# What does "charge" on capacitor really mean?

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

1. ### Email InvalidGuest

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

Example:

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. ### John PopelishGuest

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 MyersGuest

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. ### Email InvalidGuest

Let me try another example:

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

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
"ground".

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. ### John PopelishGuest

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. ### Robert C MonsenGuest

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...

Regards,
Bob Monsen