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Electromags - Chapter 4

Units (SI) is used, Q is measured in coulombs (C), R is in meters (m),...

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LESSON CONTENT

Experimental Law of Coulomb (Hayt & Buck, 2001)

Records from at least 600 B. show evidence of the knowledge of static electricity. The Greeks were responsible for the term electricity, derived from their word for amber, and they spent many leisure hours rubbing a small piece of amber on their sleeves and observing how it would then attract pieces of fluff and stuff. However, their main interest lay in philosophy and logic, not in experimental science, and it was many centuries before the attracting effect was considered to be anything other than magic or a “life force.”

Dr. Gilbert, physician to Her Majesty the Queen of England, was the first to do any true experimental work with this effect, and in 1600 he stated that glass, sulfur, amber, and other materials, which he named, would “not only draw to themselves straws and chaff, but all metals, wood, leaves, stone, earths, even water and oil.”

Shortly thereafter, an officer in the French Army Engineers, Colonel Charles Coulomb, performed an elaborate series of experiments using a delicate torsion balance, invented by himself, to determine quantitatively the force exerted between two objects, each having a static charge of electricity. His published result bears a great similarity to Newton’s gravitational law (discovered about a hundred years earlier). Coulomb stated that the force between two very small objects separated in a vacuum or free space by a distance, which is large compared to their size, is proportional to the charge on each and inversely proportional to the square of the distance between them, or

where Q 1 and Q 2 are the positive or negative quantities of charge, R is the separation, and k is a proportionality constant. If the International System of Units (SI) is used, Q is measured in coulombs (C), R is in meters (m), and the force should be newtons (N). This will be achieved if the constant of proportionality k is written as

The new constant ℰ 0 is called the permittivity of free space and has magnitude, measured in farads per meter (F/m),

(1)

The quantity ℰ 0 is not dimensionless, for Coulomb’s law shows that it has the label C 2 /N・m 2. We will later define the farad and show that it has the dimensions C 2 /N・ m; we have anticipated this definition by using the unit F/m in equation (1).

Coulomb’s law is now

(2)

The coulomb is an extremely large unit of charge, for the smallest known quantity of charge is that of the electron (negative) or proton (positive), given in SI units as 1. 602 × 10 − 19 C; hence a negative charge of one coulomb represents about 6 × 1018 electrons. Coulomb’s law shows that the force between two charges of one coulomb each, separated by one meter, is 9 × 109 N, or about one million tons. The electron has a rest mass of 9. 109 × 10 − 31 kg and has a radius of the order of magnitude of 3. 8 × 10 − 15 m. This does not mean that the electron is spherical in shape, but merely describes the size of the region in which a slowly moving electron has the greatest probability of being found. All other known charged particles, including the proton, have larger masses and larger radii, and occupy a probabilistic volume larger than does the electron.

In order to write the vector form of (2), we need the additional fact (furnished also by Colonel Coulomb) that the force acts along the line joining the two charges and is repulsive if the charges are alike in sign or attractive if they are of opposite sign.

Let the vector r 1 locate Q 1 , whereas r 2 locates Q 2. Then the vector R 12 = r 2 − r 1 represents the directed line segment from Q 1 to Q 2 , as shown in Figure 2. The vector F 2 is the force on Q 2 and is shown for the case where Q 1 and Q 2 have the same sign. The vector form of Coulomb’s law is

(3)

where a 12 = a unit vector in the direction of R 12 , or

(4)

Example:

We illustrate the use of the vector form of Coulomb’s law by locating a charge of Q 1 = 3 × 10 − 4 C at M(1, 2 , 3) and a charge of Q 2 = − 10 − 4 C at N(2, 0 , 5) in a vacuum. We desire the force exerted on Q 2 by Q 1. (Hayt & Buck, 2001)

Solution:

= 4 𝐀 𝐀 1 𝐀 0

𝐀𝐀 22 𝐀 21

444444444444444 (8⋅85 44444444444444 410 −12)(3) 2 𝐀 21 =-6(-0 + 0 + 0) =6(0 - 0 - 0) direction

magnitude F 1 = 4 – 1 – 4

b) If Q 2 is the reference F 2 = -F 1 F 2 = 6(-0 + 0 + 0) F 2 = -4 + 1 + 4

The force expressed by Coulomb’s law is a mutual force, for each of the two charges experiences a force of the same magnitude, although of opposite direction. We might equally well have written

(5)

Electric Field Intensity (Hayt & Buck, 2001)

If we now consider one charge fixed in position, say Q 1 , and move a second charge slowly around, we note that there exists everywhere a force on this second charge; in other words, this second charge is displaying the existence of a force field that is associated with charge, Q 1. Call this second charge a test charge Qt. The force on it is given by Coulomb’s law,

Coulomb’s law is linear, for if we multiply Q 1 by a factor n, the force on Q 2 is also multiplied by the same factor n. It is also true that the force on a charge in the presence of several other charges is the sum of the forces on that charge due to each of the other charges acting alone.

Note: If there are three charges involve use the formula shown below in finding forces

F 1 = F 21 + F 31 ; F 21 = 4 𝐀𝐀𝐀 1 𝐀𝐀 22 𝐀 21 , F 31 = 4 𝐀𝐀𝐀 10 𝐀𝐀 2312 𝐀 31

0 21

F 2 = F 12 + F 32 ; F 12 = 4 𝐀𝐀𝐀 10 𝐀𝐀 2122 𝐀 12 , F 32 = 4 𝐀𝐀𝐀 10 𝐀𝐀 2322 𝐀 32

F 3 = F 13 + F 23 ; F 13 = 4 𝐀𝐀𝐀 1 𝐀𝐀 22 𝐀 13 , F 23 = 4 𝐀𝐀𝐀 10 𝐀𝐀 2232 𝐀 23

0 13

Writing this force as a force per unit charge gives the electric field intensity, E 1 arising from Q 1 :

(6) E 1 is interpreted as the vector force, arising from charge Q 1 , that acts on a unit positive test charge. More generally, we write the defining expression:

(7)

in which E, a vector function, is the electric field intensity evaluated at the test charge location that arises from all other charges in the vicinity—meaning the electric field arising from the test charge itself is not included in E.

The units of E would be in force per unit charge N/C (newtons per coulomb). Again anticipating a new dimensional quantity, the volt (V), having the label of joules per coulomb (J/C), or newton-meters per coulomb (N ・ m/C), we measure electric field intensity in the practical units of volts per meter (V/m).

Now, we dispense with most of the subscripts in (6), reserving the right to use them again any time there is a possibility of misunderstanding. The electric field of a single point charge becomes:

(8)

We remember that R is the magnitude of the vector R, the directed line segment from the point at which the point charge Q is located to the point at which E is desired, and aR is a unit vector in the R direction.

We arbitrarily locate Q 1 at the center of a spherical coordinate system. The unit vector aR then becomes the radial unit vector ar , and R is r. Hence

(9)

The field has a single radial component, and its inverse-square-law relationship is quite obvious.

If we consider a charge that is not at the origin of our coordinate system, the field no longer possesses spherical symmetry, and we might as well use rectangular coordinates. For a charge Q located at the source point r’ = x’ax + y’ay + z’az, as illustrated in Figure 2, we find the field at a general field point r = xax+ yay +zaz by expressing R as r – r’, and then

(10)

Earlier, we defined a vector field as a vector function of a position vector, and this is emphasized by letting E be symbolized in functional notation by E(r).

Because the coulomb forces are linear, the electric field intensity arising from two point charges, Q 1 at r 1 and Q 2 at r 2 , is the sum of the forces on Qt caused by Q 1 and Q 2 acting alone, or

(11)

Example: 1) In order to illustrate the application of (11), we find E at P(1, 1 , 1) caused by four identical 3nC (nanocoulomb) charges located at P 1 (1, 1 , 0), P 2 (−1, 1 , 0), P 3 (−1,−1, 0), and P 4 (1,−1, 0), as shown in Figure 2. (Hayt & Buck, 2001)

Solution:

We find that r = ax + ay + az , r 1 = ax + ay , and thus r − r 1 = az. The magnitudes are: |r − r 1 | = 1, |r , |r − r 3 | = 3, and |r.

Because Q/ 4 πℰ 0 = 3 × 10−9/(4π × 8 × 10−12) = 26・m, we may now use (11) to

obtain

o r

Calculate E at M(3,-3,2) in free space caused by a) a charge Q 1 = 2μC at P 1 (0,0,0) b) a charge Q 2 = 3μC at P 2 (-1,2,3) c) a charge Q 1 = 2μC at P 1 (0,0,0) and Q 2 = 3μC at P 2 (-1,2,3)

|R1E|= 4.

electron or if we care little that the mass of the water actually increases in small but finite steps as each new molecule is added.

This is really no limitation at all, because the end results for electrical engineers are almost always in terms of a current in a receiving antenna, a voltage in an electronic circuit, or a charge on a capacitor, or in general in terms of some large-scale macroscopic phenomenon. It is very seldom that we must know a current electron by electron.

We denote volume charge density by ρν, having the units of coulombs per cubic meter (C/m 3 ).

The small amount of charge ΔQ in a small volume Δν is

(12)

and we may define ρ+ mathematically by using a limiting process on (12),

(13)

𝐀𝐀

𝐀𝐀 =

𝐀𝐀

The total charge within some finite volume is obtained by integrating throughout that volume,

(14)

Only one integral sign is customarily indicated, but the differential dν signifies integration throughout a volume, and hence a triple integration.

Cartesian Coordinate:

𝐀𝐀

Cylindrical Coordinate:

𝐀𝐀 ØØ

Spherical Coordinate:

𝐀𝐀 𝐀𝐀𝐀𝐀 𝐀𝐀𝐀𝐀 𝐀Ø

Electric Field Intensity:

𝐀𝐀 𝐀𝐀 = 2

444444444444444 𝐀𝐀

𝐀𝐀𝐀

∫ 𝐀𝐀 = ∫ 2 𝐀 𝐀

4 𝐀𝐀𝐀 𝐀

Example:

As an example of the evaluation of a volume integral, we find the total charge contained in a 2-cm length of the electron beam shown in Figure 2. (Hayt & Buck,

Solution:

From the illustration, we see that the charge density is

The volume differential in cylindrical coordinates is given in circular coordinate system (IM no. 1); therefore,

We integrate first with respect to Ø because it is so easy,

and then with respect to z, because this will simplify the last integration with respect to ρ,

If we sum the contributions of all the volume charge in a given region and let the volume element Δν approach zero as the number of these elements becomes infinite, the summation becomes an integral,

(15)

This is again a triple integral, and we shall do our best to avoid actually performing the integration. The significance of the various quantities under the integral sign of (15) might stand a little review. The vector r from the origin locates the field point where E is being determined, whereas the vector r’ extends from the origin to the source point where ρv (r’)dν’ is located. The scalar distance between the source point and the field point is |r – r’|, and the fraction (r – r’)/|r – r’| is a unit vector directed from source point to field point. The variables of integration are x’, y’, and z’ in rectangular coordinates.

Example: 2) Find the total charge inside the volume indicated; ρv= 10z 2 e-0𝐀y -1<x<2, 0<y<1, 3<z<3. Solution:

QT= 119 coulombs

Try this! Find QT, where ρv= ρ 2 z 2 sin0Ø; 0<ρ<0, 0<Ø< , 2<z<4. (Ans. 1 mC)

Field of a Line Charge (Hayt & Buck, 2001)

Up to this point we have considered two types of charge distribution, the point charge and charge distributed throughout a volume with a density ρν C/m 3. If we now consider a filament like distribution of volume charge density, such as a charged conductor of very small radius, we find it convenient to treat the charge as a line charge of density ρL C/m. We assume a straight-line charge extending along the z axis in a cylindrical coordinate system from −∞ to ∞, as shown in Figure 2. We desire the electric field intensity E at any and every point resulting from a uniform line charge density ρL.

and

Integrating by integral tables or change of variable, z’ = ρ cot θ, we have

and

or finally,

(16)

where: E= Electric Field Intensity (Volt/meter ; Newton/Coulomb) 𝐀𝐀= line charge density (Coulomb/meter) 𝐀= R= distance in meter from the pt. of observation to the line charge 𝐀𝐀= 𝐀𝐀 = direction of E or the field

We note that the field falls off inversely with the distance to the charged line, as compared with the point charge, where the field decreased with the square of the distance. Moving ten times as far from a point charge leads to a field only 1 percent the previous strength, but moving ten times as far from a line charge only reduces the field to 10 percent of its former value. An analogy can be drawn with a source of illumination,

Because only theE component is present, we may simplify:

for the light intensity from a point source of light also falls off inversely as the square of the distance to the source. The field of an infinitely long fluorescent tube thus decays inversely as the first power of the radial distance to the tube, and we should expect the light intensity about a finite-length tube to obey this law near the tube. As our point recedes farther and farther from a finite-length tube, however, it eventually looks like a point source, and the field obeys the inverse-square relationship. Before leaving this introductory look at the field of the infinite line charge, we should recognize the fact that not all line charges are located along the z axis. As an example, let us consider an infinite line charge parallel to the z axis at x = 6, y = 8, shown in Figure 2. We wish to find E at the general field point P(x, y, z). We replace ρ in (16) by the radial distance between the line charge and point, P, R=

, and let ap be aR. Thus,

Therefore,

We again note that the field is not a function of z.

Example: 1) On the line described by x=2m, y=4m, there is a uniform charge distribution of density 𝐀𝐀 = 20nC/m. Determine the electric field E at (-2,-1,4)m.

Solution:

where

Since the line charges lie along x and y axes. E at PA is equal to the sum of E at x-axis and E at y-axis.

E 1 = EX + EY ; at point E 1 , x=y, therefore EX=EY RX= 𝐀x= 4az 𝐀 (8 𝐀 𝐀 10 10 − 9−12 )(4) (0. 66666666666666 6 𝐀)

E 1 = 22 + 22

E 1 = 45az

b.) PB (0,3,4) E 2 = EX + EY

Find EX;

RX= 𝐀x= 3ay+4az |RX|=

aRx

EX EX= 10 + 14

Find EY; RY= 𝐀y= 4az |R |= √ aRy

EY= 222222222222222 (8 10 10−9−12)(4) (0. 66666666666666 6 𝐀)

EY= 22

Therefore, E 2 = 10 + 14 + 22 E 2 = 10 + 36 V/m

An infinitely long uniform charge is located at y=3, z=5. If 𝐀𝐀 = 30nC/m. Find E at the origin.

0 a z

Y = a z

|RX |=

a Rx = a z

E X = E Y = 2

Solution: E at the origin:

R= ρ= (0-3)ay + (0-5)az R= -3ay – 5az

|R|= aR= − 3333333333333 3 3𝐀− 5555555555555 5 5 𝐀 5. = -0 – 0

E= 2 (8 10 10 −−129 )(5) (−0 44444444444444 4 𝐀 − 0 77777777777777 7 𝐀)

E= -47 – 79 V/m

Try this! Using the given in item 3, find E at P(0,6,1). (Ans. 64 – 86 V/m)

Field of a Sheet of Charge (Hayt & Buck, 2001)

Another basic charge configuration is the infinite sheet of charge having a uniform density of ρS C/m 2. Such a charge distribution may often be used to approximate that found on the conductors of a strip transmission line or a parallel-plate capacitor. Static charge resides on conductor surfaces and not in their interiors; for this reason, ρS is commonly known as surface charge density. The charge-distribution family now is complete—point, line, surface, and volume, or Q, ρL, ρS, and ρν.

Let us place a sheet of charge in the yz plane and again consider symmetry (Figure 2). We see first that the field cannot vary with y or with z, and then we see that the y and z components arising from differential elements of charge symmetrically located with respect to the point at which we evaluate the field will cancel. Hence only Ex is present, and this component is a function of x alone. We are again faced with a choice of many methods by which to evaluate this component, and this time we use only one method and leave the others as exercises for a quiet Sunday afternoon.

Let us use the field of the infinite line charge (16) by dividing the infinite sheet into differential-width strips. One such strip is shown in Figure 2. The line charge density,

R

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Electromags - Chapter 4

Course: Electronics Engineering (CR 061)

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LESSON CONTENT

1. Experimental Law of Coulomb (Hayt & Buck, 2001)

Records from at least 600 B.C. show evidence of the knowledge of static electricity. The Greeks

were responsible for the term electricity, derived from their word for amber, and they spent many

leisure hours rubbing a small piece of amber on their sleeves and observing how it would then attract

pieces of fluff and stuff. However, their main interest lay in philosophy and logic, not in experimental

science, and it was many centuries before the attracting effect was considered to be anything other

than magic or a “life force.”

Dr. Gilbert, physician to Her Majesty the Queen of England, was the first to do any true

experimental work with this effect, and in 1600 he stated that glass, sulfur, amber, and other

materials, which he named, would “not only draw to themselves straws and chaff, but all metals,

wood, leaves, stone, earths, even water and oil.”

Shortly thereafter, an officer in the French Army Engineers, Colonel Charles Coulomb,

performed an elaborate series of experiments using a delicate torsion balance, invented by himself,

to determine quantitatively the force exerted between two objects, each having a static charge of

electricity. His published result bears a great similarity to Newton’s gravitational law (discovered

about a hundred years earlier). Coulomb stated that the force between two very small objects

separated in a vacuum or free space by a distance, which is large compared to their size, is

proportional to the charge on each and inversely proportional to the square of the distance between

them, or

where Q1 and Q2 are the positive or negative quantities of charge, R is the

separation, and k is a proportionality constant. If the International System of

Units (SI) is used, Q is measured in coulombs (C), R is in meters (m), and the force should be

newtons (N). This will be achieved if the constant of proportionality k is written as

The new constant ℰ0 is called the permittivity of free space and has magnitude, measured in farads

per meter (F/m),

(1)

The quantity ℰ0 is not dimensionless, for Coulomb’s law shows that it has the label C2/N・m2.

We will later define the farad and show that it has the dimensions C2/N・ m; we have anticipated this

definition by using the unit F/m in equation (1).

Coulomb’s law is now

(2)