However, if we consider
the capacitor as a whole, no electrons have been put into the capacitor.
None have been removed. The same number of electrons are in a "charged"
capacitor as in a capacitor which has been totally "discharged." Yes, a
certain amount of charge has been forced to flow momentarily during "charging," and a rising potential difference
has appeared. But the current is directed through the capacitor, and the
incoming electrons force other electrons to leave at the same time. Every
bit of charge that's injected into one terminal must be forced out of the
other terminal at the same time. The amount of charge inside the capacitor
never changes. The net charge on each plate is cancelled by the opposite
charge on the other plate. Capacitors are never "charged" with electric
Think about this:
When "charging" a capacitor, a momentary current causes the voltage to rise. Volts times electron-flow equals energy-flow ( V x I = P). Therefore during a momentary current through a capacitor, there is a joules-per-second transfer of energy from the power supply into the capacitor.
Similar trouble is caused when we say that we "charge" a battery. We
charge a battery with some energy in the form of stored chemical fuel, but
we pump electric charge through the battery and none of it builds up
inside. Fuel-chemicals build up inside. Charge doesn't.
It's all terribly confusing. What are students to think if we
tell them that "charging a battery" does not store any charge,
yet charge must flow through the battery if we want to charge it! Ugh.
The word "charge" has far too many meanings. In science this is always a
Very Bad Thing.
Another, less misleading situation is similar: think of the word "charge"
as applied to gunpowder. A charge is placed in an old cannon, followed by
a cannonball. It would be silly to assume that, because we've "charged"
the cannon, the cannon now has an electrical charge. But whenever we
state that we've "charged" a capacitor, we do assume that an electrical
charge has been stored inside. This is just as silly as mistaking
gunpowder for electrostatic charges. Charging a capacitor is like
charging a cannon; in both situations we are inserting energy, not
Here's yet another way to visualize it. Whenever we "charge" a capacitor,
the path for current is through the capacitor and back out again. The
extra electrons on one plate force electrons to leave the other plate, and
vice versa. Visualize a capacitor as being like a belt-driven wind-up
motor. If we shove the rubber belt along, the spring-motor inside the
capacitor winds up. If next we let the rubber belt go free, the wound-up
spring inside the motor drives the belt in the other direction, and the
spring becomes "discharged." But no quantity of "belt" is stored inside
this motor. The belt flows through it, and we wouldn't want to label
this motor as a "machine which accumulates rubber." Yet this is exactly
what we say whenever we state that a capacitor "stores charge."
One more try. Capacitors store charge in the same way that resistors
store charge, and inductors store charge. Inductors are full of mobile
electrons, inductors are devices for storing charge!!!! Not. A capacitor
is not a bucket for electrons. Instead it greatly resembles a length of
wire. But it's wire which, whenever you run a current along it, the total
charge inside the wire stays constant, but a voltage (and a charge
imbalance) appears at the two ends.
My favorite capacitor analogy is a heavy hollow iron sphere which is
completely full of water and is divided in half with a flexible rubber
plate through its middle. Hoses are connected to the two halves of the
sphere, where they act as connecting wires. The rubber plate is an
analogy for the dielectric. The two regions of water symbolize the
Imagine that the rubber plate is flat and undistorted at the start. If I
connect a pump to the two hoses and turn it on for a moment, the pump will
pull water from one half of the iron sphere and simultaneously force it
into the other. This will bend the rubber divider plate more and more.
The more the plate bends, the higher the back-pressure the plate exerts,
and finally the pressure-diff will grow strong enough that the pump will
stall. Next I seal off the hose connections and remove the pump. I now
have created a "charged" hydraulic capacitor.
Now think: in this analogy, water corresponds to electric charge. How
much water have I put into my iron sphere? None! The sphere started out
full, and for every bit of water that I took out of one side, I put an
equal amount into the other at the same time. Same as when running a
current through a conductor. When the pump pushed water into one side,
this extra water also forced some water out of the other side.
No water passed through the rubber, instead there was some rubber-current
in the divide. Even so, essentially I drove a water
current through my hydraulic capacitor, and this current pushed on the
rubber plate and bent it sideways. Where is the energy stored? Not
in the water, but in the potential energy of the stretched rubber plate.
The rubber plate is an analogy to the electrostatic field in the
dielectric of a real capacitor.
It would be misleading to say "this iron sphere is a device for
accumulating water", or "this sphere can be charged with water, and the
stored water can be retrieved during discharge." Both statements are
wrong. No water was injected into the sphere while it was being
"charged." (And when I wind up an old watch, am I "storing steel"
inside, putting more iron into its spring? Lol.
Imagine that I now connect a single length of pre-filled hose between the
two halves of the capacitor. As soon as the last connection is complete,
the forces created by the bent rubber plate will drive a sudden immense
spurt of water through this already-full hose. Water from one side will
be pushed into the other side, and the rubber plate will relax. I've
discharged my hydraulic capacitor. How much water has been removed from
the sphere? None! A momentary current has flowed through the sphere
device, and the rubber plate is back to the middle again, and the water
has become a bit warm through friction against the surfaces of the hose.
The stored energy has been "discharged," but no water has escaped. The
hydraulic capacitor has lost its energy, but still contains the same
amount of water.
I never really understood capacitors until I started trying to construct
proper water-analogies for them. I then discovered that my electronics
and physics classes had sent me down a dead-end path with their garbage
about "capacitors store electric charge." Since my discovery, I've gained
significantly more expertise in circuit design, which leads me to a sad
thought. Maybe the more skilled of electrical engineers and scientists
gain their extreme expertise not through classroom learning.
gain expertise in spite of our K-12 classroom learning. Maybe the experts
experts only because they have fought free of the wrong parts of grade
school science, while the rest of us are still living under the yoke of the many
electricity misconceptions we were taught in early grades.
[Hey^2!!! I just found that Oliver Lodge was building mechanical analogies for Maxwell's descriptions of EM fields and circuits... and for an 1880 lecture he built a capacitor hydraulic analogy consisting of a water-filled globe containing a smaller water-filled balloon.]
Extra notes:Capacitors store just as much charge as coils do! In both devices the total amount of charge stays constant. Both capacitors and inductors are components for storing electromagnetic energy. They're two sides of the same EM phenomenon: a coil stores energy in a volume containing a magnetic field, while a capacitor does something similar with electric fields. Coils are "discharged" by interrupting a large current and collapsing the b-field, while capactors are discharged by shorting-out a large voltage and collapsing the e-field. Neither stores any "electricity" (unless by the word 'electricity' you mean magnetic field?) Of course you can place a coil atop an insulating platform, then use a VandeGraaff generator to give your coil a large net-charge! You can do the same with a big electrolytic capacitor too. :)
Bill Beaty here again. Two points: First, the heated topic about dielectric being an insulator, and currents being impossible between capacitor plates ...seems to be about VACUUM CAPACITORS, not capacitors in general. Modern capacitors are quite different, and inside their dielectric is a large electron current. They aren't vacuum capacitors. Relative Permittivity can be seen as a ratio between the small Maxwell's displacement current in the dielectric, versus the larger dielectric polarization current (electron flow.) The dielectric constant in modern ceramic capacitors is above 2,000, so the vast majority of the current is from moving electrons in the ferroelectric ceramic. That's a genuine charge flow. The Maxwell displacement current is insignificant: well below 1%. Second: it might help to ask whether, down at the micro level within any conductor, is there any current BETWEEN the charge carriers? If there is, then there's certainly a current between the carrier-filled plates of a vacuum capacitor. Or said differently: if we have a current-sensor, and a charged particle approaches and passes it, does our sensor indicate an extremely brief pulse, where the pulse-width is associated with the diameter of the charged particle? Or instead does our sensor see each moving particle as being large and "fuzzy," where the measured current is in the fields surrounding each particle, and the current extends forwards and back from the particle location? A clamp-on inductive sensor (Rogowski coil) doesn't detect charges or their motions, instead it detects changing flux-linkage. A clamp-on sensor would report that a charged particle is indeed large and fuzzy, and the current exists in the empty spaces between the flowing charges. Current is not just on the particles' surfaces where the charge actually resides. So, a clamp-on sensor would 'see' currents in the capacitor's empty gap. Two points? Third point (I lied!), suppose we construct a capacitor where the dielectric takes the form of a long rod: much thicker than the diameter of the capacitor plates. Use a long narrow lead-zirconate- titanate PZT rod with capacitor plates attached to its circular ends. Now apply some 27MHz amperes. Do you insist that the current within the rod is zero? Really? It's not. Suppose we obtain a coil-shaped spiral rod of PZT. A "ferroelectric coil." If we apply some RF amperes through this spiralled capacitor, we'll certainly detect a strong magnetic field surrounding the device. If the current is supposedly zero within all capacitor dielectrics as some people angrily insist, how can we explain this?
Capacitors: physicists vs. engineers
> For the charge to flow in a circuit, there has to be a closed path. > Right, sir! Please correct me if I am wrong.
No, there must be a closed path ONLY if the charges must flow continously
in the circuit forever (pure DC). But if the charges flow briefly, or if
they flow back and forth (AC), then an open path is sometimes acceptable.
For example, with a metal rod 1 meter in length, it is possible to create
a standing wave of charges oscillating lengthwise in the rod with
frequency 150MHz. Inside the rod, the charges move back and forth (this
resembles the motions of the compressible air inside one pipe of a church
organ.) The charges within the rod are acting like a compressible fluid.
At far lower frequencies, the electric current can still be large if the
ends of an open circuit are joined to the plates of a capacitor. On the
metal surfaces between the plates of the capacitor, the quantities of
charge carriers behave as a "compressible fluid", while the charges within
a wire behave as an "incompressible fluid."
> In the capacitor, the charge flows from one plate to the other. Let's > assume that air is the dielectric between the 2 plates of the capacitor. > So, there is no physical conducting path between the 2 plates. So, > it's like an open circuit. How can we say that the charges move from > one plate to the other?
Two ideas are necessary:
1. Metals are always full of movable charges even when they are electrically neutral. They contain a "fluid" made of electrons, and a "solid" made of protons (made of the nuclei of metal atoms.)See the "water analogy" diagram above. The rubber barrier represents the dielectric. Yet electrical forces can reach across the dielectric, just as physical forces can be transmitted through a rubber barrier.
If you forcibly inject electrons into one plate of a capacitor, these
electrons will repel the electrons hidden within the other uncharged
plate, and these other electrons will be forced out of the capacitor's
other terminal. However, if the other capacitor terminal is connected to
nothing, then those electrons cannot leave the other plate. They will
continue to repel the electrons in the first plate, and for this reason
very few electrons can be forced into the first plate. The rule for most
capacitors is: the current in both capacitor terminals is always the
same. This means: if charge is injected into one capacitor plate, then
an equal amount of charge is pushed out of the other capacitor plate, and
if equal charge cannot leave the second plate, then we cannot force
into the first plate.
If we ignore the space between the plates, the capacitor SEEMS to be a
kind of conductor. However, it is a strange "conductor" where the two
terminals are connected by electrical forces which reach across a narrow
space. It is not a true conductor, but neither is it a simple insulator.
> Sir, I have asked the same to my undergraduate professors, but they find > it too trivial to answer. Thus, I have never been able to understand > this concept.
I think this concept is explained very badly in physics textbooks. In
physics, a single metal sphere is considered to be a "capacitor," and two
metal spheres with a large separation between them is also a "capacitor."
Yet conceptually these are very different from capacitors used in electric
With single spheres and with pairs of separate spheres, the flow of charge
into a sphere need not be equal to flows of charge out of the other
sphere or elsewhere. The entire universe serves as an extra unseen
capacitor plate. Yet in electronics, "capacitor" means only one thing:
"two closely-spaced plates." As far as electronic equipment is concerned,
capacitors only have two plates, and the current in one terminal must
always equal the current in the other terminal. This is not true for two
distant conductors, but the "general case" in textbooks is much
harder to understand. Physics textbooks examine the general case where
the currents in the capacitor terminals are not equal, a distant ground is
part of the system, and the electrostatic coupling between the plates is
not very strong. Conventional capacitors in electronic equipment are
different: the electrostatic coupling between their plates is extremely
strong, the electrostatic coupling to the earth is essentially zero, and
the current in the terminals connected to the two plates is always equal.
The charge on their plates must be equal and opposite, always. But
in physics textbooks, you can have three coulombs on one metal sphere, and
two coloumbs on another.
Here is an "electrical engineering" way to understand it. Suppose the
capacitance between two parallel plates is X microfarads. Suppose we
force some charge into one plate, but we do not connect the other plate to
anything. In this case, the lines of electric force extend outwards from
both plates to the earth. Important question: what is the capacitance of
the "capacitor" formed by one plate and the distant earth? It is small.
very small, possibly millions or even billions of times smaller
than the number of microfarads existing between the two plates.
Therefore, if we connect both capacitor terminals to a circuit and force
charges "through" this capacitor, the voltage across the plates will rise
slowly, but if we instead try to force the same current into only
one plate (and leave the second plate disconnected), then the
voltage between the capacitor and the earth will rise millions or billions
of times faster for the same current. In other words, a two-plate
capacitor has immensely more capacitance than a one-plate capacitor. In
electronics we usually ignore the "one-plate" phenomena entirely, and
imagine that capacitors are a strange kind of two-terminal conductor.
Here is a model which exposes the problem. Rather than imagining a
capacitor to be like two metal spheres, imagine it to be like one sphere
which is divided into two parts, where the gap between the two parts is
extremely small, and the capacitance value between the two parts is far
larger than the capacitance value measured between each part and the
____ ____ __/ || \__ / || \ / || \ "Engineer's Capacitor" | || | | || | A spilt metal sphere with | || | a very narrow gap between | || | the two halves | || | | || | | || | \ || / \__ || __/ \____||____/Above is a model of an "engineer's" capacitor, as opposed to the two spheres of the "physicist's" capacitor. Because of the narrow gap, it behaves very differenly than two separate spheres.
______ ______ / \ / \ The "Physicist's Capacitor" / \ / \ | | | | Two separate metal spheres | | | | | | | | | | | | \ / \ / \______/ \______/If we dump some excess charge onto one half of the engineer's capacitor, something strange occurs: half of the charge apparantly migrates to the other half of the device! The twin hemispheres behave as a single conductive sphere because any excess charge on one half of the gap will induce equal and opposite excess charge on the other half of the gap, and this leaves excess alike charges on the second half (which then distributes itself uniformly across that surface.) In physics terms, the capacitance between the two hemispheres is very large when compared to the capacitance measured from either hemisphere to distant ground.
The split-sphere device behaves almost as if it has a conductive
connection between its two halves. This connection is as real as the
connection between the ends of a resistor... yet it is made out of
electrostatic forces in space, rather than being made out of carbon.
In my experience, the the two spheres of the "physicist's
capacitor" cause distortion in physicists' concept of what the electronic
component called "capacitor" really is.
Analogy: if we were talking about coils, and coils were
just as badly, then how would physicists
visualize electric transformers? They'd be very familiar with induced
currents between pairs of distant air-core coils, yet they would be almost
entirely unaware of of the strange effects arising with closely-coupled
coils of an iron core transformer. Power transformers behave very
differently than two widely-spaced air-core inductors. And the above
split-sphere "engineer's capacitor" is analogous to an iron-core
transformer, while the "two-sphere capacitor" of the physics textbooks is
analogous to a pair if widely separated coils. Whether we're dealing with
capacitor plates, whenever the coupling between
the two parts approaches 100%, everything starts acting very differently.
On Wimshurst machines and capacitors:
Do you believe that the energy in a capacitor is trapped permanently in
the dielectric? Many people do. Their belief is caused by a
famously misleading experiment called "Dissectable Leyden Jar." It's an
experiment which involves high
voltage and corona discharge. The effect it purports to prove does not
occur in capacitors at lower voltages.
First charge up a Leyden jar using a Wimshurst Machine (or other source of
high voltage.) Now, carefully remove the inner metal from the jar. Now
remove the outer metal. Discharge everything, then hand the parts around
the classroom. Next, put the parts together again, connect the two metal
cylinders, and BANG!, there is a loud discharge.
Doesn't this prove that the energy in a capacitor is stored in the
Whenever you take apart a Leyden jar or other high voltage capacitor,
there is a corona effect which makes very strange things occur. When you
electrify a Leyden jar, and then you pull the inner metal cylinder out of
the jar, the capacitance value drops, and this makes the potential
difference skyrocket to enormous levels. The potential tries to become
huge but it cannot, because instead it creates corona along the metal
edges, and and it leaks the excess charge into the air. This corona
allows the opposite electrical charges to "paint" themselves onto both
sides of the dielectric "jar" surface. So, if you pull a leyden jar
apart, the sharp edges of the metal plates sweep along and transfer a
large percentage of the separated charges from the metal plates to the
glass surfaces. The energy is still there! It's still stored as a field
in the dielectric, but those separated charges are not on the mental
plates anymore. Instead they are now TRAPPED ON THE GLASS SURFACE!
Strange idea, huh? A capacitor with no plates, just a dielectric.
Now reassemble the Leyden jar: momentarily touch each metal plate to
ground, and put it back together again. You'll find that it's still
strongly electrified! The trapped
charges on the glass surface can still induce equal charges on the
adjacent metal plates. Touch the two terminals with your fingers and
BOOM!, the momentary current in your muscles will throw you across the
This strange effect leads many people to claim that the energy in a
capacitor is permanently trapped in the dielectric, and that it is not
stored in the electric field. This is wrong.
In order to properly perform the take-apart capacitor experiment, you must
execute the entire demonstration inside a big tank full of oil. This
prevents the corona discharges from spewing charges from the edges of the
plate onto the dielectric.
Or, perform the whole experiment at 1.0 volts, not at 10,000 volts. (Use
an electrometer to measure the voltage.) You'll find that the dielectric
doesn't store energy anymore. In order for the charges to spray onto the
dielectric, the voltage must be high.
Or, use high voltage but do this instead: before doing anything, take
apart the leyden jar. Now, lay the metal parts on a plastic sheet and use
a Wimshurst machine to charge them up. Next, use plastic tongs to
assemble the leyden jar. (The voltage across the plates will be very
low.) NOW perform the leyden jar dissection. It shouldn't work anymore,
since the initial voltage is low enough that it will prevent corona
discharges from painting any charges on the dielectric.