"Static Electricity" means "High Voltage"
|Info on spark gaps from Jim Lux:|| ||
3. Doorknob spark, relative pain
Use an adjustable, calibrated High Voltage DC power supply to charge your body to various voltages ( EXTREME DANGER! Limit the current to below a few hundred uA by using a series resistor chain. If you don't know how to use DC high-volt power supplies safely, then don't mess with them. IF YOU DO THIS WRONG, IT CAN KILL YOU.) Use the power supply to charge yourself to 3KV, 5KV, 7KV, and touch a grounded knob each time. No electrometer is needed, yet you've just verified the length, sound, and pain-level of the fingertip spark which appears when your body had been forced to take on certain high voltages. This procedure is a great way to learn the types of sparks produced by various voltages on your body, so the next time you get zapped by a car-door, you can say "YOW!, the ouchiness-factor of that spark indicates a human body voltage on the order of seven kilovolts." :)
4. Hand-waving AC field magnitude
Crudely measure your body-voltage by using an oscilloscope to compare it to a high-voltage power supply. First, connect an oscilloscope's input to a metal plate. Scuff on a carpet (during a low-humidity day), wave your hand near the metal plate, and see how much the trace on the oscilloscope is deflected. Adjust the scope's vertical gain to give a fairly large deflection during your hand-waving. (Don't TOUCH the plate while your body is charged, or you might blow out the input amplifier on the scope!)
Next, use a DC high-voltage power supply with a large current-limiting series resistor to charge your body to 5,000 volts. ( EXTREME DANGER! You MUST limit the current to below a few hundred microamps by using a series resistor chain. If you don't know how to handle high-volt DC power supplies safely, then don't mess with them. IF YOU DO THIS WRONG, IT CAN KILL YOU.) Wave your charged hand near the metal plate while seeing how far the oscilloscope trace deflects. Adjust the HV supply voltage until the deflection is about the same as when your body was charged by the rug-scuffing. Read the power-supply voltage setting, and you will know the approximate body-voltage produced by rug-scuffing.
5. Crude and simple "field mill"
A more precise version of no. 4 above... Build a crude "field-mill" e-field sensor. TO do this, first use a small DC motor to whirl a grounded wire, then place an oscilloscope probe behind the whirling wire so it is alternately shielded and unshielded, mount the entire assembly an inch or so from a large, electrically floating brass ball, then observe the scope trace. The observed AC voltage will be proportional to any DC voltage on the isolated brass ball with respect to ground. Apply a known kilovoltage to the brass ball in order to calibrate it against the AC reading on the scope (for example, 1,000 volts applied to the brass ball will produce a certain AC voltage reading on the scope, and any other voltage applied to the brass ball will be proportional.) Now go scuff on a rug while keeping one finger on the brass ball, and see what voltage is detected by the system. (No oscilloscope? Try placing a small metal disk behind the whirling wire, then connect a digital voltmeter to the plate and to ground, then set the meter to a sensitive scale on AC volts. Calibrate the voltage reading on the meter against a known DC high voltage applied to the brass ball.)
To measure REALLY high voltages (such as those produced by a VandeGraaff machine), you can use the above technique, but use the VDG sphere itself instead of the "brass ball", and place the field-mill and the oscilloscope several feet away from the VDG sphere. Calibrate it as before. If 1KV applied artificially to the VDG sphere produces a certain waveform voltage, then 100KV will produce exactly 100 times higher AC voltage.
6. Electrometer Array
Build 300 crude electrometers which drive individual LEDs. Use them to build a "visible e-field detector panel." When charged objects approach the panel, the glowing field of LEDs darkens in a pattern around the object. Scuff on the rug, hold a hand up, and observe the darkened field around your hand. What body voltage does this imply? To find out, go grab the old 7KV supply, hold the live terminal, then wave your hand around the sensor panel. Whoa! The darkened field is larger than before. Decide that the rug-scuffing body-voltage must have been maybe half of the body voltage created by the 7KV supply, maybe 3,500 volts.
Read books which say that common "static electric" sparks first appear when the voltage on your body rises above 500VDC. Use a HV DC power supply to test this, and find that their estimate is too low, that sparks cannot be seen at all until 750V, and they are very hard to notice until the voltage on your body is above 1KV.
Calculate what happens when a charged balloon is lifted from your arm. Obtain a value of 100KV. Sounds sensible. Ordinarily a VandeGraaff Generator would be needed to make arm-hair stand up so fiercely.
Read research papers from people who measure such things. Here's one from J. Chubb Inc:
Their measurements for different clothing and various car-seats give impressively high voltages, and this occurred at humidity levels above 50%. The voltages should be MUCH higher at, say 5% R.H.!
- Nylon clothes: 21,000 volts (Yowch!)
- Wool clothes: 9,000 volts
- Cotton clothes: 7,000 volts
Here's a suggestion taken from the PHYS-L discussion group:
Touch the electrode of a foil-leaf electroscope, and simultaneously scuff on the carpet. Observe the deflection of the foils. Now connect the electroscope terminal to an adjustable HV DC power supply with the other HV supply terminal connected to ground, and adjust the voltage to duplicate the foil deflection from before. Read the voltage. The power-supply voltage is the same as your rug-scuffing body-voltage was.
I rub a balloon upon my arm hair. If I know the attraction force and the capacitance between those flat, plate-like regions of opposite charge, then I can calculate the voltage. If the attraction force is 0.1Nt (like a 10gram weight) and if it is independent of plate-separation (because the plates are closely spaced,) and if the spacing of the "capacitor plates" is initially 1mm (0.001meter), then it took an amount of energy equalling (Force*distance) to pull the attractive plates apart to 1mm, and the stored energy is
Work = force * distance.1Nt * .001meter = 1e-4 Joules. For the given attractive force between the plates, we see that 100 microjoules of electrical energy is stored by this capacitor.
OK so far? Now, what is the capacitance of two capacitor plates spaced 1mm apart and having the size of that typical contact area between the balloon and my forearm? Let's say the area is 4cm by 15cm, or .04*.15 = .006 square meters. The equation for calculating the capacitance of parallel plate capacitors (plates spaced very close) is
C = k * A/DCapacitance = k*area / distance between plates, where the dielectric constant of air is k=8.9e-12 and the lengths are in meters, so the capacitance of the balloon/arm capacitor at a 1mm gap is 53pf.
An actual 4cm x 15cm foil-plate capacitor with a paper-stack dielectric
measures 95pf on my capacitor meter, which is in the ballpark. We should
expect it to be higher than our 53pF above, since it has a paper
dielectric, not air.
The energy stored was 100 microjoules, so the voltage can be had from the equation for stored energy in capacitors:
U = 0.5 * C * V^2Energy equals one half times capacitance times voltage squared, or Voltage=sqrt(2*U/C) (U is the stored energy.) Plugging in 100 microjoules and 53 pf I get an answer of 1900 volts. At 1mm spacing, the voltage is already almost 2,000 volts!
I also get a total charge of each imbalanced region which calculates to be about 0.1 microcoulomb:
U = 0.5 * Q^2 / CForce * distance = 1/2Q^2/C, or Q = sqrt(2CFd) = 1e-7coulombs. Capacitor voltage is always V = Q/C, and capacitance varies inversely with plate-spacing, therefor voltage varies directly with plate spacing. Double the plate spacing of charged, insulated plates, and we double the voltage.
U = F * d
Q = sqrt(2*C*F*d)
plate spacing V(capacitor) 1mm 1920v 5mm 9600v V = (1e-7)*D/(.006)/(8.9e-12) 1cm 19200v 5cm 95800vWHOA! 100,000 volts at a couple of inches spacing?!! However, this is reasonable, since it normally takes a VandeGraaff Generator to make arm-hair stand up so painfully rigid. In the gap between the balloon and my arm-hair, we have electric field strength which is easily the same as the field strength at the surface of the sphere of a VDG machine. Also, the capacitor equation stops working correctly as we exceed about 1cm, when the plate spacing becomes large in comparison to the shortest side of each plate. Maybe the voltage at 5cm is really only 50,000 volts. Only!!!!!!!
The capacitor voltage varies as the square-root of the attraction force,
so what would happen if this force was smaller? If the attraction
between my arm and the balloon was only 0.01Nt (1 gram weight), then the
voltage would start at 606V at 1mm, and go up to 6060V at 1cm. Not as
huge, but still pretty impressive.