electromagnetic field theory, Lecture notes of Electronics

Download electromagnetic field theory and more Lecture notes Electronics in PDF only on Docsity! Chapter 4 ELECTROSTATIC FIELDS Take risks: if you win, you will be happy; if you lose you will be wise. —PETER KREEFT 4.1 INTRODUCTION Having mastered some essential mathematical tools needed for this course, we are now prepared to study the basic concepts of EM. We shall begin with those fundamental con- cepts that are applicable to static (or time-invariant) electric fields in free space (or vacuum). An electrostatic field is produced by a static charge distribution. A typical example of such a field is found in a cathode-ray tube. Before we commence our study of electrostatics, it might be helpful to examine briefly the importance of such a study. Electrostatics is a fascinating subject that has grown up in diverse areas of application. Electric power transmission, X-ray machines, and lightning protection are associated with strong electric fields and will require a knowledge of elec- trostatics to understand and design suitable equipment. The devices used in solid-state electronics are based on electrostatics. These include resistors, capacitors, and active devices such as bipolar and field effect transistors, which are based on control of electron motion by electrostatic fields. Almost all computer peripheral devices, with the exception of magnetic memory, are based on electrostatic fields. Touch pads, capacitance keyboards, cathode-ray tubes, liquid crystal displays, and electrostatic printers are typical examples. In medical work, diagnosis is often carried out with the aid of electrostatics, as incorpo- rated in electrocardiograms, electroencephalograms, and other recordings of organs with electrical activity including eyes, ears, and stomachs. In industry, electrostatics is applied in a variety of forms such as paint spraying, electrodeposition, electrochemical machining, and separation of fine particles. Electrostatics is used in agriculture to sort seeds, direct sprays to plants, measure the moisture content of crops, spin cotton, and speed baking of bread and smoking of meat.12 'For various applications of electrostatics, see J. M. Crowley, Fundamentals of Applied Electrostat- ics. New York: John Wiley & Sons, 1986; A. D. Moore, ed., Electrostatics and Its Applications. New York: John Wiley & Sons, 1973; and C. E. Jowett, Electrostatics in the Electronics Environment. New York: John Wiley & Sons, 1976. 2An interesting story on the magic of electrostatics is found in B. Bolton, Electromagnetism and Its Applications. London: Van Nostrand, 1980, p. 2. 103 104 Electrostatic Fields We begin our study of electrostatics by investigating the two fundamental laws gov- erning electrostatic fields: (1) Coulomb's law, and (2) Gauss's law. Both of these laws are based on experimental studies and they are interdependent. Although Coulomb's law is ap- plicable in finding the electric field due to any charge configuration, it is easier to use Gauss's law when charge distribution is symmetrical. Based on Coulomb's law, the concept of electric field intensity will be introduced and applied to cases involving point, line, surface, and volume charges. Special problems that can be solved with much effort using Coulomb's law will be solved with ease by applying Gauss's law. Throughout our discussion in this chapter, we will assume that the electric field is in a vacuum or free space. Electric field in material space will be covered in the next chapter. 4.2 COULOMB'S LAW AND FIELD INTENSITY Coulomb's law is an experimental law formulated in 1785 by the French colonel, Charles Augustin de Coulomb. It deals with the force a point charge exerts on another point charge. By a point charge we mean a charge that is located on a body whose dimensions are much smaller than other relevant dimensions. For example, a collection of electric charges on a pinhead may be regarded as a point charge. Charges are generally measured in coulombs (C). One coulomb is approximately equivalent to 6 X 1018 electrons; it is a very large unit of charge because one electron charge e = -1.6019 X 10~19C. Coulomb's law states that the force /•' between two point charges (?, and Q2 is: 1. Along the line joining them 2. Directly proportional to the product QtQ2 of the charges 3. Inversely proportional to the square of the distance R between them.' Expressed mathematically, F = R2 (4.1) where k is the proportionality constant. In SI units, charges <2i and Q2 are in coulombs (C), the distance R is in meters (m), and the force F is in newtons (N) so that k = 1/4TTS0. The constant so is known as the permittivity of free space (in farads per meter) and has the value 8.854 X 10~12 = -^ r -F /m = 9 X 109 m/F 47T£n (4.2) 3Further details of experimental verification of Coulomb's law can be found in W. F. Magie, A Source Book in Physics. Cambridge: Harvard Univ. Press, 1963, pp. 408^20. 4.2 COULOMB'S LAW AND FIELD INTENSITY 107 For N point charges Qu Q2,. . . , QN located at r b r2,. . . , rN, the electric field in- tensity at point r is obtained from eqs. (4.8) and (4.10) as or t, — 6i(r - 4xejr - fepr - r)2 4irso|r - r2 A 2J 4TT£O £TX r - - rN) 4TT£O r - (4.12) EXAMPLE 4.1 Point charges 1 mC and - 2 mC are located at (3, 2, -1 ) and (—1, —1,4), respectively.Calculate the electric force on a 10-nC charge located at (0, 3, 1) and the electric field in- tensity at that point. Solution: QQk - rk) A=i,247reo|r - rk\ Q / 10"3[(0, 3, 1) - (3,2,-1)] 2.10'3[(0, 3, 1) - (-1,-1,4)] 47re0 I |(0,3, 1 ) - ( 3 , 2 , - 1 ) | 3 10"3 • 10 • 10"9 r (-3,1,2) - ( - 1 , - 1 , 4)|3 4TT = 9 - 1 0 F = - i At that point, 10 - 9 2(1,4,-3) (9 + 1 + 4yu (1 + 16 + 9)3/2 36TT - 2 [ ( - 3 , 1,2) + ( -2 , -8 ,6) 14Vl4 26V26 - 3.817ay + 7.506azmN E = Q = (-6.507, -3.817, 7.506) 10 10 • 10"9 E = -650.7ax - 381.7a. + 750.6azkV/m PRACTICE EXERCISE 4.1 Point charges 5 nC and —2 nC are located at (2,0, 4) and ( -3 ,0 , 5), respectively. (a) Determine the force on a 1-nC point charge located at (1, —3, 7). (b) Find the electric field E at (1, - 3 , 7). Answer: (a) -1.004a* - 1.284a,, + 1.4aznN, (b) -1.004ax - 1.284a,+1.4a2V/m. 108 Electrostatic Fields EXAMPLE 4.2 Two point charges of equal mass m, charge Q are suspended at a common point by two threads of negligible mass and length t. Show that at equilibrium the inclination angle a of each thread to the vertical is given by Q = 16x eomg£ sin a tan a If a is very small, show that Solution: Consider the system of charges as shown in Figure 4.3 where Fe is the electric or coulomb force, T is the tension in each thread, and mg is the weight of each charge. At A or B T sin a = Fe T cos a = mg Hence, But Hence, or sin a _ Fe 1 Q2 cos a mg mg 4ireor r = 2€ sin a Q cos a = I6irejng€2 sin3 a Q2 = I6irsomg(2 sin2 a tan a as required. When a is very small tan a — a — sin a. Figure 4.3 Suspended charged particles; for Example 4.2. 4.2 COULOMB'S LAW AND FIELD INTENSITY • 109 and so or '2_3Q2 = I6wsomgtla a = Ql 16ireomg€ PRACTICE EXERCISE 4.2 Three identical small spheres of mass m are suspended by threads of negligible masses and equal length € from a common point. A charge Q is divided equally between the spheres and they come to equilibrium at the corners of a horizontal equi- lateral triangle whose sides are d. Show that Q2 = where g = acceleration due to gravity. Answer: Proof. r21-l/2 EXAMPLE 4.3 A practical application of electrostatics is in electrostatic separation of solids. For example,Florida phosphate ore, consisting of small particles of quartz and phosphate rock, can be separated into its components by applying a uniform electric field as in Figure 4.4. Assum- ing zero initial velocity and displacement, determine the separation between the particles after falling 80 cm. Take E = 500 kV/m and Qlm = 9 /xC/kg for both positively and neg- atively charged particles. Figure 4.4 Electrostatic separation of solids; for Example 4.3. Phosphate Quartz 112 Electrostatic Fields dQ = psdS^Q = psdS (surface charge) 4 dQ = pv dv —> Q = \ pv dv (volume charge) (4.13b) (4.13c) The electric field intensity due to each of the charge distributions pL, ps, and pv may be regarded as the summation of the field contributed by the numerous point charges making up the charge distribution. Thus by replacing Q in eq. (4.11) with charge element dQ = pL dl, ps dS, or pv dv and integrating, we get E = E = E = PLdl 4-jrsJt2 PsdS A-weJi2 pvdv (line charge) (surface charge) (volume charge) (4.14) (4.15) (4.16) It should be noted that R2 and a^ vary as the integrals in eqs. (4.13) to (4.16) are evaluated. We shall now apply these formulas to some specific charge distributions. A. A Line Charge Consider a line charge with uniform charge density pL extending from A to B along the z-axis as shown in Figure 4.6. The charge element dQ associated with element dl = dz of the line is dQ = pLdl = pL dz (0,0,2)7-^ (0,0, z') dEz dE Figure 4.6 Evaluation of the E field due to i l i n e 4.3 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS • 113 and hence the total charge Q is Q = (4.17) The electric field intensity E at an arbitrary point P(x, y, z) can be found using eq. (4.14). It is important that we learn to derive and substitute each term in eqs. (4.14) to (4.15) for a given charge distribution. It is customary to denote the field point4 by (x, y, z) and the source point by (x', y', z'). Thus from Figure 4.6, dl = dz' R = (x, y, Z) - (0, 0, z') = xax + yay + (z - z')az or R = pap + (z - z') az = 2 R (z - z')az R2 R (z~ z'f]213/2 Substituting all this into eq. (4.14), we get PLE = N 213/24ireo J [p2 + (Z - z' To evaluate this, it is convenient that we define a, au and a2 as in Figure 4.6. R = [p2 + (z - z'f]m = p sec a z' = OT - p tan a, dz' = —p sec2 a da Hence, eq. (4.18) becomes —pL [ai p sec2 a [cos a a,, + sin a az] daE = 4iren p2 sec2 a PL [cos a a , + sin a a j da Thus for a finite line charge, E = PL [- (sin a2 - sin aOa,, + (cos a2 - cos a{)az] (4.18) (4.19) (4.20) 4The field point is the point at which the field is to be evaluated. 114 Electrostatic Fields As a special case, for an infinite line charge, point B is at (0, 0, °°) and A at (0, 0, -co) so that al = x/2, a2 = —x/2; the z-component vanishes and eq. (4.20) becomes E = PL (4.21) Bear in mind that eq. (4.21) is obtained for an infite line charge along the z-axis so that p and ap have their usual meaning. If the line is not along the z-axis, p is the perpendicular distance from the line to the point of interest and ap is a unit vector along that distance di- rected from the line charge to the field point. B. A Surface Charge Consider an infinite sheet of charge in the xy-plane with uniform charge density ps. The charge associated with an elemental area dS is dQ = Ps dS and hence the total charge is Q= PsdS (4.22) From eq. (4.15), the contribution to the E field at point P(0, 0, h) by the elemental surface 1 shown in Figure 4.7 is JE = dQ (4.23) Figure 4.7 Evaluation of the E field due to an infinite sheet of charge. 4.3 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS 117 It is convenient to evaluate the integral in eq. (4.29) in terms of R and r'. Hence we express cos d', cos a, and sin 6' dd' in terms of R and r', that is, cos a = z2- 2 Z h R 2 - 2zR Yr'2- r'2 R2 2zr' Differentiating eq. (4.31b) with respect to 0' keeping z and r' fixed, we obtain RdR sin 6' dd' = zr' Substituting eqs. (4.30) to (4.32) into eq. (4.29) yields E = d<t>'4xeo J , , = o J r , = 0 J ^ ^ ra rz+r' dr' r'=0 JR = z-r' r' ro £i 1 1 IA 4r'2dr'=- -2 - 47reo z z V3 or (4.31a) (4.31b) (4.32) (4.33) This result is obtained for E at P(0, 0, z). Due to the symmetry of the charge distribution, the electric field at P(r, 9, <j>) is readily obtained from eq. (4.33) as E = Q ar (4.34) which is identical to the electric field at the same point due to a point charge Q located at the origin or the center of the spherical charge distribution. The reason for this will become obvious as we cover Gauss's law in Section 4.5. EXAMPLE 4.4 A circular ring of radius a carries a uniform charge pL C/m and is placed on the xy-planewith axis the same as the z-axis. (a) Show that E(0, 0, h) = pLah 2eo[h2 + a2}13/2 "z 118 Electrostatic Fields (b) What values of h gives the maximum value of E? (c) If the total charge on the ring is Q, find E as a -> 0. Solution: (a) Consider the system as shown in Figure 4.9. Again the trick in finding E using eq. (4.14) is deriving each term in the equation. In this case, dl = a d4>, R = R R = a ( - haz = \a2 1.211/2 R R or a« R -aap /?2 |R|3 [a2 + h2}13/2 Hence E = PL (-aap + By symmetry, the contributions along ap add up to zero. This is evident from the fact that for every element dl there is a corresponding element diametrically opposite it that gives an equal but opposite dEp so that the two contributions cancel each other. Thus we are left with the z-component. That is, pLahaz 4vso[h2 + a2]13/2 d<t> = pLahaz 2so[h2 a2f2 as required. Figure 4.9 Charged ring; for Example 4.4. I 4.3 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS • 119 (b) dh 2eo For maximum E, = 0, which implies that dh [h2 + a2]3 a2 - 2hz = 0 or h = ± (c) Since the charge is uniformly distributed, the line charge density is Q so that Asa or in general PL = E = Qh 213/2 H z E = —~si A-KSJI2 4ireor which is the same as that of a point charge as one would expect. PRACTICE EXERCISE 4.4 A circular disk of radius a is uniformly charged with ps C/m2. If the disk lies on the z = 0 plane with its axis along the z-axis, (a) Show that at point (0, 0, h) _h \ (b) From this, derive the E field due to an infinite sheet of charge on the z = 0 plane. (c) If a <3C h, show that E is similar to the field due to a point charge. Answer: (a) Proof, (b) — a,, (c) Proof 2en 122 Electrostatic Fields and E3 = PL2ireop where ap (not regular ap but with a similar meaning) is a unit vector along LP perpendicu- lar to the line charge and p is the length LP to be determined from Figure 4.10(b). Figure 4.10(b) results from Figure 4.10(a) if we consider plane y = 1 on which E3 lies. From Figure 4.10(b), the distance vector from L to P is R = P = = VTo, R Hence, lOir • 10 - 9 2 T T - 10- 9 10 Vio"* Vio' (a, - 3az) 36TT = 187r(ax - 3a,) Thus by adding Eu E2, and E3, we obtain the total field as E = -162Trax + 270ira, - 54x3, V/m Note that to obtain ar, ap, or a«, which we always need for finding F or E, we must go from the charge (at position vector r') to the field point (at position vector r); hence ar, ap, or an is a unit vector along r — r'. Observe this carefully in Figures 4.6 to 4.10. PRACTICE EXERCISE 4.6 In Example 4.6 if the line x = 0, z = 2 is rotated through 90° about the point (0, 2, 2) so that it becomes x = 0, y = 2, find E at (1, 1, -1 ) . Answer: -282.7a.* + 564.5a, V/m. 4.4 ELECTRIC FLUX DENSITY The flux due to the electric field E can be calculated using the general definition of flux in eq. (3.13). For practical reasons, however, this quantity is not usually considered as the most useful flux in electrostatics. Also, eqs. (4.11) to (4.16) show that the electric field in- tensity is dependent on the medium in which the charge is placed (free space in this chapter). Suppose a new vector field D independent of the medium is defined by D = eoE (4.35) II 4.4 ELECTRIC FLUX DENSITY H 123 We define electric flux f in terms of D using eq. (3.13), namely, = \D-dS (4.36) In SI units, one line of electric flux emanates from +1 C and terminates on - 1 C. There- fore, the electric flux is measured in coulombs. Hence, the vector field D is called the elec- tric flux density and is measured in coulombs per square meter. For historical reasons, the electric flux density is also called electric displacement. From eq. (4.35), it is apparent that all the formulas derived for E from Coulomb's law in Sections 4.2 and 4.3 can be used in calculating D, except that we have to multiply those formulas by eo. For example, for an infinite sheet of charge, eqs. (4.26) and (4.35) give (4.37) (4.38) Note from eqs. (4.37) and (4.38) that D is a function of charge and position only; it is in- dependent of the medium. and for a volume charge distribution, eqs. (4.16) and (4.35) give D = , Pvdv EXAMPLE 4.7 Determine D at (4, 0, 3) if there is a point charge —5TT mC at (4, 0, 0) and a line charge3TT mC/m along the y-axis. Solution: Let D = DQ + DL where D e and DL are flux densities due to the point charge and line charge, respectively, as shown in Figure 4.11: Q (r - r') = eoE = Q 4mR/ 47r|r - r ' where r - r ' = (4, 0, 3) - (4, 0, 0) = (0, 0, 3). Hence, Also In this case - 3 / DQ -5TT • 10^(0, 0, 3) 4TT|(0,0, 3) |3 2= -0.138 az mC/m (4, 0, 3) - (0, 0, 0) (4, 0, 3) * |(4,0, 3) - (0,0, 0)| 5 p = |(4,0,3) - (0,0,0)| = 5 124 Electrostatic Fields = 3?rC/m --y Q = -SirC Figure 4.11 Flux density D due to a point charge and an infinite line charge. Hence, Thus 3TT 2TT(25) 3az) = 0.24ax + 0.18azmC/m2 D = DG + DL = 240a* + 42a2 /xC/m2 PRACTICE EXERCISE 4.7 A point charge of 30 nC is located at the origin while plane y = 3 carries charge 10nC/m2. Find D at (0,4, 3). Answer: 5.076a,, + 0.0573az nC/m2. 4.5 GAUSS'S LAW—MAXWELL'S EQUATION Gauss's5 law constitutes one of the fundamental laws of electromagnetism. Gauss's law stales thai the loial electric Mux V through any closed surface is equal to the total charge enclosed by that surface. Karl Friedrich Gauss (1777-1855), a German mathematician, developed the divergence theorem of Section 3.6, popularly known by his name. He was the first physicist to measure electric and mag- netic quantities in absolute units. For details on Gauss's measurements, see W. F. Magie, A Source Book in Physics. Cambridge: Harvard Univ. Press, 1963, pp. 519-524. <r if 4.6 APPLICATIONS OF GAUSS'S LAW • 127 Since D is everywhere normal to the Gaussian surface, that is, D = D^n applying Gauss's law (V = genciosed) gives Q = i> D • dS = Dr $> dS = Dr Aitr (4.44) where § dS = surface. Thus LQ / £ = 0 r2 sin 6 dd dcf> = 4irr2 is the surface area of the Gaussian (4.45)11 as expected from eqs. (4.11) and (4.35). B. Infinite Line Charge Suppose the infinite line of uniform charge pL C/m lies along the z-axis. To determine D at a point P, we choose a cylindrical surface containing P to satisfy symmetry condition as shown in Figure 4.14. D is constant on and normal to the cylindrical Gaussian surface; that is, D = Dpap. If we apply Gauss's law to an arbitrary length € of the line PJ = Q = = Dp 2irp€ (4.46) where § dS = 2irp€ is the surface area of the Gaussian surface. Note that J D • dS evalu- ated on the top and bottom surfaces of the cylinder is zero since D has no z-component; that means that D is tangential to those surfaces. Thus D as expected from eqs. (4.21) and (4.35). 2irp (4.47) Figure 4.14 Gaussian surface about an infinite line line charge P /C/m Gaussian surface 128 Electrostatic Fields C. Infinite Sheet of Charge Consider the infinite sheet of uniform charge ps C/m2 lying on the z = 0 plane. To deter- mine D at point P, we choose a rectangular box that is cut symmetrically by the sheet of charge and has two of its faces parallel to the sheet as shown in Figure 4.15. As D is normal to the sheet, D = Dzaz, and applying Gauss's law gives Ps | dS = Q = <f> D • dS = Dz dS + dS op ^bottom (4.48) Note that D • dS evaluated on the sides of the box is zero because D has no components along ax and ay. If the top and bottom area of the box each has area A, eq. (4.48) becomes and thus PsA = DZ(A+ A) Ps (4.49) or (4.50) as expected from eq. (4.25). D. Uniformly Charged Sphere Consider a sphere of radius a with a uniform charge pv C/m3. To determine D everywhere, we construct Gaussian surfaces for eases r < a and r > a separately. Since the charge has spherical symmetry, it is obvious that a spherical surface is an appropriate Gaussian surface. Infinite sheet of charge ps C/m2 Figure 4.15 Gaussian surface about an infinite line sheet of charge. Gaussian surface 4.6 APPLICATIONS OF GAUSS'S LAW H 129 For r < a, the total charge enclosed by the spherical surface of radius r, as shown in Figure 4.16 (a), is Gene = \Pvdv = pAdv = pA I I r2 sin 6 drdd d<j> (4.51) and V = <P D • dS = Dr $ dS = Dr = Dr4xr2 Hence, TP = <2enc g i y e s D r 4xr 2 = rlsm.6ded<$> =o (4.52) or 0 < r « a (4.53) For r > a, the Gaussian surface is shown in Figure 4.16(b). The charge enclosed by the surface is the entire charge in this case, that is, while G e n e = \ p v d v = p v \ d v = p = pv - ira sinO drdd =o = cb D - dS = Dr4irr2 (4.54) (4.55) Gaussian surface II f I \ Figure 4.16 Gaussian surface for a uniformly charged sphere when: (a) r & a and (b) r £ a. (a) (b) 132 • Electrostatic Fields A charge distribution with spherical symmetry has density '' P°r rv , - — r. EXAMPLE 4.9 = 1 R ' r> R Determine E everywhere. Solution: The charge distribution is similar to that in Figure 4.16. Since symmetry exists, we can apply Gauss's law to find E. y (a) For r < R So<PE-dS = gene = P r r IT r 2ir eoEr 4 x r = Qenc = pv r sin 9 d<t> dB dr Jo Jo Jo i . 2 Por , PoT-r4= 4-Trr — dr = R R or (b) For r > R, = Qenc = r rir r2ic 0 J0 pvr sin 6 d<j) dd dr = I — 4irr2dr+ 0 • 4wr2 dr Jo R ]R or PRACTICE EXERCISE 4.9 A charge distribution in free space has pv = 2r nC/m3 for 0 £ r £ 10 m and zero otherwise. Determine E at r = 2 m and r = 12 m. Answer: 226ar V/m, 3.927ar kV/m. 4.7 ELECTRIC POTENTIAL 133 4.7 ELECTRIC POTENTIAL From our discussions in the preceding sections, the electric field intensity E due to a charge distribution can be obtained from Coulomb's law in general or from Gauss's law when the charge distribution is symmetric. Another way of obtaining E is from the electric scalar po- tential V to be defined in this section. In a sense, this way of rinding E is easier because it is easier to handle scalars than vectors. Suppose we wish to move a point charge Q from point A to point B in an electric field E as shown in Figure 4.18. From Coulomb's law, the force on Q is F = QE so that the work done in displacing the charge by d\ is dW = - F • d\ = -QE • d\ (4.58) The negative sign indicates that the work is being done by an external agent. Thus the total work done, or the potential energy required, in moving Q from A to B is (4.59) Dividing W by Q in eq. (4.59) gives the potential energy per unit charge. This quantity, denoted by VAB, is known as the potential difference between points A and B. Thus (4.60) Note that 1. In determining VAB, A is the initial point while B is the final point. 2. If VAB is negative, there is a loss in potential energy in moving Q from A to B; this implies that the work is being done by the field. However, if VAB is positive, there is a gain in potential energy in the movement; an external agent performs the work. 3. VAB is independent of the path taken (to be shown a little later). 4. VAB is measured in joules per coulomb, commonly referred to as volts (V). Origin Figure 4.18 Displacement of point charge Q in an electrostatic field E. 134 Electrostatic Fields As an example, if the E field in Figure 4.18 is due to a point charge Q located at the origin, then E so eq. (4.60) becomes 4iren Q Q 2 rA 4ireor Q [ l d r &r (4.61) (4.62a) or vAB = vB-vA (4.62b) where VB and VA are the potentials (or absolute potentials) at B and A, respectively. Thus the potential difference VAB may be regarded as the potential at B with reference to A. In problems involving point charges, it is customary to choose infinity as reference; that is, we assume the potential at infinity is zero. Thus if VA = 0 as rA —» °° in eq. (4.62), the po- tential at any point (rB —> r) due to a point charge Q located at the origin is V = Q 4irenr (4.63) Note from eq. (4.62a) that because E points in the radial direction, any contribution from a displacement in the 6 or </> direction is wiped out by the dot product E • d\ = E cos 8 dl = E dr. Hence the potential difference VAB is independent of the path as asserted earlier. The potential ;il an\ poim is the pulomial dit'tcrcntx" helwecn thai poim and a chosen poinl in which the potential is /em. In other words, by assuming zero potential at infinity, the potential at a distance r from the point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point. Thus V = - E • dl (4.64) If the point charge Q in eq. (4.63) is not located at the origin but at a point whose po- sition vector is r', the potential V(x, y, z) or simply V(r) at r becomes V(r) = Q4iren r — r' (4.65) 4.7 ELECTRIC POTENTIAL 137 where VQ and VL are the contributions to V at that point due to the point charge and the line charge, respectively. For the point charge, VQ= -SE-d\= ~ Q ar • dr ar Q + c,4xeor For the infinite line charge, VL = - I E • d\ = - PL 2?r£op dp ap PL 2irsc In p + C2 Hence, V = - PL2ireo lnp + Q 4irenr + C where C = Cx + C2 = constant, p is the perpendicular distance from the line y = 1, z = 1 to the field point, and r is the distance from the point charge to the field point. (a) If V = 0 at O(0, 0, 0), and V at A(5, 0, 1) is to be determined, we must first determine the values of p and r at O and A. Finding r is easy; we use eq. (2.31). To find p for any point (x, y, z), we utilize the fact that p is the perpendicular distance from (x, y, z) to line y = 1, z = 1, which is parallel to the x-axis. Hence p is the distance between (x, y, z) and (x, 1, 1) because the distance vector between the two points is perpendicular to ax. Thus p = |(x, y, z) ~ (x, 1, 1)| = V(y - I)2 + (z - I)2 Applying this for p and eq. (2.31) for r at points O and A, we obtain P o = | ( 0 , 0 , 0 ) - (0,l r o = |(0,0,0) - ( - 3 pA= |(5,0, 1 ) - ( 5 , 1 rA = |(5,0, 1) - ( - 3 = \Tl )| = 5 = 1 )| = 9 Hence, 0 PA - 2 • 10"9 \fl -In + J 1 ro r/ 5 • 10~9 2TT 10- 9 1 4TT 10- 9 36TT " 36TT 0 - V, = - 3 6 In V 2 + 45 ( - - - 138 Electrostatic Fields or VA = 36 In V 2 - 4 = 8.477 V fnote H T I T ! g Stant C by Subtracti"g one another and that it does not matter which one is subtracted from which. (b) If V = 100 at 5(1, 2, 1) and Vat C(-2, 5, 3) is to be determined, we find PB= 1(1,2,1) - (1,1,1)| = 1 rB = |(1, 2, 1) - (-3, 4, 0)| = V2T Pc= K-2,5,3) - (-2,1,1)| = V20 rc= |(-2,5,3) - (-3,4,0)| = 2xeo or 361n = -50.175 V Vc = 49.825 V 21 J (c) To find the potential difference between two points, we do not need a potential refer- ence if a common reference is assumed. = Vc - VB = 49.825 - 100 = -50.175 V as obtained in part (b). PRACTICE EXERCISE 4.11 A point charge of 5 nC is located at the origin. If V = 2 V at (0, 6, -8 ) , find (a) The potential at A(-3, 2,6) (b) The potential at B(\, 5, 7) (c) The potential difference VAB Answer: (a) 3.929 V, (b) 2.696 V, (c) -1.233 V. 4.8 RELATIONSHIP BETWEEN E AND V—MAXWELL'S EQUATION 139 4.8 RELATIONSHIP BETWEEN E AND V— MAXWELL'S EQUATION As shown in the previous section, the potential difference between points A and B is inde- pendent of the path taken. Hence, VAB that is, VBA + VAB = $ E • d\ = 0 or (4.73) This shows that the line integral of E along a closed path as shown in Figure 4.19 must be zero. Physically, this implies that no net work is done in moving a charge along a closed path in an electrostatic field. Applying Stokes's theorem to eq. (4.73) gives E • d\ = (V X E) • dS = 0 or V X E = 0 (4.74) Any vector field that satisfies eq. (4.73) or (4.74) is said to be conservative, or irrotational, as discussed in Section 3.8. Thus an electrostatic field is a conservative field. Equation (4.73) or (4.74) is referred to as Maxwell's equation (the second Maxwell's equation to be derived) for static electric fields. Equation (4.73) is the integral form, and eq. (4.74) is the differential form; they both depict the conservative nature of an electrostatic field. From the way we defined potential, V = — / E • d\, it follows that dV = -Edl= -Ex dx - Eydy - Ez dz Figure 4.19 Conservative nature of an electrosta- tic field. 142 Electrostatic Fields or 45 W = — Q = 28.125 Method 2: Since Vis known, this method is a lot easier. W=-Q - VA) = 1 0 ( j ^ sin 90° cos 60° - y sin 30° cos 120°) • 1(T6 - 28.125 ̂ J as obtained before PRACTICE EXERCISE 4.12 Given that E = (3*2 + v) a, + Aa, W/m, find the work done in moving a - 2 MC charge from (0, 5, 0) to (2, - 1 , 0) by taking the path (a) (0,5,0) ~>(2, 5,0) -> (2, - 1 , 0 ) (b) y = 5 - 3x Answer: (a) 12 mJ, (b) 12 mJ. 4.9 AN ELECTRIC DIPOLE AND FLUX LINES An electric dipole is formed when two poim charges of equal magnitude but oppo- site sign are separated by a small distance. The importance of the field due to a dipole will be evident in the subsequent chapters Consider the dipole shown in Figure 4.20. The potential at point P(r, 6, 0) is given by r2\ 4TTEO (4.77) where r, and r2 are the distances between P and +Q and P and -Q, respectively If r » d,r2- r, = d cos 6, r2rx - r2, and eq. (4.77) becomes V = Q dcosd (4.78) 4.9 A N ELECTRIC DIPOLE AND FLUX LINES Hi 143 Figure 4.20 An electric dipole. dcosd Since d cos 6 = d • ar, where d = daz, if we define as the dipole moment, eq. (4.78) may be written as (4.79) (4.80) Note that the dipole moment p is directed from — Q to +Q. If the dipole center is not at the origin but at r', eq. (4.80) becomes V(r) = p (r - r') 47ren|r - r ' (4.81) The electric field due to the dipole with center at the origin, shown in Figure 4.20, can be obtained readily from eqs. (4.76) and (4.78) as E = - V V = - _ Qd cos 0 27T£nr3 ay l ay QJ sin 6 or E = (2 cos 6 ar + sin 6 ae) (4.82) where p = |p| = Qd. 144 Electrostatic Fields Notice that a point charge is a monopole and its electric field varies inversely as r2 while its potential field varies inversely as r [see eqs. (4.61) and (4.63)]. From eqs. (4.80) and (4.82), we notice that the electric field due to a dipole varies inversely as r3 while its potential varies inversely as r2. The electric fields due to successive higher-order multi- poles (such as a quadrupole consisting of two dipoles or an octupole consisting of two quadrupoles) vary inversely as r4, r5, r6,. . . while their corresponding potentials vary in- versely as r3, r4, r5, . . . . The idea of electric flux lines (or electric lines of force as they are sometimes called) was introduced by Michael Faraday (1791-1867) in his experimental investigation as a way of visualizing the electric field. An electric flux line is an imaginary path or line drawn in such a way thai its direc- tion at any poinl is the direction of Ihc electric field at that point. In other words, they are the lines to which the electric field density D is tangential at every point. Any surface on which the potential is the same throughout is known as an equipoten- tial surface. The intersection of an equipotential surface and a plane results in a path or line known as an equipotential line. No work is done in moving a charge from one point to another along an equipotential line or surface (VA - VB = 0) and hence E-dl (4.83) on the line or surface. From eq. (4.83), we may conclude that the lines of force or flux lines (or the direction of E) are always normal to equipotential surfaces. Examples of equipotential surfaces for point charge and a dipole are shown in Figure 4.21. Note from these examples that the direction of E is everywhere normal to the equipotential flux line Figure 4.21 Equipotential surfaces for (a) a point charge and (b) an electric dipole. 4.10 ENERGY DENSITY IN ELECTROSTATIC FIELDS 147 If, instead of point charges, the region has a continuous charge distribution, the sum- mation in eq. (4.87) becomes integration; that is, WE = -\pLVdl (line charge) W£ = - | psV dS (surface charge) WE = — I pvV dv (volume charge) Since pv = V • D, eq. (4.90) can be further developed to yield WE = ~ j(V-D)Vdv But for any vector A and scalar V, the identity V • VA = A • VV + V(V • A) or (V • A)V = V • VA - A • VV holds. Applying the identity in eqs. (4.92) to (4.91), we get WE = - (V • VD) dv (D • VV) dv (4.88) (4.89) (4.90) (4.91) (4.92) (4.93) By applying divergence theorem to the first term on the right-hand side of this equation, we have \ 1 WE = - 4> (VD) • dS (D • VV) dv (4.94) From Section 4.9, we recall that V varies as 1/r and D as 1/r2 for point charges; V varies as 1/r2 and D as 1/r3 for dipoles; and so on. Hence, VD in the first term on the right-hand side of eq. (4.94) must vary at least as 1/r3 while dS varies as r2. Consequently, the first integral in eq. (4.94) must tend to zero as the surface S becomes large. Hence, eq. (4.94) reduces to (4.95)WE= - - (D • VV) dv = | | (D • E) dv and since E = - VV and D = eoE (4.96) 148 • Electrostatic Fields From this, we can define electrostatic energy density wE (in J/m ) as dW* 1 _ „ 1 i _ D2 wE = dv 2 2eo (4.97) so eq. (4.95) may be written as WE = wE dv (4.98) EXAMPLE 4.14 Three point charges - 1 nC, 4 nC, and 3 nC are located at (0, 0, 0), (0, 0, 1), and (1, 0, 0), respectively. Find the energy in the system. Solution: w = w, + w2 + w3 = 0 + Q2V21 + G3 V32) -a- 4TT£O Q\ 1(1,0,0) - (0,0,0)| |(l,0,0) - (0,0,l)| 4ir 10" - 4 - 3 36TT = 91-^= - 7 | nJ = 13.37 nJ Alternatively, W = 2 • 2 Qi 2 2 L4TS O (1 ) = 9( ^= - 7 ) nJ = 13.37 nJ as obtained previously. 4.10 ENERGY DENSITY IN ELECTROSTATIC FIELDS 149 PRACTICE EXERCISE 4.14 Point charges <2, = 1 nC, Q2 = - 2 nC, Q3 = 3 nC, and Q4 = - 4 nC are posi- tioned one at a time and in that order at (0, 0, 0), (1,0, 0), (0, 0, -1 ) , and (0, 0, 1), respectively. Calculate the energy in the system after each charge is positioned. Answer: 0, - 1 8 nJ, -29.18 nJ, -68.27 nJ. EXAMPLE 4.15 A charge distribution with spherical symmetry has density "po, 0 < r < R Pv = 0, r>R Determine V everywhere and the energy stored in region r < R. Solution: The D field has already been found in Section 4.6D using Gauss's law. (a) Forr » R,E = -^^ar. 3e</ Once E is known, V is determined as 3eor + C,, R Since V(r = oo) = o,Ci = 0. (b) For r =£ /?, E = — ar. 3eo Hence, por 6eo From part (a) V(r = R) = . Hence, 3e0 d\= -— | rdr3eo + C, 6eo 2en 152 Electrostatic Fields 7. The total work done, or the electric potential energy, to move a point charge Q from point A to B in an electric field E is W = - Q \ E - d l 8. The potential at r due to a point charge Q at r ' is V(r) = Q 47rso|r - r ' | C where C is evaluated at a given reference potential point; for example C = 0 if V(r —> oo) = 0. To determine the potential due to a continuous charge distribution, we replace Q in the formula for point charge by dQ = pL dl, dQ = ps dS, or dQ = pv dv and integrate over the line, surface, or volume, respectively. 9. If the charge distribution is not known, but the field intensity E is given, we find the potential using V=-\E-dl W 10. The potential difference VAB, the potential at B with reference to A, is VAB= - J -dl = -=VB-VA 11. Since an electrostatic field is conservative (the net work done along a closed path in a static E field is zero), E • dl = 0 or V X E = 0 (second Maxwell's equation to be derived) 12. Given the potential field, the corresponding electric field is found using . E = -VV 13. For an electric dipole centered at r ' with dipole moment p, the potential at r is given by V(r) = P • (r - r') 47rco|r - r ' |3 14. D is tangential to the electric flux lines at every point. An equipotential surface (or line) is one on which V = constant. At every point, the equipotential line is orthogonal to the electric flux line. REVIEW QUESTIONS ^ 153 15. The electrostatic energy due to n point charges is WE = ~ 2 QkVk 1 For a continuous volume charge distribution, = - \D-Edv = - | eo\E\zdv REVIEW QUESTIONS 4.1 Point charges Q, = 1 nC and Q2 = 2 nC are at a distance apart. Which of the following statements are incorrect? (a) The force on Ql is repulsive. (b) The force on Q2 is the same in magnitude as that on Qx. (c) As the distance between them decreases, the force on Ql increases linearly. (d) The force on Q2 is along the line joining them. (e) A point charge Q3 = — 3 nC located at the midpoint between Q{ and Q2 experiences no net force. 4.2 Plane z = 10 m carries charge 20 nC/m2. The electric field intensity at the origin is (a) -10a ,V/m (b) -187razV/m (c) -727razV/m (d) -360irazV/m 4.3 Point charges 30 nC, -20 nC, and 10 nC are located at ( -1 ,0 ,2) , (0,0,0), and (1,5, — 1), respectively. The total flux leaving a cube of side 6 m centered at the origin is: (a) - 2 0 nC (b) 10 nC (c) 20 nC (d) 30 nC (e) 60 nC 4.4 The electric flux density on a spherical surface r = b is the same for a point charge Q located at the origin and for charge Q uniformly distributed on surface r = a(a < b). (a) Yes (b) No (c) Not necessarily 154 Electrostatic Fields 4.5 The work done by the force F = 4ax - 3ay + 2az N in giving a 1 nC charge a displace- f lO + 2 7 iay - az m sment o (a) 103 nJ (b) 60 nJ (c) 64 nJ (d) 20 nJ 4.6 By saying that the electrostatic field is conservative, we do not mean that (a) It is the gradient of a scalar potential. (b) Its circulation is identically zero. (c) Its curl is identically zero. (d) The work done in a closed path inside the field is zero. (e) The potential difference between any two points is zero. 4.7 Suppose a uniform electric field exists in the room in which you are working, such that the lines of force are horizontal and at right angles to one wall. As you walk toward the wall from which the lines of force emerge into the room, are you walking toward (a) Points of higher potential? (b) Points of lower potential? (c) Points of the same potential (equipotential line)? 4.8 A charge Q is uniformly distributed throughout a sphere of radius a. Taking the potential at infinity as zero, the potential at r = b < a is (a) - (b) - (c) - (d) - Q 0 4irsor2 a Q = 4ireor2 " Q dr dr - Qr 4irena dr dr 4.9 A potential field is given by V = 3x2y - yz- Which of the following is not true? (a) At point (1, 0, - 1), V and E vanish. (b) x2y = 1 is an equipotential line on the xy-plane. (c) The equipotential surface V = — 8 passes through point P(2, —1,4). (d) The electric field at P is 12a^ - 8a,, - az V/m. (e) A unit normal to the equipotential surface V = —8 at P is —0.83a^ + 0.55aj,+ 0.07a7. +Q -2 -2Q ) 2 0 -2 +2Q 2 -e PROBLEMS Figure 4.24 For Problem 4.14. 157 *4.15 State Gauss's law. Deduce Coulomb's law from Gauss's law thereby affirming that Gauss's law is an alternative statement of Coulomb's law and that Coulomb's law is im- plicit in Maxwell's equation V • D = pv. 4.16 Determine the charge density due to each of the following electric flux densities: (a) D = %xyax + 4x\ C/m2 (b) D = p sin <t> ap + 2p cos + 2z2az C/m2 (c) D = 2 cos 6 ar + sin 0 C/m2 4.17 Let E = xyax + x2ay, find (a) Electric flux density D. (b) The volume charge density pv. 4.18 Plane x + 2y = 5 carries charge ps = 6 nC/m2. Determining E at ( - 1 , 0, 1). 4.19 In free space, D = 2v2a,t + 4xy - az mC/m2. Find the total charge stored in the region l < x < 2 , l < y < 2 , - 1 < z < 4. 4.20 In a certain region, the electric field is given by D = 2p(z + l)cos </> ap - p(z + l)sin 0 a0 + p2 cos <t> az /^C/m2 (a) Find the charge density. \ . (b) Calculate the total charge enclosed by the volume 0 < p < 2, 0 < <t> < x/2, 0 < z < 4. (c) Confirm Gauss's law by finding the net flux through the surface of the volume in (b). *4.21 The Thomson model of a hydrogen atom is a sphere of positive charge with an electron (a point charge) at its center. The total positive charge equals the electronic charge e. Prove 158 Electrostatic Fields that when the electron is at a distance r from the center of the sphere of positive charge, it is attracted with a force F = where R is the radius of the sphere. 4.22 Three concentric spherical shells r = 1, r = 2, and r = 3 m, respectively, have charge distributions 2, - 4 , and 5 (a) Calculate the flux through r = 1.5 m and r = 2.5 m. (b) Find D at r = 0.5, r = 2.5, and r = 3.5 m. 4.23 Given that j Determine D everywhere. 4.24 Let / Up nC/m3, \0, 1 <P < 2 otherwise mC/m3, 0, 1 < r < 4 r> 0 (a) Find the net flux crossing surface r = 2 m and r = 6 m. (b) Determine D at r = 1 m and r = 5 m. 4.25 Find the work done in carrying a 5-C charge from P(l, 2, -4 ) to R(3, - 5 , 6) in an elec- tric field E = ax + z \ + 2yzaz V/m 4.26 Given that the electric field in a certain region is E = (z + 1) sin 0 a, + (z + 1) cos a0 + p sin <£ az V/m determine the work done in moving a 4-nC charge from (a) A(l,0, 0)toB(4, 0,0) (b) S(4, 0, 0) to C(4, 30°, 0) (c) C(4, 30°, 0)toD(4, 30°, -2 ) (d) AtoD 4̂ 27 In an electric field E = 20r sin 6 ar + lOr cos 6 ae V/m, calculate the energy expended in transferring a 10-nC charge (a) From A(5, 30°, 0°) to B(5, 90°, 0°) (b) From A to C( 10, 30°, 0°) (c) FromAtoD(5,30°, 60°) (d) From A to £(10, 90°, 60°) PROBLEMS 159 4.28 Let V = xy2z, calculate the energy expended in transfering a 2-^C point charge from ( 1 , - 1 , 2) to (2, 1,-3). 4.29 Determine the electric field due to the following potentials: (a) V= x2 + 2y2 + 4z2 (b) V = sin(x2 + y2 + z2)1'2 (c) V = p2(z + l)sin (j> (d) V = e'r sin 6 cos 2<t> 4.30 Three point charges gi = 1 mC, Q2 = - 2 m C , and Q3 = 3 mC are, respectively, located at (0, 0, 4), ( -2 , 5, 1), and (3, - 4 , 6). (a) Find the potential VP at P(-1, 1, 2). (b) Calculate the potential difference VPQ if Q is (1, 2, 3). 4.31 In free space, V = x2y(z + 3) V. Find (a) Eat (3, 4, - 6 ) (b) the charge within the cube 0 < x,y,z < 1. 4.32 A spherical charge distribution is given by r<a 2 j 0, r> a Find V everywhere. 4.33 To verify that E = yzax + xzay + xyaz V/m is truely an electric field, show that (a) V X E = 0 (b) j>L E • d\ = 0, where L is the edge of the square defined \ayO<x,y<2,z= 1 • 4.34 (a) A total charge Q = 60 fiC is split into two equal charges located at 180° intervals around a circular loop of radius 4 m. Find the potential at the center of the loop. (b) If Q is split into three equal charges spaced at 120° intervals around the loop, find the potential at the center. Q (c) If in the limit pL = — , find the potential at the center. 4.35 For a spherical charge distribution Pv = Po(a2 - r2) , 0, r < a r > a (a) Find E and V for r> a (b) Find E and V for r < a (c) Find the total charge (d) Show that E is maximum when r = 0.145a.