will pump Q = C*V coulombs of charge into the
    capacitor. At that point, you can say that the capacitor is carrying
    a charge of Q.

    If you now disconnect the voltage source, the charge on the capacitor
    has no place to go, so it just sits there.

    Note that this does NOT mean that either of the terminals of the capacitor
    'have' charge Q.

    If you now create a path between the nodes of the capacitor, then
    charge will flow to equalize the voltage on both sides of the capacitor.
    The amount of charge that will flow is equal to C * V (which is the amount
    you needed to add to bring it from ground to V.)

    Thus, the charge, Q, really means that it will take that much charge
    flowing in or out of one of the terminals to equalize the terminals
    of the cap.

    For your situation above, removing the ground on b does nothing, because
    no charge will flow from one side of a cap to another.

    Hope this helps...

    Bob Monsen
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