Chapter2 7th

CHAPTER 2 2.1. Four 10nC positive charges are located in the z = 0 plane at the corners of a square 8cm on a side. A fifth 10nC positive charge is located at a point 8cm distant from the other charges. Calculate the magnitude of the total force on this fifth charge for ǫ = ǫ0 : Arrange the charges in the xy plane at locations (4,4), (4,-4), (-4,4), and (-4,-4). Then the fifth charge √ will be on the z axis at location z = 4 2, which puts it at 8cm distance from the other four. By symmetry, the force on the fifth charge will be z-directed, and will be four times the z component of force produced by each of the four other charges. q2 (10−8 )2 4 4 √ × F =√ × = = 4.0 × 10−4 N 2 4πǫ0 d2 2 4π(8.85 × 10−12 )(0.08)2 2.2. Two point charges of Q1 coulombs each are located at (0,0,1) and (0,0,-1). (a) Determine the locus of the possible positions of a third charge Q2 where Q2 may be any positive or negative value, such that the total field E = 0 at (0,1,0): The total field at (0,1,0) from the two Q1 charges (where both are positive) will be E1 (0, 1, 0) = 2Q1 Q1 cos 45◦ ay = √ ay 2 4πǫ0 R 4 2πǫ0 √ where R = 2. To cancel this field, Q2 must be placed on the y axis at positions y > 1 if Q2 > 0, and at positions y < 1 if Q2 < 0. In either case the field from Q2 will be E2 (0, 1, 0) = −|Q2 | ay 4πǫ0 and the total field is then
 Q1 |Q2 | √ Et = E1 + E2 = − =0 4πǫ0 4 2πǫ0 Therefore |Q2 | Q √1 = ⇒ y = 1 ± 21/4 (y − 1)2 2  |Q2 | Q1 where the plus sign is used if Q2 > 0, and the minus sign is used if Q2 < 0. (b) What is the locus if the two original charges are Q1 and −Q1 ? √ In this case the total field at (0,1,0) is E1 (0, 1, 0) = −Q1 /(4 2πǫ0 ) az , where the positive Q1 is located at the positive z (= 1) value. We now need Q2 to lie along the line x = 0, y = 1 in order to cancel the field from the positive and negative Q1 charges. Assuming Q2 is located at (0, 1, z), the total field is now −Q1 |Q2 | =0 az + Et = E1 + E2 = √ 4πǫ0 z 2 4 2πǫ0  or z = ±21/4 |Q2 |/Q1 , where the plus sign is used if Q2 < 0, and the minus sign if Q2 > 0. 1 2.3. Point charges of 50nC each are located at A(1, 0, 0), B(−1, 0, 0), C(0, 1, 0), and D(0, −1, 0) in free space. Find the total force on the charge at A. The force will be:
 (50 × 10−9 )2 RCA RDA RBA F= + + 4πǫ0 |RCA |3 |RDA |3 |RBA |3 where RCA = ax − ay , RDA = ax + ay , and RBA = 2ax . The magnitudes are |RCA | = |RDA | = and |RBA | = 2. Substituting these leads to 
(50 × 10−9 )2 1 2 1 √ + √ + F= ax = 21.5ax µN 4πǫ0 2 2 2 2 8 √ 2, where distances are in meters. 2.4. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with one charge at the origin, and with the three nearest charges at (a, 0, 0), (0, a, 0), and (0, 0, a). Find an expression for the total vector force on the charge at P (a, a, a), assuming free space: The total electric field at P (a, a, a) that produces a force on the charge there will be the sum of the fields from the other seven charges. This is written below, where the charge locations associated with each term are indicated:   Enet (a, a, a) = q 4πǫ0 a2 a + a + a  ay + az ax + az ax + ay  x  y z √ √ √ √ + + + + a + a + a  x y z       3 3 2 2 2 2 2 2     (0,a,a) (a,a,0) (a,0,a) (0,0,0) (a,0,0) (0,a,0) (0,0,a) The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain 
1 1.90 q 2 1 q2 √ √ + + 1 (a F(a, a, a) = qEnet (a, a, a) = + a + a ) = (ax + ay + az ) x y z 4πǫ0 a2 3 3 4πǫ0 a2 2 in which the magnitude is |F| = 3.29 q 2 /(4πǫ0 a2 ). 2.5. Let a point charge Q1 25 nC be located at P1 (4, −2, 7) and a charge Q2 = 60 nC be at P2 (−3, 4, −2). a) If ǫ = ǫ0 , find E at P3 (1, 2, 3): This field will be
 10−9 25R13 60R23 E= + 4πǫ0 |R13 |3 |R23 |3 √ √ where R13 = −3ax + 4ay − 4az and R23 = 4ax − 2ay + 5az . Also, |R13 | = 41 and |R23 | = 45. So
 10−9 25 × (−3ax + 4ay − 4az ) 60 × (4ax − 2ay + 5az ) + E= 4πǫ0 (41)1.5 (45)1.5 = 4.58ax − 0.15ay + 5.51az b) At what point on the y axis is Ex = 0? P3 is now  at (0, y, 0), so R13 = −4ax +  (y + 2)ay − 7az 2 and R23 = 3ax + (y − 4)ay + 2az . Also, |R13 | = 65 + (y + 2) and |R23 | = 13 + (y − 4)2 . Now the x component of E at the new P3 will be:
 25 × (−4) 60 × 3 10−9 + Ex = 4πǫ0 [65 + (y + 2)2 ]1.5 [13 + (y − 4)2 ]1.5 2 To obtain Ex = 0, we require the expression in the large brackets to be zero. This expression simplifies to the following quadratic: 0.48y 2 + 13.92y + 73.10 = 0 which yields the two values: y = −6.89, −22.11 2.6. Three point charges, each 5 × 10−9 C, are located on the x axis at x = −1, 0, and 1 in free space. a) Find E at x = 5: At a general location, x,
 1 1 1 q ax + + E(x) = 4πǫ0 (x + 1)2 x2 (x − 1)2 At x = 5, and with q = 5 × 10−9 C, this becomes E(x = 5) = 5.8 ax V/m. b) Determine the value and location of the equivalent single point charge that would produce the same field at very large distances: For x >> 1, the above general field in part a becomes . E(x >> 1) = 3q ax 4πǫ0 x2 Therefore, the equivalent charge will have value 3q = 1.5 × 10−8 C, and will be at location x = 0. c) Determine E at x = 5, using the approximation of (b). Using 3q = 1.5 × 10−8 C and x = 5 in . the part b result gives E(x = 5) = 5.4 ax V/m, or about 7% lower than the exact result. 2.7. A 2 µC point charge is located at A(4, 3, 5) in free space. Find Eρ , Eφ , and Ez at P (8, 12, 2). Have
 2 × 10−6 RAP 2 × 10−6 4ax + 9ay − 3az EP = = 65.9ax + 148.3ay − 49.4az = 4πǫ0 |RAP |3 4πǫ0 (106)1.5 Then, at point P , ρ = √ 82 + 122 = 14.4, φ = tan−1 (12/8) = 56.3◦ , and z = z. Now, Eρ = Ep · aρ = 65.9(ax · aρ ) + 148.3(ay · aρ ) = 65.9 cos(56.3◦ ) + 148.3 sin(56.3◦ ) = 159.7 and Eφ = Ep · aφ = 65.9(ax · aφ ) + 148.3(ay · aφ ) = −65.9 sin(56.3◦ ) + 148.3 cos(56.3◦ ) = 27.4 Finally, Ez = −49.4 V/m 2.8. A crude device for measuring charge consists of two small insulating spheres of radius a, one of which is fixed in position. The other is movable along the x axis, and is subject to a restraining force kx, where k is a spring constant. The uncharged spheres are centered at x = 0 and x = d, the latter fixed. If the spheres are given equal and opposite charges of Q coulombs: a) Obtain the expression by which Q may be found as a function of x: The spheres will attract, and so the movable sphere at x = 0 will move toward the other until the spring and Coulomb forces balance. This will occur at location x for the movable sphere. With equal and opposite forces, we have Q2 = kx 4πǫ0 (d − x)2 3 √ from which Q = 2(d − x) πǫ0 kx. b) Determine the maximum charge that can be measured in terms of ǫ0 , k, and d, and state the separation of the spheres then: With increasing charge, the spheres move toward each other until they just touch at xmax = d − 2a. Using the part a result, we find the maximum measurable  charge: Qmax = 4a πǫ0 k(d − 2a). Presumably some form of stop mechanism is placed at x = x− max to prevent the spheres from actually touching. c) What happens if a larger charge is applied? No further motion is possible, so nothing happens. 2.9. A 100 nC point charge is located at A(−1, 1, 3) in free space. a) Find the locus of all points P (x, y, z) at which Ex = 500 V/m: The total field at P will be: EP = 100 × 10−9 RAP 4πǫ0 |RAP |3 where RAP = (x+1)ax +(y−1)ay +(z −3)az , and where |RAP | = [(x+1)2 +(y−1)2 +(z −3)2 ]1/2 . The x component of the field will be Ex =
 100 × 10−9 (x + 1) = 500 V/m 4πǫ0 [(x + 1)2 + (y − 1)2 + (z − 3)2 ]1.5 And so our condition becomes: (x + 1) = 0.56 [(x + 1)2 + (y − 1)2 + (z − 3)2 ]1.5 b) Find y1 if P (−2, y1 , 3) lies on that locus: At point P , the condition of part a becomes  3 3.19 = 1 + (y1 − 1)2 from which (y1 − 1)2 = 0.47, or y1 = 1.69 or 0.31 2.10. A positive test charge is used to explore the field of a single positive point charge Q√at P (a, b, c). If the test charge is placed at the origin, the force on it is in the direction 0.5 ax − 0.5 3 ay , and when the test charge is moved to (1,0,0), the force is in the direction of 0.6 ax − 0.8 ay . Find a, b, and c: We first construct the field using the form of Eq. (12). We identify r = xax + yay + zaz and r′ = aax + bay + caz . Then E= Q [(x − a) ax + (y − b) ay + (z − c) az ] 3/2 4πǫ0 [(x − a)2 + (y − b)2 + (z − c)2 ] (1) Using (1), we can write the two force directions at the two test charge positions as follows: at (0, 0, 0) : √ [−a ax − b ay − c az ] = 0.5 a − 0.5 3 ay x (a2 + b2 + c2 )1/2 (2) at (1, 0, 0) : [(1 − a) ax − b ay − c az ] = 0.6 ax − 0.8 ay ((1 − a)2 + b2 + c2 )1/2 (3) 4 √ We observe immediately that c = 0. Also, from (2) we find that b = −a 3, and therefore √ a2 + b2 = 2a. Using this information in (3), we write for the x component:  1−a (1 − a)2 + b2 =√ 1−a = 0.6 1 − 2a + 4a2 or 0.44a2 + 1.28a − 0.64 = 0, so that a= −1.28 ±  (1.28)2 + 4(0.44)(0.64) = 0.435 or − 3.344 0.88 The corresponding b values are respectively −0.753 and 5.793. So the two possible P coordinate sets are (0.435, −0.753, 0) and (−3.344, 5.793, 0). By direct substitution, however, it is found that only one possibility is entirely consistent with both (2) and (3), and this is P (a, b, c) = (−3.344, 5.793, 0) 2.11. A charge Q0 located at the origin in free space produces a field for which Ez = 1 kV/m at point P (−2, 1, −1). a) Find Q0 : The field at P will be
 Q0 −2ax + ay − az EP = 4πǫ0 61.5 Since the z component is of value 1 kV/m, we find Q0 = −4πǫ0 61.5 × 103 = −1.63 µC. b) Find E at M (1, 6, 5) in cartesian coordinates: This field will be: EM
 −1.63 × 10−6 ax + 6ay + 5az = 4πǫ0 [1 + 36 + 25]1.5 or EM = −30.11ax − 180.63ay − 150.53az . c) Find E at M (1, 6, 5) in cylindrical coordinates: At M , ρ = 80.54◦ , and z = 5. Now √ 1 + 36 = 6.08, φ = tan−1 (6/1) = Eρ = EM · aρ = −30.11 cos φ − 180.63 sin φ = −183.12 Eφ = EM · aφ = −30.11(− sin φ) − 180.63 cos φ = 0 (as expected) so that EM = −183.12aρ − 150.53az . √ d) Find E at M (1, 6, 5) in spherical coordinates: At M , r = 1 + 36 + 25 = 7.87, φ = 80.54◦ (as before), and θ = cos−1 (5/7.87) = 50.58◦ . Now, since the charge is at the origin, we expect to obtain only a radial component of EM . This will be: Er = EM · ar = −30.11 sin θ cos φ − 180.63 sin θ sin φ − 150.53 cos θ = −237.1 5 2.12. Electrons are in random motion in a fixed region in space. During any 1µs interval, the probability of finding an electron in a subregion of volume 10−15 m2 is 0.27. What volume charge density, appropriate for such time durations, should be assigned to that subregion? The finite probabilty effectively reduces the net charge quantity by the probability fraction. With e = −1.602 × 10−19 C, the density becomes ρv = − 0.27 × 1.602 × 10−19 = −43.3 µC/m3 10−15 2.13. A uniform volume charge density of 0.2 µC/m3 is present throughout the spherical shell extending from r = 3 cm to r = 5 cm. If ρv = 0 elsewhere: a) find the total charge present throughout the shell: This will be Q=  0 2π  0 π  .05 .03
.05 r3 0.2 r2 sin θ dr dθ dφ = 4π(0.2) = 8.21 × 10−5 µC = 82.1 pC 3 .03 b) find r1 if half the total charge is located in the region 3 cm < r < r1 : If the integral over r in part a is taken to r1 , we would obtain
r r3 1 4π(0.2) = 4.105 × 10−5 3 .03 Thus
3 × 4.105 × 10−5 + (.03)3 r1 = 0.2 × 4π 1/3 = 4.24 cm 2.14. The charge density varies with radius in a cylindrical coordinate system as ρv = ρ0 /(ρ2 + a2 )2 C/m3 . Within what distance from the z axis does half the total charge lie? Choosing a unit length in z, the charge contained up to radius ρ is

 ρ  1  2π  ρ πρ0 1 −1 ρ0 ′ ′ = 2 1− ρ dρ dφdz = 2πρ0 Q(ρ) = ′2 2 2 2(a2 + ρ′ 2) 0 a 1 + ρ2 /a2 0 (ρ + a ) 0 0 The total charge is found when ρ → ∞, or Qnet = πρ0 /a2 . It is seen from the Q(ρ) expression that half of this occurs when ρ = a. 2.15. A spherical volume having a 2 µm radius contains a uniform volume charge density of 1015 C/m3 . a) What total charge is enclosed in the spherical volume? This will be Q = (4/3)π(2 × 10−6 )3 × 1015 = 3.35 × 10−2 C. b) Now assume that a large region contains one of these little spheres at every corner of a cubical grid 3mm on a side, and that there is no charge between spheres. What is the average volume charge density throughout this large region? Each cube will contain the equivalent of one little sphere. Neglecting the little sphere volume, the average density becomes ρv,avg = 3.35 × 10−2 = 1.24 × 106 C/m3 (0.003)3 6 2.16. Within a region of free space, charge density is given as ρv = ρ0 r/a C/m3 , where ρ0 and a are constants. Find the total charge lying within: a) the sphere, r ≤ a: This will be  Qa =  2π π a 0 0 0  ρ0 r 2 r sin θ dr dθ dφ = 4π a  a 0 ρ0 r3 dr = πρ0 a3 a b) the cone, r ≤ a, 0 ≤ θ ≤ 0.1π: Qb =  2π 0  0.1π  a 0 0 ρ0 r 2 ρ0 a3 r sin θ dr dθ dφ = 2π [1 − cos(0.1π)] = 0.024πρ0 a3 a 4 c) the region, r ≤ a, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤ 0.2π. Qc =  0.2π 0  0 0.1π  0 a ρ0 r 2 r sin θ dr dθ dφ = 0.024πρ0 a3 a  0.2π 2π  = 0.0024πρ0 a3 2.17. A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If ǫ = ǫ0 : a) Find E at P (1, 2, 3): This will be EP = RP ρl 2πǫ0 |RP |2 where RP = (1, 2, 3) − (1, −2, 5) = (0, 4, −2), and |RP |2 = 20. So EP =
 16 × 10−9 4ay − 2az = 57.5ay − 28.8az V/m 2πǫ0 20 b) Find E at that point in the z = 0 plane where the direction of E is given by (1/3)ay − (2/3)az : With z = 0, the general field will be
 (y + 2)ay − 5az ρl Ez=0 = 2πǫ0 (y + 2)2 + 25 We require |Ez | = −|2Ey |, so 2(y + 2) = 5. Thus y = 1/2, and the field becomes: Ez=0 =
 ρl 2.5ay − 5az = 23ay − 46az 2πǫ0 (2.5)2 + 25 2.18. An infinite uniform line charge ρL = 2 nC/m lies along the x axis in free space, while point charges of 8 nC each are located at (0,0,1) and (0,0,-1). a) Find E at (2,3,-4). The net electric field from the line charge, the point charge at z = 1, and the point charge at z = −1 will be (in that order): Etot
 1 2ρL (3 ay − 4 az ) q(2 ax + 3 ay − 5 az ) q(2 ax + 3 ay − 3 az ) + + = 4πǫ0 25 (38)3/2 (22)3/2 7 Then, with the given values of ρL and q, the field evaluates as Etot = 2.0 ax + 7.3 ay − 9.4 az V/m b) To what value should ρL be changed to cause E to be zero at (0,0,3)? In this case, we only need scalar addition to find the net field: E(0, 0, 3) = Therefore  1 2ρL 1 + =− q 4 16 3
q q ρL + =0 + 2 2πǫ0 (3) 4πǫ0 (2) 4πǫ0 (4)2 ⇒ ρL = − 15 q = −0.47q = −3.75 nC/m 32 2.19. A uniform line charge of 2 µC/m is located on the z axis. Find E in cartesian coordinates at P (1, 2, 3) if the charge extends from a) −∞ < z < ∞: With the infinite line, we know that the field will have only a radial component in cylindrical coordinates (or x and y components in cartesian). The field from an infinite line on the z axis is generally E = [ρl /(2πǫ0 ρ)]aρ . Therefore, at point P : EP = ρl RzP (2 × 10−6 ) ax + 2ay = = 7.2ax + 14.4ay kV/m 2πǫ0 |RzP |2 2πǫ0 5 where RzP is the vector that extends from the line charge to point P , and is perpendicular to the z axis; i.e., RzP = (1, 2, 3) − (0, 0, 3) = (1, 2, 0). b) −4 ≤ z ≤ 4: Here we use the general relation EP =  ρl dz r − r′ 4πǫ0 |r − r′ |3 where r = ax + 2ay + 3az and r′ = zaz . So the integral becomes (2 × 10−6 ) EP = 4πǫ0  4 −4 ax + 2ay + (3 − z)az dz [5 + (3 − z)2 ]1.5 Using integral tables, we obtain: EP = 3597
(ax + 2ay )(z − 3) + 5az (z 2 − 6z + 14) 4 V/m = 4.9ax + 9.8ay + 4.9az kV/m −4 The student is invited to verify that when evaluating the above expression over the limits −∞ < z < ∞, the z component vanishes and the x and y components become those found in part a. 8 2.20. The portion of the z axis for which |z| < 2 carries a nonuniform line charge density of 10|z| nC/m, and ρL = 0 elsewhere. Determine E in free space at: a) (0,0,4): The general form for the differential field at (0,0,4) is dE = ρL dz (r − r′ ) 4πǫ0 |r − r′ |3 where r = 4 az and r′ = z az . Therefore, r − r′ = (4 − z) az and |r − r′ | = 4 − z. Substituting ρL = 10|z| nC/m, the total field is E(0, 0, 4) =  2 10−8 |z| dz az = 4πǫ0 (4 − z)2 −2 = 10−8 4π × 8.854 × 10−12  2 10−8 z dz az − 4πǫ0 (4 − z)2  0 10−8 z dz az 2 0 −2 4πǫ0 (4 − z) 
2
0  4 4 ln(4 − z) + − ln(4 − z) + az 4−z 0 4 − z −2 = 34.0 az V/m b) (0,4,0): In this case, r = 4 ay and r′ = z az as before. The field at (0,4,0) is then E(0, 4, 0) =  2 −2 10−8 |z| dz (4 ay − z az ) 4πǫ0 (16 + z 2 )3/2 Note the symmetric limits on the integral. As the z component of the integrand changes sign at z = 0, it will contribute equal and opposite portions to the overall integral, which will cancel completely (the z component integral has odd parity). This leaves only the y component integrand, which has even parity. The integral therefore simplifies to E(0, 4, 0) = 2  2 0 2
−2 × 10−8 ay 1 4 × 10−8 z dz ay √ = 18.98 ay V/m = π × 8.854 × 10−12 4πǫ0 (16 + z 2 )3/2 16 + z 2 0 2.21. Two identical uniform line charges with ρl = 75 nC/m are located in free space at x = 0, y = ±0.4 m. What force per unit length does each line charge exert on the other? The charges are parallel to the z axis and are separated by 0.8 m. Thus the field from the charge at y = −0.4 evaluated at the location of the charge at y = +0.4 will be E = [ρl /(2πǫ0 (0.8))]ay . The force on a differential length of the line at the positive y location is dF = dqE = ρl dzE. Thus the force per unit length acting on the line at postive y arising from the charge at negative y is F=  0 1 ρ2l dz ay = 1.26 × 10−4 ay N/m = 126 ay µN/m 2πǫ0 (0.8) The force on the line at negative y is of course the same, but with −ay . 2.22. Two identical uniform sheet charges with ρs = 100 nC/m2 are located in free space at z = ±2.0 cm. What force per unit area does each sheet exert on the other? The field from the top sheet is E = −ρs /(2ǫ0 ) az V/m. The differential force produced by this field on the bottom sheet is the charge density on the bottom sheet times the differential area there, multiplied by the electric field from the top sheet: dF = ρs daE. The force per unit area is then just F = ρs E = (100 × 10−9 )(−100 × 10−9 )/(2ǫ0 ) az = −5.6 × 10−4 az N/m2 . 9 2.23. Given the surface charge density, ρs = 2 µC/m2 , in the region ρ < 0.2 m, z = 0, and is zero elsewhere, find E at: a) PA (ρ = 0, z = 0.5): First, we recognize from symmetry that only a z component of E will be present. Considering a general point z on the z axis, we have r = zaz . Then, with r′ = ρaρ , we obtain r − r′ = zaz − ρaρ . The superposition integral for the z component of E will be: Ez,PA ρs = 4πǫ0  2π  0.2 2πρs z ρ dρ dφ =− z 2 2 1.5 (ρ + z ) 4πǫ0 0 0 
ρs 1 1 √ √ = − z 2ǫ0 z 2 + 0.04 z2  1  z 2 + ρ2 0.2 0 With z = 0.5 m, the above evaluates as Ez,PA = 8.1 kV/m. b) With z at −0.5 m, we evaluate the expression for Ez to obtain Ez,PB = −8.1 kV/m. 2.24. For the charged disk of Problem 2.23, show that: a) the field along the z axis reduces to that of an infinite sheet charge at small values of z: In general, the field can be expressed as 
ρs z 1− √ Ez = 2ǫ0 z 2 + 0.04 . At small z, this reduces to Ez = ρs /2ǫ0 , which is the infinite sheet charge field. b) the z axis field reduces to that of a point charge at large values of z: The development is as follows:  

  z z 1 ρs ρs . ρs = Ez = 1−  1− √ 1− = 2ǫ0 2ǫ0 2ǫ0 1 + (1/2)(0.04)/z 2 z 2 + 0.04 z 1 + 0.04/z 2 where the last approximation is valid if z >> .04. Continuing:  0.04ρs π(0.2)2 ρs . ρs  Ez = 1 − [1 − (1/2)(0.04)/z 2 ] = = 2ǫ0 4ǫ0 z 2 4πǫ0 z 2 This the point charge field, where we identify q = π(0.2)2 ρs as the total charge on the disk (which now looks like a point). 2.25. Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC at P (2, 0, 6); uniform line charge density, 3nC/m at x = −2, y = 3; uniform surface charge density, 0.2 nC/m2 at x = 2. The sum of the fields at the origin from each charge in order is:

 
(12 × 10−9 ) (−2ax − 6az ) (0.2 × 10−9 )ax (3 × 10−9 ) (2ax − 3ay ) E= − + 4πǫ0 (4 + 36)1.5 2πǫ0 (4 + 9) 2ǫ0 = −3.9ax − 12.4ay − 2.5az V/m 10 2.26. An electric dipole (discussed in detail in Sec. 4.7) consists of two point charges of equal and opposite magnitude ±Q spaced by distance d. With the charges along the z axis at positions z = ±d/2 (with the positive  positive z location), the electric field in spherical coordinates is given by  charge at the E(r, θ) = Qd/(4πǫ0 r3 ) [2 cos θar + sin θaθ ], where r >> d. Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude q: a) at (0,0,z): Here, θ = 0, ar = az , and r = z. Therefore F(0, 0, z) = qQd az N 4πǫ0 z 3 b) at (0,y,0): Here, θ = 90◦ , aθ = −az , and r = y. The force is F(0, y, 0) = −qQd az N 4πǫ0 y 3 2.27. Given the electric field E = (4x − 2y)ax − (2x + 4y)ay , find: a) the equation of the streamline that passes through the point P (2, 3, −4): We write dy Ey −(2x + 4y) = = dx Ex (4x − 2y) Thus 2(x dy + y dx) = y dy − x dx or 2 d(xy) = 1 1 d(y 2 ) − d(x2 ) 2 2 So C1 + 2xy = 1 2 1 2 y − x 2 2 or y 2 − x2 = 4xy + C2 Evaluating at P (2, 3, −4), obtain: 9 − 4 = 24 + C2 , or C2 = −19 Finally, at P , the requested equation is y 2 − x2 = 4xy − 19 b) a unit vector specifying the direction √ of E at Q(3, −2, 5): Have EQ = [4(3) + 2(2)]ax − [2(3) − 4(2)]ay = 16ax + 2ay . Then |E| = 162 + 4 = 16.12 So aQ = 16ax + 2ay = 0.99ax + 0.12ay 16.12 11 2.28 A field is given as E = 2xz 2 ax + 2z(x2 + 1)az . Find the equation of the streamline passing through the point (1,3,-1): Ez x2 + 1 x2 + 1 dz = = ⇒ zdz = dx ⇒ z 2 = x2 + 2 ln x + C dx Ex xz x At (1,3,-1), the expression is satisfied if C = 0. Therefore, the equation for the streamline is z 2 = x2 + 2 ln x. 2.29. If E = 20e−5y (cos 5xax − sin 5xay ), find: a) |E| at P (π/6, 0.1, 2): Substituting this point, we obtain EP = −10.6ax − 6.1ay , and so |EP | = 12.2. b) a unit vector in the direction of EP : The unit vector associated with E is (cos 5xax − sin 5xay ), which evaluated at P becomes aE = −0.87ax − 0.50ay . c) the equation of the direction line passing through P : Use dy − sin 5x = = − tan 5x ⇒ dy = − tan 5x dx dx cos 5x Thus y = 1 5 ln cos 5x + C. Evaluating at P , we find C = 0.13, and so y= 1 ln cos 5x + 0.13 5 2.30. For fields that do not vary with z in cylindrical coordinates, the equations of the streamlines are obtained by solving the differential equation Eρ /Eφ = dρ(ρdφ). Find the equation of the line passing through the point (2, 30◦ , 0) for the field E = ρ cos 2φ aρ − ρ sin 2φ aφ : Eρ dρ −ρ cos 2φ dρ = = = − cot 2φ ⇒ = − cot 2φ dφ Eφ ρdφ ρ sin 2φ ρ Integrate to obtain
C 2 ln ρ = ln sin 2φ + ln C = ln sin 2φ  ⇒ ρ2 = C sin 2φ √ At the given point, we have 4 = C/ sin(60◦ ) ⇒ C = 4 sin 60◦ = 2 3. Finally, the equation for √ the streamline is ρ2 = 2 3/ sin 2φ. 12