Unit Systems [Electrostatics (1958)]

9. Basic Systems

Although the English system of weights and measures is more familiar to the layman, virtually all scientific work is carried on in units that stem from the metric system. In this country, mechanical and civil engineers still use the foot-pound-second (fps) system, but other branches of engineering have adopted metric units more or less unanimously.

In the metric system, the fundamental unit of length is the meter; originally intended to be one-ten-millionth of the distance from the North Pole to the equator around a great circle, the meter is now redefined as the distance between two scratches on a platinum iridium bar located in the Bureau of Standards in the United States. Since a meter contains 39.37 inches, it is approximately a yard in length. One distinct advantage of the metric system is that it is decimally organized. (Fundamental units are multiplied or divided by factors of ten to form new, conveniently-handled sub- and multiple-units.) In the chart that follows, the standard prefixes are italicized; these prefixes are invariably used to mean the same thing regardless of the units to which they are assigned.

Fundamental length unit = 1 meter

1 micrometer ... 10^-6 meter

1 millimeter ... 10^-3 meter

1 centimeter .. .10^-2 meter

1 kilometer .... .10^3 meters

1 mega-meter .. .10^6 meters (µ.m)

(mm)

(cm)

(km)

(Mm)

If a weightless cube measuring 1 cm on each side is filled with water, the volume of the water is one cubic centimeter (1 cm^3 or 1 cc) . This volume of water is then said to constitute the basic unit of mass, i.e., one gram (1 g or 1 gm). Mass is not to be confused with weight; mass is a constant for any given body of matter (within relativistic limits) and is determined by the number, kind, and concentration of the molecules that go into the structure of the body. It is often said that mass of a body is its material content. Weight, on the other hand, is a force produced by locating a given mass within a gravitational field. Thus weight of a given constant mass is not, in itself, constant. For example, a man who weighs 180 lb on the earth would weigh only 30 lb on the moon, although his material content does not change as a result of the trip. Thus a gram is de fined as the mass of one cubic centimeter of distilled water. Subdivisions and multiples of the gram may be expressed by means of the same prefixes as described for the meter. That is, 1 microgram = 10^-6 grams, 1 milligram = 10^-3 grams, and 1 kilogram = 10^3 grams.

Intervals of time are always compared by the motions of bodies.

Two intervals are defined as equal, when a body moving under exactly the same circumstances moves equally far in both cases. One of the simplest natural units of time is the period of rotation of the earth determined by observing successive noonday positions of the sun. The interval between two successive noons is called the solar day, and the basic unit of time selected for the metric system (as well as the English system, incidentally) is the mean solar second or the 86,400th part of a mean solar day. For the purposes of this guide, it shall be acceptable to conclude that there are 60 mean solar seconds in one minute and 360 degrees mean solar seconds in one hour.

Many lengths measured in everyday life can be conveniently stated in centimeters (cm), masses in grams (gm), and intervals in seconds (sec). Thus, the metric system stated in terms of these basic units is often referred to as the cgs system (centimeter-gram-second).

10. Force in the CGS System

A clear understanding of the meaning of force is possible only from a consideration of Newton's Second Law of Motion. In this famous law, Newton stated that the acceleration taken on by a body under the action of an applied force (push or pull) is directly proportional to the force and inversely proportional to the mass of the body. Algebraically stated:

a= Kf m (2)

in which acceleration is measured in cm/sec^2 and mass is measured in grams. K is a constant that can be made to become unity if the unit of force is correctly chosen.

Solving for Kf: Kf=ma (3)

Using the cgs units for mass and acceleration, the product ma must be expressed in gm=cm/sec^2 (mass in gm X acceleration in cm/sec^2). This "long-winded" unit is known by a simpler name, the dyne. If we are willing to express force in dynes, then Equation 3 may be rewritten:

K x dynes = dynes (4)

Hence, the constant K is now equal to I and may be dropped from the equation, and Newton's Second Law may be stated:

f = ma (5)

... where f is in dynes, m is in grams, and acceleration is in cm/sec^2. Under these conditions, a dyne may now be defined as the force required to impart an acceleration of one cm/sec^2 to a mass of one gm.

11. Weight Expressed in CGS Units

Weight is a special case of a force acting on a mass in a gravitational field. On the surface of the earth, a freely falling body accelerates under the action of gravity at the rate of approximately 980 cm/sec^2. (This figure is slightly different in different geographical locations and is taken merely as a rough average.) It is customary to assign the letter "g" to this acceleration, and, with weight symbolized as "w," the Second Law becomes:

W = mg (6)

Because our units must be consistent if K is to remain 1, w must be expressed in dynes if mass (m) is in gm and acceleration (g) is in cm/sec^2.

This immediately leads to the conclusion that one gm of mass on the earth's surface weighs 980 dynes. This conclusion is reached when substitution of known numbers is made in Equation 6:

w = 1 gm X 980 cm/sec^2 (7)

or ... w (of 1 gm mass) = 980 gm-cm/sec^2 = 980 dynes (8)

The foregoing discussion points out the fact that any weight obtained in grams is actually that number multiplied by 980, ex pressed in dynes.

The dyne is an important unit in electrostatics and will be related to Coulomb's Law shortly.

12. The MKS System of Units

The mks system of weights and measures differs from the cgs system in the relative of the units. The following table of conversions will assist you in understanding the changes that must be made:

1 kg 1000 gm 1 m 100 cm

Like the gram, the kilogram is a unit of mass. Weight is measured in newtons. We may substitute these quantities into Equation 6 as follows: w=mg or weight in newtons = mass in kilograms X g in meters/sec^2

Since g is 980 cm/sec^2, g is also expressible as 9.8 m/sec^2.

Thus, the weight of a 1-kg mass is:

w (1 kg mass) = 9.8 kg-m/sec^2 = 9.8 newtons (9)

Because many textbooks express Coulomb's Law in mks units, it is just as important to be capable of handling this law in mks units as it is to be able to state it in terms of cgs units.

13. Units for Coulomb's law

In the statement of Coulomb's Law in Section 7 and in its equation (Equation 1), units were not assigned.

If units in the cgs system are desired, then we are constrained to measure force in dynes, and distance in cm. K is set equal to I for a vacuum, but the magnitude of the unit charges- q and q'- remain undefined. To keep the equation free of difficult-to-remember con version factors, the electrostatic unit of charge magnitude is defined: One electrostatic unit of charge (esu) is that charge magnitude that will repel another identical unit charge with a force of one dyne, when the distance between them is one centimeter and when the charges are in a vacuum.

Substituting into Equation 1, this definition results in: 1 esu charge x 1 esu charge 1 dyne = K X (1 cm)^2 (10)

The force between two unit charges in air differs so little from that in a vacuum that K is also considered unity for air.

As has been shown, the suitable choice of units makes it possible to phrase Coulomb's Law in very simple terms. As the reader will recognize, the choice of the esu is dictated in the first place by Coulomb's Law, its selection being based on the dyne as a force unit and the centimeter as a distance unit. The esu unit of charge is a measure of quantity; by measurement it has been shown to be the equivalent of the charge on approximately 2.1 X 10exp9 (over two thou sand million) electrons.

Another very common unit of electrical quantity is the coulomb.

It is a much larger unit of charge, equivalent to 3 X 10exp9 esu or about 6.3 x 10 exp 18 electrons. If q and q' are expressed in coulombs, r in meters, and force in newtons (as required by the mks system), Coulomb's Law becomes more complex, numerically:

(11)

The quantity in the parenthesis is the value of K required to convert the basic definition of Coulomb's Law in the cgs system (in which the unit of charge is the esu, force is in dynes, and distance in cm) to a true relationship in the mks system where F is in newtons, q in coulombs, and r in meters.

14. Practical Electrical Units

Although practical electrical units (volts, ohms, amperes, coulombs, etc.) are employed more often in computations involving cur rent electricity, they are often encountered in electrostatic problems as well.

At the basis of this system is the practical unit of electrical quantity, the coulomb. The coulomb, of course, measures the same factor as the electrostatic unit charge (esu) and, as stated earlier, it is a much larger unit, being 3 X 10exp9 greater in value.

In accordance with the Law of Conservation of Energy, work must be done to move any quantity of electricity from one electric potential to another. Work as a physical term, and energy are virtually synonymous (energy is the capacity for doing work), and it is more customary to speak of electrical energy rather than the work done.

The mechanical definition of work is the product of force applied and distance moved in the direction of the force or:

W=Fs (12)

…where f is force in dynes, s is distance in cm, and W is work in dyne-cm. The dyne-cm is called an erg. The practical unit of electrical energy is called the joule. One joule contains 10^7 ergs.

The ratio of electrical energy to charge appears very often in calculations and practical problems. The ratio defines the practical unit of electrical potential energy or simply "potential," the volt. In terms of the quantities discussed above, the volt may be defined as follows: one volt of potential difference exists between two points if one joule of work must be done to move one coulomb of charge from one point to the other.

Rate of flow of charge is commonly called current intensity. The unit of measure for current intensity is the ampere, which is defined as a flow of I coulomb of charge past a given point per second. (This flow will under definite conditions deposit silver at the rate of .001118 grams per second from a specified solution of silver nitrate.)

I= Q t

where I is in amperes, Q is in coulombs, and t is in seconds.

(13)

The resistance to the flow of an electric current offered by a conductor is measured in ohms. One ohm is defined as that resistance through which one ampere passes when the applied potential is one volt. (An ohm may also be legally defined as the resistance at zero degrees centigrade of a column of mercury of uniform cross section, having a length of 10exp6. 3 cm and a mass of 14.45 grams; this is also approximately the resistance of 1000 ft. of number 10 soft-drawn annealed copper wire at zero degrees centigrade.)

QUIZ

1. Convert the following to centimeters, using exponential notation wherever necessary. (a) 30 meters, (b) 3 X 10^4 micrometers, (c) 8.3 kilometers, (d) 0.0006 mega-meters.

2. Explain the difference between mass and weight. Of the two quantities, which one would be expected to vary (for a given object) if a person were to take this object from the equator to the north pole?

3. State the equation for Newton's Second Law of Motion. If mass is expressed in grams, acceleration in cm/sec/sec, and K is unity, what unit is the appropriate unit for the force?

4. What is the weight in dynes of a 1-gram mass? Explain how you obtain this figure.

5. Prove that a newton is equal to a kg-m/sec^2. How many dynes are there in 1 newton?

6. Define an esu in terms of two identical unit charges.

7. Define the joule. How many ergs are required to make up one joule?

8. Define the volt, the ohm, and the ampere.

9. Relate the volt, the ohm, and the ampere in a single equation. By what name is this relationship known?

10. What current in amperes flows through a 10 ohm electric broiler coil when the line voltage is 120 volts?