This one is OLD! For an expanded version (with better diagrams) see: http://amasci.com/emotor/cap1.html Date: Sat, 8 Jun 1996 14:17:53 -0700 (PDT) From: "Bill Beaty" To: ELECTRON@CCTR.UMKC.EDU Subject: Re: light bulbs On Thu, 6 Jun 1996, Rader William E wrote: > On a more serious note (for every one out there), I would like to know > how a capacitor REALLY works. I have read as many texts as you have but > don't you believe them ! > If you have access to a high voltage generator (Wimherst sic ) and a > take-apartable Layden jar please play(i mean experiment) with these and > tell me how a capaciator really works I have beefs with textbook explanations of capacitors too. "Capacitors store charge." No! Flat out wrong! Wait and hear me out, I'm not insane. ;) When we "charge" a conventional metal-plate capacitor, we take electrons from one plate and put them onto the other. This creates opposite areas of imbalanced charge: one plate has less electrons and excess protons, and the other plate has more electrons than protons. However, if we consider the capacitor as a whole, no electrons have been put into the capacitor. None have been removed. The same amount of electrons is in a "charged" capacitor as in a capacitor which has been totally "discharged." Yes, a current has been forced to flow momentarily from one plate to the other during charging, and a rising potential difference has appeared. But the current is directed THROUGH the capacitor, and incoming electrons force other electrons to leave. Every bit of charge that's injected into one terminal must be forced out of the other terminal at the same time. Think about this: when "charging" a capacitor, momentary current gives a voltage 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 to the capacitor. Therefor, during "charging", ENERGY is placed in the capacitor. But "charge" is not. When we "charge" a capacitor, we give it a charge of energy. Because we use the word "charge" to refer both to electric charges and ALSO to a quantity of energy, capacitor explanations are impossible to understand. "Charging" a capacitor means injecting electrical energy into the device. Similar trouble is caused when we say that we "charge" a battery. We charge a battery with energy in the form of stored chemical fuel, but we pump electric charge THROUGH the battery and none builds up inside. It's terribly confusing: charging a battery does not store charge, yet charge must flow through the battery if we want to charge it! The word "charge" has too many meanings, and 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 when we say that we "charge" a capacitor, we DO assume that an electrical charge has been stored, and this is just as silly as mistaking gunpowder for electrons. Charging a capacitor is like charging a cannon, in both situations we are inserting energy, not electrical charge. When we "charge" a capacitor, the path for current is THROUGH the capacitor and back out. 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 we let the rubber belt go free, the wound-up spring within the motor drives the belt in the other direction, and the spring becomes "discharged." But no quantity of "belt" is stored inside this motor, and we wouldn't want to label this device as "machine which accumulates rubber." Yet this is exactly what we imply if we state that a capacitor "stores charge." 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. _________ _ / | | \ _ / _____| |_____ \ / / | | \ \ THICK IRON SPHERE, RUBBER PLATE | | | | | | DOWN THE MIDDLE, BOTH SIDES FULL | | | | | | OF WATER, w/CONNECTING HOSES | | | | | | | | | | | | \// \_____| |_____/ \\/ //\ _ | | _ /\\ ___// \____|_|____/ \\____ ---- ---- 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 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 it exerts, and finally the pressure will grow strong enough that the pump will stall. If 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. When the pump pushed water into one side, this also pushed water out of the other side. 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. Would it 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"? Yes, very misleading. No water was injected into the sphere while it was being "charged." If I now connect a single length of pre-filled hose between the two halves of the capacitor, the forces created by the bent rubber plate will drive a sudden immense spurt of water through this 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 discharged? None! A momentary current has flowed through the 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 has 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. Instead they gain expertise in spite of classroom learning. Maybe the experts are experts because they have fought free of their classroom learning, while the rest of us are still living under the yoke of the electricity misconceptions we were taught. --------------------------------------- On Wimshurst machines and capacitors: There is a corona effect which makes take-apart Leyden jars behave very strangely. If you electrify (not "charge" !) a Leyden jar, and then pull the inner metal cylinder out of the jar, the capacitance value drops, and this makes the potential difference skyrocket. The potential tries to become huge but it cannot, because instead it creates corona along the metal edges and leaks excess charge into the air. This corona allows opposite electrical charges to "paint" themselves on either side 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 stored as a field in the dielectric, but the separated charges are no longer on the metal plates, they are now TRAPPED ON THE GLASS SURFACE! Strange idea, huh? A capacitor with no plates, just a dielectric. If you now short out the two metal plates of the leyden jar, then put the whole thing back together, you'll find that it's still 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 the momentary current will throw you across the room. This strange effect leads many people to claim that the energy in a capacitor is stored 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. Also, it is much better to use a parallel-plate capacitor with totally shielded or rounded edges, and to pull the plates away from each other rather than sliding them along the dielectric as is done with the take-apart leyden jar. This can prevent the "painting of charge" phenomena from transfering charges to the opposite surfaces of the dielectric. .....................uuuu / oo \ uuuu........,............................. William Beaty voice:206-762-3818 bbs:206-789-0775 cserv:71241,3623 EE/Programmer/Science exhibit designer http://amasci.com/ Seattle, WA 98117 billbeskimo.com SCIENCE HOBBYIST web page