Electrostatic Voltmeter  page:

Surface  DC Voltmeter

Most Common Use of the Surface DC Voltmeter

[The subject of static electricity encompasses many details. To solve a specific static problem, you will only need to read about a few specific details. For this reason, the next four paragraphs of text below are a very brief (but complete) description of all the uses of this meter. There are words and phrases in the text that aren’t defined in a standard dictionary; these words and phrases appear as links. Click on them to jump to the definition in the glossary near the bottom of this page. At the end of each glossary definition, you can click "back" to return to where you were. In addition to unfamiliar words, certain concepts are mentioned in the text, but not all these concepts may be familiar. A number in square brackets appears immediately after a new concept is mentioned. Click on the number to jump to the concept’s explanation at the bottom of this page. Then click "back" to return. This way, you only need to click the glossary definitions or explanations in square brackets that you need. If you’re looking at a printed copy of this, the glossary’s word definitions are listed at the end of this paper, and they are in alphabetical order. The concept explanations are listed just before the definitions, and are in numerical order.]

Other Uses of the Static DC Voltmeter

Because of the difference between conductors and insulators, the meter will read voltage when measuring a conductive surface, but it will read the surface’s electric field when measuring an insulator (for an explanation of conductors vs. insulators, see [1]). When measuring these quantities, the reading depends on how far the sensor disk (at the top of the meter) is held from the surface that is being tested. The meter’s reading also depends on how large the test surface is; that is, how much of the horizon (as seen by the sensor disk) is covered by the surface that is being tested. (See formulas [2]).

In some environments, air ions are either deliberately introduced -- by using an ionizer -- or they may be accidentally present, such as from nearby sparking of high-voltage sources. These ions can either help remove surface static electricity or add more, depending on the situation [3]. If "accidental" ions are present, the meter can be used to measure how many excess ions per second are hitting a given area ("excess" meaning either more negative than positive or more positive than negative, so that when excess ions are present, they will charge surfaces either negative or positive). If you use the optional "charge plate detector" kit, you can also create a charged surface to measure how rapidly a neutral ionizer will discharge that surface. (See [4] on how to measure both accidental and deliberately introduced ions.)

There are several types of anti-static coatings. They may be sprayed onto surfaces, painted on, or molded into surfaces. Some materials that we normally think of as insulators, such as glass or paper, are actually slightly conductive. On the other hand, most plastics are excellent insulators (surface conductivity of materials: [5]). The meter can be used to determine the approximate number of ohms per square of an anti-static surface [6].

When a surface carries charge, a second surface may be attracted to or repelled by the first surface depending on whether the surfaces are insulators or conductors, and whether the charge polarity of the two surfaces is the same or opposite. The meter’s readings can be interpreted to determine the approximate force per area ("pressure") between two surfaces [7].

An insulating sheet may have a certain amount of negative charge distributed on one surface, and some positive charge on the other. This will create a voltage difference between the front and back surfaces. You can measure this voltage difference, but it requires a special technique [8].

Explanations

[1] The Difference Between Conductors and Insulators

On a conductive surface, charges can freely move, whereas on an insulator any charges are stuck where they sit. Because of this difference, the voltage on a conductive surface is the same at one point as at every other point of the surface, and if you bring a small grounded conductor near the surface, the charges will rearrange themselves so that the number of charges per area will be much more in the region which is nearest the small grounded conductor. This ‘bunching up’ of charges in that region will rob some charges from the other parts of the large conductive surface, but it will not significantly change the voltage of the surface if the "small" grounded conductor is much smaller than the (large) conductive surface.

In contrast, an insulator with a large number of excess charges on its surface will be at high voltage. However, that voltage is not constant across the surface, but is usually higher near the center of the surface and lower at the edges. Furthermore, if you bring a small grounded (grounded = "zero voltage") conductor near the charged insulator, the surface voltage in the region which is closest to the grounded conductor will decrease considerably, but unlike a conductive surface, charges will not ‘bunch up’ there, because the imbedded charges on the surface cannot move.

Therefore, a conductive surface has a well defined voltage, but a charged insulating surface does not, because the voltage is usually not the same at every point on an insulating surface. Instead, a charged insulator has a well defined surface electric field which is proportional to how many excess charges per unit area that are stuck on the surface. Because these charges can't move around in response to external factors, they maintain a constant charge-per-area, and hence they will produce an electric field that is the same everywhere on the sheet, provided that the amount of charge per unit area is the same everywhere on the sheet. (Note however that if a grounded conductor is brought near a charged insulator sheet, there will be charges of the opposite polarity that will accumulate on the face of the grounded conductor. This can contribute some additional field, so that the electric field at the surface of the insulator is the sum of two sources: the electric field generated by the charges on the insulator surface, and the field generated by any nearby grounded conductor. At most, this conductor will double the electric field compared to the electric field generated by the insulator sheet alone.

In comparison, on a metal sheet that is at a high voltage, charges are distributed on the sheet in a particular way. If a small, grounded metal disk is brought near the center of the high voltage metal sheet, many more charges will accumulate in that region and there will be a strong electric field between that large sheet and the small disk. Here is the reason: suppose the large sheet is at +100 volts with respect to the earth and the small disk is connected to earth ground (0 volts). Suppose also that the small disk is brought to within 5 cm in elevation above the large sheet. Then the average electric field that exists in the region between the small disk and the large sheet is equal to the voltage difference (100 volts) divided by the distance between them (5 cm). This can be written as ‘electric field’ = 20 volts/cm (V/cm), which is the same as 2000 volts/meter (V/m). Other areas of the large conductive sheet may have a much weaker electric field, because these other areas are much farther from any room walls (which are generally grounded) or from the floor. The electric field which exists at the surface of a charged metal sheet (at voltage "V"), is roughly given by ‘electric field’ = V/D, where D is the distance to the nearest grounded object. Therefore the electric field is high at the area of the large metal surface which is nearest the small grounded disk, and is much lower elsewhere. This is because the voltage is 100 volts everywhere on the large sheet, but it is 0 volts on the small disk. In the region of air directly under the small disk, the voltage gradient is such that it must go from 100 V to 0V in a distance of only 5 cm, so the electric field is high there (20 V/cm). At other areas of the large sheet far from the small disk, the voltage in space changes much more slowly with distance, going from 100 V to 0V in perhaps a few hundred centimeters.

In summary, there is an electric field at every point on any large metal sheet if that sheet is held at high voltage (with respect to ground). This field is not of the same strength everywhere. It can be made much stronger in a small area of the sheet if a grounded metal object (such as a small disk connected to ground through a long wire) is brought near the metal sheet. The electric field there can be made as strong as required, by bringing the disk closer and closer to the metal sheet. This is important to know, because a spark in air will occur if the electric field exceeds one million V/m. (There is one effect that limits sparking, however. There must be at least about ½ millimeter of air gap to allow a spark to initiate efficiently. To get an electric field of one million V/m across a 1/2mm gap, you need 500 Volts. Lower voltages than this don’t cause the air to break down and spark directly, so if you apply only 250 volts between two contacts, for example, there is no air gap distance that can initiate a spark. A spark can be sustained at lower than 500 volts, however, provided that something else initiates the spark, such as the touching together of the contacts and subsequent vaporization of some metal.)

On a large sheet of insulating material, if excess charges have been deposited on the surface (with the same 'amount of excess charge per unit area' everywhere on the surface), then the electric field produced by the surface will be the same everywhere, pointing perpendicular to (or normal to) the surface. If a small, grounded metal disk is brought close to the sheet of insulator, charges will rapidly flow through the ground wire and will accumulate on the small disk. These charges are the opposite polarity of the sheet's charge. If the small disk is held close to the insulating sheet, then the amount of charge per area on the small disk will be equal to the amount of charge per area on the large sheet, but of opposite polarity. If the small disk is held very close (much closer than its diameter), the small disk and the large (charged) insulating sheet will each contribute approximately the same amount of electric field to the space between the sheet and the disk. That is, the total electric field in this space is twice as much as what the insulating sheet alone would produce.

The voltage (with respect to ground) at any point on the insulator sheet can be approximately calculated; it's roughly the ‘electric field’ multiplied by the ‘distance from the nearest ground to the insulator sheet’. (Although the exact calculation is the integral of the electric field's dot product with the "dr" vector.) Under a grounded disk at nearby distance "L" from the charged insulating sheet, the voltage is approximately V = 2EL, where "E" is the fraction of the electric field that is produced by the charged insulating sheet, and therefore "2E" is the total electric field, because the nearby small disk has effectively doubled the field. Clearly, if V = 2EL is the voltage of the sheet just under the disk, then by pulling the disk farther away to twice the distance (2L), the voltage will increase to twice as much. If the (grounded) disk is instead placed onto the charged insulating surface (L = 0), then the voltage there is zero. Clearly, the greater the distance between the charged insulator sheet and the nearest grounded object, the higher the voltage of that particular area of the insulator sheet. This effect is important because if you pull a charged insulator away from a grounded surface, it becomes more likely to spark to a grounded needle or point that is held at a fixed distance from the insulator. Charged insulators will actually be attracted to a grounded surface (because the grounded surface will accumulate the opposite charge). Therefore, when you separate a charged insulator from a grounded surface, you are doing work, and thus adding electrical energy to the charged insulator. This is one way of explaining why the insulator becomes more likely to spark.   [back]

[2] Formulas for Interpretation of Surface Voltage and Charge

All the expressions below are intended to be used with standard international (SI) units. These units include volts (V), coulombs (C) (= amp • seconds), and meters (m), and of course seconds (sec.).

If a flat insulator surface has a certain number of excess positive charges per unit surface area, expressed by the letter S, which is in units of C/m² (coulombs per square meter), then the electric field "E" points perpendicular to the surface (and away from the surface, when the charges are positive). The strength of the electric field is given by

E = 5.7 X 10¹⁰ S. The equation in the previous sentence means that you must multiply 5.7 x 10¹⁰ by the value of S, the surface charge per area, measured in C/m², and you will then obtain "E", the electric field in V/m. Therefore if the charge per area is 10^-9 C/m², for example, then E = 57 V/m ("V/m" is the SI unit of electric field). If the flat surface is an insulator sheet with total charge per area of S ("total charge per area of S" adds together the charge on the front plus back surfaces of the sheet), then there is also an electric field on the back side, of the same strength as that on the front side. This back-side field also points away from the surface and it doesn't matter what percent of the total charge is on the back surface vs. the front surface.

If a grounded (through an earth wire) metal sheet is brought very near the charged insulator sheet, and is parallel with it, the electric field that exists between this grounded metal sheet and the insulator sheet will be twice as much as without the grounded metal sheet (that is, E = 11.4 x 10¹⁰ S which is usually written 1.14 x 10¹¹ S), while the electric field on the back side of the insulator sheet will be reduced to zero. It is as though all the electric field on the back side turned around and reinforced the electric field coming out of the front side (the side with the metal sheet just over it), thus doubling the strength of the front-side field. (Remember the "E" vs. "2E" explanation in [1].)

This doubling occurs because the grounded metal sheet, when brought close to the insulator sheet, will acquire the same surface charge per area of S, but of the opposite polarity. This charge will have flowed in from the wire that is connected to ground. Thus in the close space between the insulator and metal, the insulator produces an electric field of 5.7 x 10¹⁰ S, pointing away from the insulator, and the metal produces the same strength of field, pointing toward the metal. Thus these two fields reinforce each other for a total field of 1.14 x 10¹¹ S. However, on the back side of the insulator (opposite from the side where the metal sheet is) the insulator produces an electric field of 5.7 x 10¹⁰ S pointing away from the insulator, while the metal’s field (which passes right through any insulator without being reduced in strength) is

5.7 x 10¹⁰ S pointing toward the metal (and toward the insulator). Therefore the two fields cancel there, and the only region with a nonzero electric field is the region between the metal and insulator sheet.

As can be seen, when you bring a grounded sheet (or other grounded object) close to a charged sheet, the action can alter the electric field. The amount of alteration is much greater if the charged sheet is metal (as opposed to insulator) that is held (by being connected for example to a high voltage supply through a wire) at a fixed voltage "V". In that case, when another grounded (V=0) metal sheet is brought a small distance "L" away, the electric field in between is E = V/L, by definition. Each sheet will contribute ½ of this electric field, so the electric field contributed by each sheet is V/(2L), and the surface charge per area S on each sheet is the same (but of opposite sign), and can be determined by solving for S if we know that V/(2L) = 5.7 x 10¹⁰ S. Note that if you reduce L in this equation to one tenth, for example, that action will multiply S by ten. This means that the closer the two metal sheets are held together, the more surface charge per area S will be on each surface. Therefore the high voltage supply that is holding one sheet at voltage V, and the ground that is connected to the other sheet, must supply more and more charge to their respective sheet as L (the distance between them) is reduced.

The Surface DC Voltmeter reads a number proportional to the amount of charge on its sensor disk, which is itself proportional to the value of S there. As was shown in the previous paragraph, if a grounded conductor is moved toward a metal sheet (a sheet which is held at fixed voltage), the quantity S (in coulombs per square meter) on the grounded conductor will increase. Therefore, S on the sensor disk will increase as the sensor disk is brought closer and closer to a metal sheet held at voltage V. However, as mentioned earlier, if the sensor disk is instead brought closer and closer to a uniformly charged insulator sheet, then the value S on the disk won’t change much, because the electric field produced by the insulator doesn’t change much with distance (at least it doesn’t if the sensor disk is near the insulator sheet). Suppose there is an extremely wide sheet of insulator which has uniformly-distributed surface charge per area of S, and the Surface DC Voltmeter is properly set at zero and then brought into the vicinity of the center of the insulator sheet, with the sensor disk pointed toward the insulator sheet. Then the digital display will read a certain number, and that number will not change significantly when you change the spacing between the meter and the insulating sheet (as long as the spacing remains significantly less than the width of the insulator sheet and as long as the sensor disk is pointed directly toward the sheet).

The actual electric field "E" produced by the insulator sheet can be deduced by the reading on the display "V_displayed". It turns out that E = 28.5 · V_displayed, (if V_displayed is expressed in volts), or E = (28500) • V_displayed (if expressed in kilovolts). For example, if the display reads V_displayed = 1.000 ("kilovolts") near a charged insulator sheet, then the actual electric field produced by that insulator is E = 28,500 (= 2.85 x 10⁴) V/m.

When V_displayed is "1.000", it turns out that the charge per area on the sensor disk is S = 5.04 x 10^-7 C/m². (This is of course also the ‘charge per area’ required on an insulating sheet in order to produce an electric field of E = 28.5 KV/m, as was discussed in the previous paragraph.) Therefore, the value of S on the insulator sheet is proportional to V_displayed on the meter, with a reading of V_displayed of "1.00" (KV) equivalent to S = 5.04 x 10^-7 C/m², or E = 28.5 KV/m.

If the conductive or insulator surfaces that are being measured are not very large, the meter readings will be lower than they should be. This is because the surface does not cover the entire horizon as seen by the sensor disk. Formulas can be used to correct this error.

If you are measuring the surface charge or electric field of an insulator sheet that is far from a grounded surface, and the insulator sheet is fairly small, with a diameter (or an average width) of "D", and you are measuring a distance "L" away from the insulator sheet, then multiply the V_displayed by the factor 1 + 4L²/D². Note that if the diameter of the insulator sheet (D) is much greater than the distance (L) between the sensor disk and the sheet, then L/D is almost zero, so the above factor is just one (unity). In this case the sheet completely dominates the view as seen by the sensor-disk. But the factor is greater than 1 if L and D are roughly the same or if L is greater than D. In this case the sheet only covers part of the sensor disk’s "view", and the factor is greater than one. The distances "L" and "D" can be any units, such as inches, as long as both are expressed in the same units. Therefore, if the insulator sheet is not large enough to cover the sensor disks entire "view", from horizon to horizon, then you should multiply V_displayed by the

‘Correction Factor For Insulators’ = 1 + 4L²/D²

to find out what an infinitely large insulator sheet (with the same value of S) would have read. (The previous paragraphs assumed an infinitely large insulator sheet when it was stated that a meter reading of "1.000" corresponds to S = 5.04 x 10^-7 c/m² and to E = 28.5 kV/m.) Then you can calculate S and E for the insulator sheet.

If you are instead measuring the voltage V of a conductive sheet, there is one more factor to consider. On a large conductive sheet held at voltage V, as the sensor is pulled away, the number V_displayed will read significantly less and less, even if the sheet is so large that it always completely covers the sensor disk’s view. When the sensor disk is much less than L = 1" away the reading decreases as L^-1 (one over L), but if the disk is more than L = 1" away, the reading decreases more gradually, at about L^-1/2 (one over the square root of L). This decrease is independent of the horizon effect (which would be the factor 1 + 4L²/D² if the sheet were an insulator, but is different for a conductor). There is also a horizon effect factor for a small conductive sheet, but this is the square root of the equivalent factor for an insulator.

The sensitivity, which falls off as L^-1 close to a conductive sheet and L^-1/2 far from it, is given by ‘Relative Sensitivity’ = (1/[2.5L])^{[(L + .5) / 2L + .5]}, where L is in inches. (Please see the formula for ‘Correction Factor’ below for the final form of the correction.) This is the number 1/2.5L raised to the power (L + .5) / (2L + .5). Note that (L + .5) / (2L + .5) is between ½ and 1, no matter what positive value of L is selected; at L = 0, it is 1, and if L is very large, it is ½. This relative sensitivity, which applies to a very large conductive sheet, is a larger number for L< 1", and a smaller number for L>1". The ‘relative sensitivity’ can be evaluated, for example, at L = 1" (where ‘relative sensitivity’ is .577) and at L = 5" (where it is .266). Note that .266/.577 = .461; therefore, if the V_displayed was 1.000 at a distance of 1", it will be 0.461 at 5". The reading at 5" can be corrected to what it would have read at 1", simply by multiplying the 5" reading (of 0.461) by .577/.266, and this principle can be applied to readings of L<1" also. In general, multiply the reading (if taken at a distance L that is>not 1") by the factor .577 /’Relative Sensitivity’, which is the factor (.577) • (2.5L)^{[L + .5 / 2L + .5]}. Note that the "1 / 2.5L" became just "2.5L", which is the inverse. Remember also that "L" must always be measured in inches. For a small conductive sheet, the horizon effect factor is

(1 + 4L² / D²)^1/2. Therefore the entire factor to multiply the meter’s reading V_displayed by is:

‘Correction Factor For Conductors’ = (1 + 4L² / D²)^1/2 • (.577) • (2.5L)^{[L + .5 / 2L + .5]},

where "L" is the ‘sensor disk’ – to – ‘conductive sheet’ distance, in inches; and D is the effective diameter of the conductive sheet in inches.   [back]

[3] How ions add (or remove) surface static electricity

If an air ion collides with a solid surface, it will usually exchange an electron with the surface. The air ion will then become neutral and will break apart into ordinary separate molecules of H₂O, O₂, etc. If the air ion was originally negative, it will add an electron to the surface. A positive air ion will remove an electron from the surface, leaving it positively charged.

If the surface is metal and is connected by wire to earth ground, then the ion’s charge will flow back to ground through that wire. If the surface is instead non-grounded metal, the voltage will become more and more positive (negative) as more and more positive (negative) ions collide with it. Such a metal surface can be discharged back to ground (V = 0) at any time by touching it to a ground wire. If positive (negative) ions continue to collide with a non-grounded metal surface, the voltage of the surface will continue to increase positive (negative), until the ions are so strongly repelled by the surface that the ions stop colliding and veer away. Then the surface has reached its maximum voltage for that particular situation. If the positive (negative) ion source is then positioned closer to the metal surface, or if the speed of the ions is increased, then the metal surface’s voltage will become higher. Conductive surfaces that are near negative ionizers may reach several thousand volts. If near a neutral ionizer, a surface may sometimes reach several hundred volts simply because the negative and positive ions may not be quite equal in number.

In contrast, insulator surfaces that are charged because they were rubbed or were exposed exclusively to positive (negative) ions, cannot be discharged simply be connecting a ground wire to the insulator. Charges sitting on the surface of the insulator cannot slide along the insulator as would be the case with a metal surface. To remove a single charge from an insulator surface, that charge must either be neutralized by colliding with an air ion of opposite polarity, or the single charge must be touched (within a few atomic diameters of the surface) by a grounded conductor. Once the surface charge is removed, that region will generally carry no evidence that it had ever been charged. This means that air ions do not "scar" the surface.

You might think that a charged insulator surface can be discharged simply by pressing a grounded metal surface against it. However, this usually doesn’t work. The metal surface may increase the electric field on the charged insulator, to as much as twice normal, as the metal surface is being brought closer and closer. This increases the chance that a spark will occur. If the surface carries enough charge, there will be a spark, and ions can flow through the air and discharge the area(s) near the spark(s). This will remove some of the charge. If the insulator doesn’t have enough surface charge, it will not spark, and this method would then only remove charge in the regions of the insulator where the grounded metal was only a few atomic diameters away from the insulator. The metal sheet only makes contact at a few raised areas of the insulator, so that typically less than 1% of the insulator can ever come within a few atomic diameters of the grounded metal. Therefore, over 99% of the surface may still be charged.

The insulator sheet can be discharged fully by a different technique: dip it in grounded water or wipe with a damp, grounded cloth or paper towel. This allows all areas of the insulator to come into contact with the water. The damp cloth technique is more convenient because, if done correctly, it leaves no water spots. Remember to let the final 'residual water' air-dry. (If you wipe with a dry cloth, you will charge up the insulator again).

Metal wires with sharp points, or metal strips called "tinsel", are often used to discharge insulator surfaces by grounding the wires and passing them over the insulator. The electric field at isolated, sharpened, grounded wires is usually much higher than the electric field at the surface of the nearby charged insulator. This is because the voltage change (over distance) is very large near each point, where it might go from thousands of volts one centimeter away from the point, to zero right at the point. Therefore each point will often initiate a spark and shower the nearby charged insulator with the appropriate polarity of air ions to discharge it. This discharge is only partial, however; when the electric field of the insulator becomes weak enough due to loss of charge, sparking stops.

Ions are sometimes used to make insulating objects that carry a permanent charge or produce a permanent electric field. Such objects would be like magnets, but would produce electric field instead of magnetic field. Ions may be either coated on the surface, or implanted fairly deep into the insulator surface using high velocity ions. This type of electric-field-source is called an "electret" (which may also be made by heating & cooling certain types of materials while in a strong electric field, analogous to the way magnets are produced). Electrets face a problem that magnets do not: the air is full of electric + and – charges, but as far as is known, magnetic north or south charges (poles) do not exist as separate entities. Therefore as an electret object is exposed to a typical atmospheric environment, ions start colliding with it and deposit their charges in exactly the right way as to reduce the overall electric field, eventually to zero. It doesn’t matter whether the ion charges are embedded deep in the electret (far from the surface), or whether the electret is charged just at its surface; air ions will eventually "short" the electret out. If you try to remove the extra 'ions from the air that are stuck to the surface' by dipping in water, not one of them will budge. In fact, the water dip is a good way to "short" out the electret permanently, even if unwanted ions from the air had not previously shorted the electret.

Compare this to a magnet, in which no north or south poles are freely floating in the air. If there were such single poles, the north ones would stick to the south pole of the magnet and eventually "short" it out, as would the south ones stick to the north pole of a magnet. Any electret that is manufactured will therefore have a finite life, and should be stored away from air ions.

In general, a neutral ionizer (producing both positive and negative ions) can discharge any charged surface, if given enough time (see next explanation, [4]). There must be appropriate air circulation, however. Otherwise the ions might not reach the charged surface.  [back]

[4] How to Measure Accidental and Deliberately-Produced Ions

Air ions may collide with the sensor disk. If an excess of positive (relative to negative) ions are colliding with the disk, the digital display will read a higher and higher positive number (or if it starts with a negative number, it will read a smaller and smaller negative number until it crosses past 0.000). Of course, an excess of negative ions instead would make the display read a more and more negative number.

If the display is "RESET" to zero and the sensor disk is held out in the air under normal conditions, the display will remain stable. Here’s why: Typically the air (provided that an ionizer or ion source is not present) has fewer than 2000 ions per cubic centimeter, and air which is nearly stationary cannot transport ions very rapidly anyway. Therefore, very few ions would collide with the sensor disk under typical air conditions, and the displayed number will remain stable. However, if an ion source (or electrical contacts that are sparking) is nearby, the display may begin increasing if they’re mostly positive ions, or decreasing below zero if they’re mostly negative. A total amount of charge of .33 nanocoulombs (.33 x 10^-9 C) on the sensor disk will cause a reading of "1.000 KV". It is proportional, so that 3.3 nanocoulombs (nC) will cause a reading of "10.000 KV". This amount of charge, 3.3 nC, is equal to 2.1 x 10¹⁰ ions. (One ion carries or lacks one electron.) Therefore, if you "RESET" the meter and then (after releasing the RESET button) you see the display climb to 10.000 in a certain number of seconds, then the sensor disk has had an excess charge of 3.3 nC or 2.1 x 10¹⁰ ions collide with it during those certain number of seconds. "Excess" here means the amount of positive charge minus the amount of negative. For example, if 5.3 nC of positive-charged ions and 2.0 nC of negative-charged ions hit the sensor disk, the total charge on the disk will be +3.3 nC. (Even though the display will read "10.000 KV", the sensor disk is only actually at a small fraction of that voltage – it’s only at 1 volt. This is because the disk is connected through a .0033 microfarad capacitor to ground.)

Remember that as a (non-grounded) surface is bombarded by only one polarity of air ions, that surface will eventually become so strongly charged that it will repel any more ions of that polarity. This effect will never happen to the sensor disk, however. Because the sensor disk is connected through a capacitor to ground, it always stays within a few volts of ground, (instead of the much higher voltage it would have if the capacitor weren’t there). The meter’s circuitry amplifies that signal to turn it into the correct displayed number. Therefore if you "RESET" the meter and then point the sensor disk toward an ion source, and multiply the meter’s reading in "KV" by .33, you’ll get the amount of charge in nanocoulombs that has hit a 1" grounded disk, in the amount of time since the last "RESET". For example, if a certain number of seconds after "RESET", the display has climbed to "10.000", it means 3.3 nC has accumulated on the sensor disk in that certain number of seconds (but the number of seconds doesn’t enter in the calculation). Therefore the meter can measure how much total charge is being deposited onto surfaces, in units of ‘nanocoulombs within a one-inch circle’, during whatever time interval you choose to hold the sensor disk in the vicinity of the ion source. You start the measurement time by releasing the RESET button, and end it by pressing (and continuing to press) HOLD. (Instead of using HOLD, you could end the measurement time by rapidly removing the meter from the ionized region). Multiply the displayed number by .33 to get units of nC per 1" circle. (Multiply the reading instead by .42 to get nC per square inch or by .067 to get nC/cm², or by 6.7 x 10^-7 to get C/m², which is the standard unit of "S", surface charge per area).

Clearly, you can use the meter to measure whether any significant amount of charge is accumulating on an uncharged surface due to a large number of air ions (which must first have sufficient speed to collide with the surface). Once you measure this charge per area (S) on a surface, you can calculate its electric field

(E = 5.66 x 10¹⁰ S if "S" is in C/m²), and you then know how much charge would need to be added to a given area to neutralize all its surface charge.

If you don’t want the meter reading to climb upward or downward when a large number of ions are in the air, you will need the optional "charge plate detector" kit. Snap the white ion shield over the sensor disk. If a large number of positive (negative) ions are present, the outside surface of the shield will quickly charge to a high positive (negative) voltage, which will prevent any more ions from hitting the ion shield. After the shield has charged sufficiently to repel further ions, the display will become stable. Then "RESET" the meter and use it in the normal way (the same way as it would be used without the ion shield), except that when doing the RESET, the sensor should be pointing at a grounded sheet that is at least 6" (15cm) away, or at the room’s ceiling. (If the grounded sheet is closer than 6" away, there may be some ‘charge image effect’ where the charge on the top of the ion shield will interact with the grounded sheet and cause an inaccurate zero reading.

An important test of a neutral ionizer (designed to neutralize the charge on surfaces whether that charge is positive or negative) is how fast the ionizer can reduce the surface charge down to ½ its original value. This amount of time is defined as the ‘charge half-life’ and its value depends on the distance away from the neutral ionizer and on the amount of air circulation. To measure the charge half-life, you will need the optional "charge plate detector" kit. Take the ion shield (the white plastic rectangle) and discharge it. To do this, wipe both the top and bottom surfaces with a slightly damp cloth or paper towel. It is better if the damp cloth is in contact with both sides at the same time as you wipe. This will short any positive charge on one side to negative on the other. This discharge step must also be done before using the white sheet as an ion shield in general. Then only hold the ion shield on its edges, near where two "+" or two "-" stickers are. The next step is to snap the ion shield onto the rear of the meter, but note that you have a choice of "+" facing outward or "-" facing outward. If you want to measure the charge half-life of positive charges, snap the ion shield onto the meter so that two "+" stickers face outward (away from the sensor disk). Turn the meter ON and RESET it, with the ground wire connected. (The display should read near zero now.) Then rub the ion shield ("+" side) with the "+" side (fuzzy side) of the charging foam, until the meter reads at least 5.000, and note the actual reading. Then bring the meter into the region that is to be tested and count how many seconds are required for the display to reduce to one-half its original value. This time is the charge half-life.

If it’s too fast to time accurately, then recharge the ion shield to over 5.000, and hold the meter in the ‘region to be tested’ for only one second, and rapidly remove the meter. Note this reading. Call this V₂. (Call the initial reading V₁, which was 5.000 KV or more). Then place the meter back in the 'region to be tested' at least another 5 seconds and remove the meter from that region. Call this final reading V₃. The 'charge half-life' is T_1/2 = .31/log(V₁/[V₂ - V₃]). For example, if the reading drops from 5.500 to 0.200 in 1 second and V₃ is -0.150, then the charge half-life is .31/log(5.5/[.2 - (-.15)] = .31 log[5.5/.35] = .259 sec. Note that the "log" is base ten.

If, on the other hand, too much time is required for the V_displayed to drop to 1/2, and you don't have the patience to wait that long, note the initial reading V₁, (over 5.000 KV), and then place the meter in the 'region to be tested' for 10 seconds, and then remove the meter from that region. Now note the new reading as V₂. You don't need to measure V₃ in this case. Here T_1/2 (slow discharge) = 7.2 · V₁/(V₁ - V₂). For example, if V₁ = 5.500 and V₂ = 5.471, then T_1/2 = 1370 sec.

If you want to measure the charge half-life of a negatively charged surface, remove the ion shield, wipe it with a damp cloth to discharge it, and snap it back on, now with the "-" stickers facing away from the meter. Then charge to -5.000 or more by rubbing that surface with the "-" (foam) side of the charging foam. Repeat the measurement method of charge half-life as above.

The charge half-life is a useful number because the amount of surface charge is reduced by half for every time a charge half life elapses. For example, if the charge half-life is .22 second, then while exposing the surface to neutrally ionized air, after .22 second, the charge will be down to ½. After .44 second, it will be ¼, and after .66 second, it will be 1/8. This will tell you how long a charged object must be exposed to the ionizer to achieve a certain amount of charge reduction. [back]

[5] Surface Conductivity of Materials

In general, most plastics are excellent insulators, with resistance in the range 10¹⁵ ohms per square or higher. (‘Ohms per square’ is in "Definitions"). At 10¹⁵ ohms per square, a 10 cm square sheet will leak so slowly that it will require about 15 hours to discharge to one half its original charged value. (This assumes there are no ions in the air, which would discharge the surface faster, and that one edge of the 10 cm square is connected to a wide metal connector which is grounded.) Epoxy is more conductive, and may be as low as 10¹¹ ohms per square, depending on the type. At 10¹⁵ ohms per square, plastic which is rubbed with a dissimilar material will often acquire a high voltage. This high voltage is difficult to discharge unless the plastic can be coated with a conductive film. One color of the "Krylon" brand of solvent-based spray paint is slightly conductive: it is "semi-flat black". All other colors are essentially non-conductive. A single layer of the semi-flat black is about 10¹¹ ohms per square, and may be sprayed on most plastics except polyethylene or polypropylene. When painted with this, surfaces become a slightly worse conductor at high humidity than at low, which is counter-intuitive.

Glass, paper, cardboard, wood, concrete, soil, and most rocks are much more conductive than plastics. They all become much more conductive at high humidity. Glass is typically 10¹² ohms per square at 40% humidity and 10¹¹ at 70%.  [back]

[6] Measuring Surface Resistance (Ohms Per Square)

If you want to measure the resistance of a square surface and the resistance is very high, then an amount of charge should first be applied to the surface. After the charge is applied, one edge of the square surface should be connected to ground, and you can measure how much time it takes for the center of the square to reduce its voltage to one half (T_1/2). This is the charge half-life. The number of ‘ohms per square’ is then approximately 10¹¹ T_1/2/L, where T_1/2 is measured in seconds, and L is the length of the square’s edge, in meters.

There are certain rules concerning how the measurement must be done. The first step is to charge the surface. The surface must not be electrically connected to ground or to any other voltage. If the surface to be tested is a separate piece, like a tile, place it on top of a plastic (not cardboard) box. This will be a good insulator. The piece can be charged by the "induction" technique. With this method of charging, a charged object is held near the piece that is to be charged. Then a ground wire is touched to the surface of the piece and held there so that charges can flow along the (weakly conductive) surface. It’s best to allow the charges to flow at least several seconds if the conductivity is fairly weak. Then remove the ground wire. Only after the ground wire is removed, then remove the charged object from the vicinity of the piece to be tested. The piece will now have the opposite polarity of the charged object. (For the "charged object", you can use the ion shield, by rubbing either the "+" side with the "+" side of the charging foam, or "-" with "-".)

There is a second charging method besides the "induction" technique: The plastic box that is being used as a platform can be charged directly by rubbing with one side of the foam (the best polarity depends on what type of plastic the box is made of.) Then the piece to be tested is dropped onto the plastic box. (Avoid touching the piece while it is in contact with the charged plastic box. While holding the piece, you will gradually discharge it.)

Once either the box beneath the test piece or the test piece itself is charged, use the meter to measure the charge half-life. With this ‘ohms per square’ measurement, the half-life is measured while one edge of the test piece connected to ground.

If there are no external high-voltage sources (so that it is safe to touch the system with your finger), then it is easiest to use your finger as the ground connector. To measure charge half-life, decide what the approximate diameter or average of the width and length of the test piece is. Call this distance "L". "RESET" the meter (while pointed to ground, as usual), and then point the sensor disk toward the center of the test piece, but the sensor disk should be approximately a distance "L" from the test piece, instead of one inch away. Bring your finger or a metal foil strip into position so it can be easily dropped onto one edge (this edge should be approximately length "L"), but hold it about ½" over the edge, so it does not yet make electrical contact with the test piece. Read the meter and then drop your finger or a grounded metal foil into contact with the edge. Determine how many seconds are required for the reading to go to about one half. This is T_1/2. If the reading drops very quickly so that T_1/2 is less than one second, then redo the charging step and measure the initial meter reading again. (Call this value V₁.) Then hold your finger or foil on the edge for exactly one second and then remove it from the edge. Record the reading, and call it V₂. Then ground the edge one more time for at least 5 seconds. This final reading is V₃.

Then

T_1/2 = .31 / log[(V₁) / V₂ – V₃]. (Log is base 10.)

If the reading instead drops very slowly so that T_1/2 is over 100 seconds and you don’t want to wait that long, then read the meter initially just after you ground the edge (not before). Then measure the amount of time required for the reading to drop by 1%. (For example, from 5.000 to 4.950). Call this time T_1%. Then

T_1/2 = 68T_1%

No matter which of the three methods is used to determine T_1/2, the number of ohms per square of the piece is R = 10¹¹ T_1/2 / L, where R is in ohms (per square), T_1/2 is in seconds, and L is in meters. For example, if L = 10cm (.1m), and T_1/2 = 5 sec, then

R = 5 x 10¹² ohms per square.  [back]

[7] Force Per Area (Pressure) Between Charged Surfaces

If one surface carries charge per area of S₁ and another carries S₂, then when the two surfaces are held close together (much closer than their width), there will be an attractive (or repulsive) force per area given by

Force/Area = 5.7 x 10¹⁰ S₁ · S₂ [newtons/m²]

The force is repulsive if both surfaces have the same charge and is attractive if the surfaces are of opposite charge. The units are force (in newtons) divided by area (in square meters). An example is if both S₁ and S₂ = 10^-5 coulombs/m², then

F/A = 5.7 x 10¹⁰ • 10^-5 • 10^-5 = 5.7 newtons/m²,

where a Newton is the weight of about 102 grams. Expressed in units of gram (force) per square centimeter, is Force/Area = 5.8 x 10⁸S₁S₂ [gramforce/cm²].

As long as the two surfaces are insulators, they will maintain most of their separate charges even if the surfaces touch each other. However, if one surface is conductive, something very different may happen. If a conductive surface is brought near a charged insulator surface and then the conductive surface is connected to ground, charge will flow in the grounding wire until the value of S on the conductor is the negative of the value of S on the insulator. Then the overall electric field outside the immediate area goes to zero, but if you attempt to separate the insulator from conductor, they will be attracted to each other with

Force/Area = 5.7 x 10¹⁰(S)² [newtons/m²],

and there will be a strong electric field only in the region between the insulator and the conductor. (The electric field strength is 1.1 x 10¹¹S there.) This attraction will occur after the conductor is momentarily grounded regardless of what charge the conductor originally carried. Sometimes, the conductive surface may momentarily ground itself without needing a ground wire connection. This happens if the back of the conductor creates an arc (at a sharp point) and ejects many ions into the air. Then the conductor will be attracted to the charged insulator because it carries opposite charge. This attraction can happen even if the "conductor" is only weakly conductive, like paper.

However, the behavior of weak conductors is often unpredictable. Sometimes when a weak conductor, or an insulator, is brought near a charged surface; the charged surface will emit ions that charge the weak conductor with the same polarity as the charged surface. Then the weak conductor is repelled from the charged surface. In general, when two charged surfaces are brought into contact (or near contact), and the higher charge per area from among the two is "S", then the maximum force/area between the two surfaces is

5.7 x 10¹⁰S²[newtons/m²]

or

5.8 x 10⁸S²[gramforce/cm²],

and this force may be either attractive or repulsive, depending on details of how the charge flowed during contact. Good conductors, like metal, are more likely to be attracted to a charged insulating surface after near contact. Poor conductors and insulators are more likely to be repelled. A good insulator however, will not change its value of S if exposed to another insulating surface, as long as that other surface is not very strongly charged. (If it is charged enough that it is emitting sparks, then the strongly charged insulator will, because of emitted air ions, change the value of S on the other insulator.)  [back]

[8] Measuring the Voltage Differential Through an Insulator Sheet or Film

A sheet or film of insulator may have a voltage difference across its thickness. This voltage difference is equal to the product of ‘electric field inside the insulator’ multiplied by the ‘insulator’s thickness’. (Because the sheet is not empty space and also not just air, there is a fine distinction between what is the definition of ‘electric field’ inside vs. outside the sheet. In this application, the distinction is not important). The voltage arises because there is a difference between the amount of charge per area on the top face of the sheet vs. the bottom face. If the charge per area on the top and bottom faces were equal and of the same polarity, the voltage difference would be zero.

If you simply measured the sheet by holding the Surface DC Voltmeter nearby, the sheet would act as an insulator with net charge equal to the positive charge minus the negative charge. This is not really a "voltage", because the sheet is not a conductor. Instead, the meter will display a number that is proportional to the electric field strength. Furthermore, it will display approximately the same number whether you measure from the positive surface or the negative surface. There is, however, a well-defined voltage difference between the surfaces. One way to measure it is to put a grounded metal sheet against the negative side and measure the voltage 1" from the positive side. This will give a positive number (in kilovolts). Then reverse the sheet to get (probably) a negative number, being careful not to connect both the positive and negative surfaces to ground at the same time, or to short the two surfaces together. The sum of the positive number + the magnitude of the negative number is the number of kilovolts of voltage difference through the surface. If the voltage reads positive on both sides (or negative on both sides), then you should subtract the two positive numbers (or subtract the magnitudes of the two negative numbers) from each other to get the true voltage difference across the sheet. If you know that only the top face is charged, then you only need to ground the bottom face and measure the top face. In this case, the bottom face would read zero anyway.  [back]

Definitions

    Air Ions – These are clumps of several atoms and molecules which are normally in the air (such as H₂O, CO₂, oxygen, etc.). Each clump is held together by either an excess electron(s) or a missing electron(s), giving the clump either negative or positive charge. Each clump contains a larger number of atoms than would normally be found in (uncharged) air. The most common type of negative ion (with one excess electron) is a 2-atom oxygen molecule (O₂) surrounded by several water molecules; if an electron is shot out into the air, this mix of oxygen and water will quickly surround the electron and carry it along, forming a large, negatively-charged particle. Typical positive ions contain a hydrogen atom or an ammonia molecule, along with (possibly) water, carbon dioxide, or nitrogen. When an ion collides with a solid surface or with another ion of opposite polarity, the charge that held the ion together is lost. Then the ion breaks apart into normal air (O₂, H₂O etc.).

Ions are created by energetic events, such as a high-voltage spark (generally requires over 500 volts), a hot object (generally at least visibly glowing orange-hot), combustion or smoke, or radioactive decay (a typical 5 MeV ejected alpha particle will create about 50,000 negative and 50,000 positive ions as it travels a few centimeters through air, losing energy until it comes to a stop). These various energetic events are capable of producing negative and positive ions. One common process can create only negative ions: it is the evaporation of water in the presence of oxygen (or air). When water evaporates, approximately one water molecule in every 3 x 10¹³ carries an excess electron with it. This molecule quickly associates with an oxygen (O₂) and with other water molecules to form a stable negative ion. This process is responsible for some of the electrical activity in the atmosphere.

Air ions are affected by an electric field. The speed of travel is proportional to the strength of the field. In a field of 10⁴ V/m, typical-sized ions will travel at a constant (non-accelerating) speed of 1 to 1.5 meter/sec. It is typical for indoor ions to last on average roughly one minute before colliding with an object and discharging their single charge back to ground. A highly-charged positive surface will attract negative ions, which collide with the positive surface and therefore reduce its charge. Likewise, highly negative surfaces attract positive ions. For this reason, air ions are often deliberately injected into a room to control static buildup. The air can contain large numbers of both positive and negative ions simultaneously without neutralizing each other to any great extent. Typical outdoor air in fair weather contains 200 to 1500 positive ions and about the same number of negative ions, per cubic centimeter. Indoor air usually has fewer ions, because there are charged surfaces robbing ions out of the air. When ions are deliberately introduced, it is usually 10⁵ or more per cm³, of both positive and negative. If a charged surface is nearby, ions of the opposite polarity will be attracted and will slowly neutralize the surface. The speed of neutralization will depend on the number of ions per cm³ that are present.  [back]

    Electric Field – This is an absolute quantity and it is a vector. That is, it has both magnitude and direction. It is "absolute" (unlike voltage) because it can be measured at any point in space and it has an unambiguous value, which depends neither on how it is measured nor on any relative "zero" level. Electric field acts on an electric charge in the same way that gravity acts on a mass (except that there exist negative charges, which are accelerated in the opposite direction from the way the electric field is pointed. With gravity however, there is no such thing as "negative mass"). The electric force that acts on a ‘test charge’ (which is a single electric charge put at a particular location and then released, so it will be moved by the local electric field) is the product of the strength of the electric field measured at the location of the ‘test charge’ (but not including the electric field that is self-produced by the ‘test charge’), multiplied by the amount of charge in the test charge. Electric field is usually expressed by the ratio ‘volts per meter’ (V/m). This is in units of voltage divided by length. However, if we accept the previous statement that force = ‘electric field’ multiplied by ‘charge’, then we would expect ‘electric field’ to be in units of force ¸ charge. In standard international units, force ¸ charge is in units of newtons per coulomb (Nt./C). (A newton is the weight of about 100 grams or 1/5 pound, and a coulomb is the charge in one amp · second.)

As it turns out, 1 V/m = 1 Nt./C exactly. How then, do volts and meters (length unit) relate to electric field? Actually, the Nt./C expression for electric field makes it intuitively clear that in an electric field of 1 Nt./C, that one coulomb (C) of charge will experience a force of one Newton (Nt.). With that same strength of electric field, rewritten as 1 V/m, the voltage will rise by one volt for every one meter distance you travel "upstream" against the electric field direction. As previously mentioned, if one coulomb of positive ‘test charge’ is placed in this electric field of 1 V/m, the ‘test charge’ will experience a downstream force of 1 newton. In a region of space where the electric field is 2 V/m, the downstream force on the one coulomb ‘test charge’ is 2 newtons.

Technically, the electric field is the negative of the (vector) gradient of the voltage. Electric field is emitted by a surface if that surface contains a quantity of excess (negative or positive) charge per surface area. If a sheet of insulator, (not metal) contains excess positive charges which are uniformly distributed along the sheet and are distributed with a ‘charge per area’ equal to one coulomb per square meter, then the electric field emitted by the charges on the sheet will be 5.66 x 10¹⁰ V/m. The field direction is perpendicular to the sheet, pointing away from the sheet. The field on the back side of the sheet is the same strength, but pointing 180 degrees away (in the opposite direction), again away from the sheet. (It’s actually very difficult to get one coulomb per square meter of charge onto a sheet, but the relation given above is proportional, so that a charge per area of only 10^-10 coulombs per square meter will create an electric field of only 10^-10 as much as above, or 5.66 V/m). The field doesn’t decrease much as you measure it farther and farther from the insulating sheet unless you get a distance away that is approximately equal to ½ the width of the sheet. Then the electric field strength will decrease significantly if you measure even farther away. The fact that the electric field is of approximately the same strength whether you measure it right on the surface or a distance away up to ½ the sheet’s width or diameter (although measurement should be done close to the center of the surface of the sheet), is a property of a charged insulator, on which the trapped excess charges can’t wander along the surface. A large charged sheet of metal is different; its electric field becomes significantly weaker as you move the sensor away, even if you start with the sensor very close.  [back]

    Ionizer (British: Ioniser) – A device which adds ions to the air. For anti-static purposes, a ‘neutral ionizer’ adds an approximately equal number of positive and negative ions to the air. (There are also negative ionizers which are used to clean the air by causing clumping of particulates, but negative-only ionizers can only discharge surfaces that are positively charged.) Neutral ionizers generally are one design of these two possibilities: high-voltage needle-type, or radioactive type.

The high-voltage needle ionizers have one or more metal needles that are near a grounded metal surface. The needles are connected to high-voltage AC (generally more than 3000 VAC). While the needles are in the positive phase of the AC wave, they shoot positive ions as far as 100 cm forward into the air. Then while negative, they shoot negative ions up to 100 cm away also. To distribute ions farther than about 100 cm, the air must be circulating, such as from the action of a fan. Ions only remain "alive" approximately one minute, and this must be taken into account when setting up the circulation pattern.

The radioactive ionizers are usually small brushes which are used, for example, to clean dust off photographic negatives. A small speck of radioactive polonium or americium is shielded on the sides, but shoots alpha particles forward toward the area that is just about to be wiped by the brush. The 50,000 pairs of ions produced by each alpha particle will flood the area within 5 cm of the radioactive source and will discharge all surfaces in that region.  [back]

    Ohms per square – This is a measure of surface resistivity. If a plastic square is coated with a thin layer of graphite, and then the connectors from an ohm meter are attached to two opposite edges of the square, using special wide clips that make contact all along one edge and all along the opposite edge, the ohm meter will read a specific resistance. If the square is then cut into 4 equal smaller squares (quarters), and wide clips make contact with both edges of one of these quarter squares, the resistance will be the same as with the big square. In fact, if you begin with a large sheet of slightly conductive material, and cut out any size square, the resistance measured across opposite edges will be the same, independent of the size of the square. This number is the ‘ohms per square’ of the surface. Given that R is the number of ‘ohms per square’ of a surface, and D is the length of one side of the square (in meters), then the charge half-life is approximately T_1/2 = 10^-11 • D • R. For example, if a particular material has a resistance of 10¹² ohms per square, and a sample was cut out of size 10 cm x 10cm (= .1 x .1m) in size, and you grounded the full (.1m) length of one edge, the average charge would reduce to one half in 100 seconds, which is T_1/2.   [back]

    Sensor disk – This is the round brass disk on the rear face of the Surface DC Voltmeter. This should be pointed toward the surface or object to be measured. If that surface is charged, some charges of the opposite polarity will rapidly flow through the earth grounding wire (connected to the meter) until the sensor disk has the same amount of electric charge per area (but of opposite sign) as does the surface that is being tested. The digital display then reads a signal proportional to this amount of charge. (Technically, the sign of the charge on the sensor disk is reversed.) The sensor disk is connected to ground through a capacitor, and this means that the sensor disk is at a much lower voltage than the test surface is. For example, if the display reads a magnitude of "10.000" kilovolts, the sensor disk is only at about 1 volt. The (low) voltage on the sensor disk is directly proportional to the digital reading, with about 2 volts corresponding to a full-scale reading of "19.999".

In addition to charging as a result of proximity to charged objects, the sensor disk may accumulate charges from air ions, from sparks, or from direct contact. Note that when the sensor disk is charged by proximity to a charged object, the sensor disk will then return to zero charge when removed from proximity to the object, so this kind of charging is "reversible". On the other hand, sparks (etc.) will charge it and the charge will remain until it is "RESET". Because the sensor disk is connected through a .033 microfarad capacitor to ground, and a voltage of 1 volt on the disk will cause a digital reading of "10.000", then such a reading means that the amount of charge on the disk is 1 • 3.3 x 10^-9 = 3.3 nanocoulombs, which means that at full scale (19.999), the disk will have about 6.6 nanocoulombs on it, which can be either positive or negative. When you press the RESET button, all of the accumulated charge on the disk is rapidly shorted back to ground, but this operation only works properly if the disk is not in an electric field. That's the reason that your hand (or other conductor that is at the same voltage as the ground connection to the meter) must shield the sensor disk when you do the RESET function.  [back]

    Voltage – This is a relative quantity. That is, the voltage difference between two points in space can be measured, but it’s difficult to assign an absolute voltage by itself at one particular point. A charge moving from a point in space that is at a certain voltage -- to another point at another (relative) voltage -- is a situation that is physically almost identical to a weight being moved from one altitude to another altitude; it takes work to raise a weight from a lower altitude to a higher altitude, and the amount of work is proportional to the altitude change. Similarly, it takes work to raise a positive charge from a point that is at a certain voltage to a point that’s at a higher voltage, and the amount of work required is proportional to the voltage change. Just like the assignment of an absolute altitude is arbitrary (should we call the "zero" altitude sea level, or land surface level, or the center of the earth, or something else?), the assignment of "zero" voltage is arbitrary.

Any perfect electrical conductor will have the same voltage anywhere inside itself. That is, it takes no work to move an electric charge from one part of a conductor to another part. For this reason, the earth ground is often defined as zero voltage, because the earth is a fairly good conductor. However, this "zero" is not precisely defined; there may be a small voltage difference between two different points miles apart on land. The atmosphere is constantly contributing electrical charges to the surface in a non-uniform way. If these excess charges can’t mix fast enough over the earth’s surface, there may be a slight excess, for example, of positive charges on the surface in one area. This will cause that area to have a slightly more positive voltage than the average of the earth’s surface. However, within an area as small as a building, the earth ground voltage at one location under the building is almost always within one volt of the voltage in any other location under the building.

If one region in a room has an excess of negative charges and a nearby region has excess positive charges, the positive region will have a higher (more positive) voltage, and the negative region will have lower (more negative) voltage. Any free positive charge will accelerate toward the most negative voltage while any free negative charge will accelerate toward the most positive.

To define "volt", if you slowly move a single proton from one position to a new position where the voltage is one volt higher, you’ve done 1.6 x 10^-19 joules of work. Similarly, if you move one coulomb, which is one amp-second of charge, up by one volt, then you’ve done one joule (or one watt-second) of work.  [back]