Capitulo 2 solução hayt eletromagnetismo 8ed, Manuais, Projetos, Pesquisas de Eletromagnetismo

Baixe Capitulo 2 solução hayt eletromagnetismo 8ed e outras Manuais, Projetos, Pesquisas em PDF para Eletromagnetismo, somente na Docsity! CHAPTER 2 2.1. Three point charges are positioned in the x-y plane as follows: 5nC at y = 5 cm, -10 nC at y = −5 cm, 15 nC at x = −5 cm. Find the required x-y coordinates of a 20-nC fourth charge that will produce a zero electric field at the origin. With the charges thus configured, the electric field at the origin will be the superposition of the individual charge fields: E0 = 1 4π≤0 ∑ 15 (5)2 ax − 5 (5)2 ay − 10 (5)2 ay ∏ = 1 4π≤0 µ 3 5 ∂ [ax − ay] nC/m The field, E20, associated with the 20-nC charge (evaluated at the origin) must exactly cancel this field, so we write: E20 = −1 4π≤0 µ 3 5 ∂ [ax − ay] = −20 4π≤0ρ2 µ 1√ 2 ∂ [ax − ay] From this, we identify the distance from the origin: ρ = q 100/(3 √ 2) = 4.85. The x and y coordinates of the 20-nC charge will both be equal in magnitude to 4.85/ √ 2 = 3.43. The coodinates of the 20-nC charge are then (3.43,−3.43). 2.2. Point charges of 1nC and -2nC are located at (0,0,0) and (1,1,1), respectively, in free space. Determine the vector force acting on each charge. First, the electric field intensity associated with the 1nC charge, evalutated at the -2nC charge location is: E12 = 1 4π≤0(3) µ 1√ 3 ∂ (ax + ay + az) nC/m in which the distance between charges is √ 3 m. The force on the -2nC charge is then F12 = q2E12 = −2 12 √ 3π≤0 (ax + ay + az) = −1 10.4π≤0 (ax + ay + az) nN The force on the 1nC charge at the origin is just the opposite of this result, or F21 = +1 10.4π≤0 (ax + ay + az) nN 12 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: F = (50× 10−9)2 4π≤0 ∑ RCA |RCA|3 + RDA |RDA|3 + RBA |RBA|3 ∏ where RCA = ax − ay, RDA = ax + ay, and RBA = 2ax. The magnitudes are |RCA| = |RDA| = √ 2, and |RBA| = 2. Substituting these leads to F = (50× 10−9)2 4π≤0 ∑ 1 2 √ 2 + 1 2 √ 2 + 2 8 ∏ ax = 21.5ax µN 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 4π≤0a2   ax + ay + az 3 √ 3| {z } (0,0,0) + ay + az 2 √ 2| {z } (a,0,0) + ax + az 2 √ 2| {z } (0,a,0) + ax + ay 2 √ 2| {z } (0,0,a) + ax|{z} (0,a,a) + ay|{z} (a,0,a) + az|{z} (a,a,0)   The force is now the product of this field and the charge at (a, a, a). Simplifying, we obtain F(a, a, a) = qEnet(a, a, a) = q2 4π≤0a2 ∑ 1 3 √ 3 + 1√ 2 + 1 ∏ (ax + ay + az) = 1.90 q2 4π≤0a2 (ax + ay + az) in which the magnitude is |F| = 3.29 q2/(4π≤0a2). 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 ≤ = ≤0, find E at P3(1, 2, 3): This field will be E = 10−9 4π≤0 ∑ 25R13 |R13|3 + 60R23 |R23|3 ∏ where R13 = −3ax +4ay−4az and R23 = 4ax−2ay +5az. Also, |R13| = √ 41 and |R23| = √ 45. So E = 10−9 4π≤0 ∑ 25× (−3ax + 4ay − 4az) (41)1.5 + 60× (4ax − 2ay + 5az) (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 and R23 = 3ax + (y − 4)ay + 2az. Also, |R13| = p 65 + (y + 2)2 and |R23| = p 13 + (y − 4)2. Now the x component of E at the new P3 will be: Ex = 10−9 4π≤0 ∑ 25× (−4) [65 + (y + 2)2]1.5 + 60× 3 [13 + (y − 4)2]1.5 ∏ 13 2.9b) Find y1 if P (−2, y1, 3) lies on that locus: At point P , the condition of part a becomes 3.19 = £ 1 + (y1 − 1)2 §3 from which (y1 − 1)2 = 0.47, or y1 = 1.69 or 0.31 2.10. A charge of -1 nC is located at the origin in free space. What charge must be located at (2,0,0) to cause Ex to be zero at (3,1,1)? The field from two point charges is given generally by ET = q1(r− r01) 4π≤0|r− r01|3 + q2(r− r02) 4π≤0|r− r02|3 (1) where we let q1 = −1nC and q2 is to be found. With q1 at the origin, r01 = 0. The position vector for q2 is then r02 = 2ax. The observation point at (3,1,1) gives r = 3ax + ay + az. Eq. (1) becomes 1 4π≤0 ∑ −1(3ax + ay + az) (32 + 1 + 1)3/2 + q2[(3− 2)ax + ay + az] (1 + 1 + 1)3/2 ∏ Requiring the x component to be zero leads to q2 = 35/2 113/2 = 0.43 nC 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 EP = Q0 4π≤0 ∑ −2ax + ay − az 61.5 ∏ Since the z component is of value 1 kV/m, we find Q0 = −4π≤061.5 × 103 = −1.63 µC. b) Find E at M(1, 6, 5) in cartesian coordinates: This field will be: EM = −1.63× 10−6 4π≤0 ∑ ax + 6ay + 5az [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 = √ 1 + 36 = 6.08, φ = tan−1(6/1) = 80.54◦, and z = 5. Now Eρ = EM · aρ = −30.11 cosφ− 180.63 sinφ = −183.12 Eφ = EM · aφ = −30.11(− sinφ)− 180.63 cosφ = 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 cosφ− 180.63 sin θ sinφ− 150.53 cos θ = −237.1 16 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 10−15 = −43.3 µC/m3 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 = Z 2π 0 Z π 0 Z .05 .03 0.2 r2 sin θ dr dθ dφ = ∑ 4π(0.2) r3 3 ∏.05 .03 = 8.21× 10−5 µC = 82.1 pC 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 ∑ 4π(0.2) r3 3 ∏r1 .03 = 4.105× 10−5 Thus r1 = ∑ 3× 4.105× 10−5 0.2× 4π + (.03) 3 ∏1/3 = 4.24 cm 2.14. The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge density is represented by ρv = −0.1/(ρ2 + 10−8) pC/m3 for 0 < ρ < 3 × 10−4 m, and ρv = 0 for ρ > 3× 10−4 m. a) Find the total charge per meter along the length of the beam: We integrate the charge density over the cylindrical volume having radius 3× 10−4 m, and length 1m. q = Z 1 0 Z 2π 0 Z 3×10−4 0 −0.1 (ρ2 + 10−8) ρ dρ dφ dz From integral tables, this evaluates as q = −0.2π µ 1 2 ∂ ln ° ρ2 + 10−8 ¢ ØØØ 3×10−4 0 = 0.1π ln(10) = −0.23π pC/m b) if the electron velocity is 5 × 107 m/s, and with one ampere defined as 1C/s, find the beam current: Current = charge/m× v = −0.23π [pC/m]× 5× 107 [m/s] = −11.5π × 106 [pC/s] = −11.5π µA 17 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 (0.003)3 = 1.24× 106 C/m3 2.16. Within a region of free space, charge density is given as ρv = (ρ0r/a) cos θ C/m3, where ρ0 and a are constants. Find the total charge lying within: a) the sphere, r ≤ a: This will be Qa = Z 2π 0 Z π 0 Z a 0 ρ0r a cos θ r2 sin θ dr dθ dφ = 2π Z a 0 ρ0r3 a dr = πρ0a3/2 b) the cone, r ≤ a, 0 ≤ θ ≤ 0.1π: Qb = Z 2π 0 Z 0.1π 0 Z a 0 ρ0r a cos θ r2 sin θ dr dθ dφ = π ρ0a3 4 £ 1− cos2(0.1π) § = 0.024πρ0a3 c) the region, r ≤ a, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤ 0.2π. Qc = Z 0.2π 0 Z 0.1π 0 Z a 0 ρ0r a cos θ r2 sin θ dr dθ dφ = 0.024πρ0a3 µ 0.2π 2π ∂ = 0.0024πρ0a3 2.17. A uniform line charge of 16 nC/m is located along the line defined by y = −2, z = 5. If ≤ = ≤0: a) Find E at P (1, 2, 3): This will be EP = ρl 2π≤0 RP |RP |2 where RP = (1, 2, 3)− (1,−2, 5) = (0, 4,−2), and |RP |2 = 20. So EP = 16× 10−9 2π≤0 ∑ 4ay − 2az 20 ∏ = 57.5ay − 28.8az V/m 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 Ez=0 = ρl 2π≤0 ∑ (y + 2)ay − 5az (y + 2)2 + 25 ∏ We require |Ez| = −|2Ey|, so 2(y + 2) = 5. Thus y = 1/2, and the field becomes: Ez=0 = ρl 2π≤0 ∑ 2.5ay − 5az (2.5)2 + 25 ∏ = 23ay − 46az 18 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/(2π≤0(0.8))]ay. The force on a differential length of the line at the positive y location is dF = dqE = ρldzE. Thus the force per unit length acting on the line at postive y arising from the charge at negative y is F = Z 1 0 ρ2l dz 2π≤0(0.8) ay = 1.26× 10−4 ay N/m = 126ay µN/m 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/(2≤0)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 = ρsdaE. The force per unit area is then just F = ρsE = (100× 10−9)(−100× 10−9)/(2≤0)az = −5.6× 10−4 az N/m2. 2.23. Given the surface charge density, ρs = 2µC/m2, in the region ρ < 0.2 m, z = 0. 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 r0 = ρaρ, we obtain r− r0 = zaz − ρaρ. The superposition integral for the z component of E will be: Ez,PA = ρs 4π≤0 Z 2π 0 Z 0.2 0 z ρ dρ dφ (ρ2 + z2)1.5 = −2πρs 4π≤0 z " 1p z2 + ρ2 #0.2 0 = ρs 2≤0 z ∑ 1√ z2 − 1√ z2 + 0.04 ∏ With z = 0.5 m, the above evaluates as Ez,PA = 8.1 kV/m. b) PB(ρ = 0, z = −0.5). With z at −0.5 m, we evaluate the expression for Ez to obtain Ez,PB = −8.1 kV/m. c) Show that 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 Ez = ρs 2≤0 ∑ 1− z√ z2 + 0.04 ∏ At small z, this reduces to Ez .= ρs/2≤0, which is the infinite sheet charge field. d) Show that the z axis field reduces to that of a point charge at large values of z: The development is as follows: Ez = ρs 2≤0 ∑ 1− z√ z2 + 0.04 ∏ = ρs 2≤0 " 1− z z p 1 + 0.04/z2 # .= ρs 2≤0 ∑ 1− 1 1 + (1/2)(0.04)/z2 ∏ where the last approximation is valid if z >> .04. Continuing: Ez .= ρs 2≤0 £ 1− [1− (1/2)(0.04)/z2] § = 0.04ρs 4≤0z2 = π(0.2)2ρs 4π≤0z2 This the point charge field, where we identify q = π(0.2)2ρs as the total charge on the disk (which now looks like a point). 21 2.24. a) Find the electric field on the z axis produced by an annular ring of uniform surface charge density ρs in free space. The ring occupies the region z = 0, a ≤ ρ ≤ b, 0 ≤ φ ≤ 2π in cylindrical coordinates: We find the field through E = Z Z ρsda(r− r0) 4π≤0|r− r0|3 where the integral is taken over the surface of the annular ring, and where r = zaz and r0 = ρaρ. The integral then becomes E = Z 2π 0 Z b a ρs ρ dρ dφ (zaz − ρaρ) 4π≤0(z2 + ρ2)3/2 In evaluating this integral, we first note that the term involving ρaρ integrates to zero over the φ integration range of 0 to 2π. This is because we need to introduce the φ dependence in aρ by writing it as aρ = cosφax + sinφay, where ax and ay are invariant in their orientation as φ varies. So the integral now simplifies to E = 2πρs z az 4π≤0 Z b a ρ dρ (z2 + ρ2)3/2 = ρs z az 2≤0 " −1p z2 + ρ2 #b a = ρs 2≤0 " 1p 1 + (a/z)2 − 1p 1 + (b/z)2 # az b) from your part a result, obtain the field of an infinite uniform sheet charge by taking appropriate limits. The infinite sheet is obtained by letting a→ 0 and b→∞, in which case E→ ρs/(2≤0)az as expected. 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: E = ∑ (12× 10−9) 4π≤0 (−2ax − 6az) (4 + 36)1.5 ∏ + ∑ (3× 10−9) 2π≤0 (2ax − 3ay) (4 + 9) ∏ − ∑ (0.2× 10−9)ax 2≤0 ∏ = −3.9ax − 12.4ay − 2.5az V/m 22 2.26. Radially-dependent surface charge is distributed on an infinite flat sheet in the xy plane, and is char- acterized in cylindrical coordinates by surface density ρs = ρ0/ρ, where ρ0 is a constant. Determine the electric field strength, E, everywhere on the z axis. We find the field through E = Z Z ρsda(r− r0) 4π≤0|r− r0|3 where the integral is taken over the surface of the annular ring, and where r = zaz and r0 = ρaρ. The integral then becomes E = Z 2π 0 Z ∞ 0 (ρ0/ρ) ρ dρ dφ (zaz − ρaρ) 4π≤0(z2 + ρ2)3/2 In evaluating this integral, we first note that the term involving ρaρ integrates to zero over the φ integration range of 0 to 2π. This is because we need to introduce the φ dependence in aρ by writing it as aρ = cosφax + sinφay, where ax and ay are invariant in their orientation as φ varies. So the integral now simplifies to E = 2πρs z az 4π≤0 Z ∞ 0 dρ (z2 + ρ2)3/2 = ρs z az 2≤0 " ρ z2 p z2 + ρ2 #∞ ρ=0 = ρs 2≤0z az 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 dx = Ey Ex = −(2x + 4y) (4x− 2y) Thus 2(xdy + y dx) = y dy − xdx or 2 d(xy) = 1 2 d(y2)− 1 2 d(x2) So C1 + 2xy = 1 2 y2 − 1 2 x2 or y2 − x2 = 4xy + C2 Evaluating at P (2, 3,−4), obtain: 9− 4 = 24 + C2, or C2 = −19 Finally, at P , the requested equation is y2 − 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 16.12 = 0.99ax + 0.12ay 23