A (4, 3, 2) 4ax 3ay 2az b) Give a unit vector extending from the origin to the midpoint of line AB.With A (6, and B 13 B(2, 1), we use the fact that 10, or 23 B)ax (2 23 B)ay (4 13 B)az 10 Expanding, obtain 36 8B 49 B 2 4 83 B 49 B 2 16 83 B 19 B 2 100 or B 2 8B 44 0.This is found through aAB 0.62, c) a unit vector directed from the origin to the midpoint of the line AB.Note that the midpoint, (3, 3), as determined from part c happens to have z coordinate of 3.
![]() This is the point we are looking for. A vector field is specified as G 24xyax 12(x 2 2)ay 18z2 az. Therefore Then D 1 (cos sin ) 1 2 1 cos ) sin ) cos sin2 and D 1 cos ) sin ) sin sin cos 0 Therefore 7 1 1.19b. Evaluate D at the point where 2, and z 5, expressing the result in cylindrical and cartesian coordinates: At the given point, and in cylindrical coordinates, D. Then Q RP Q Q P 5ax 4ay 6az 25 16 36 8.8 b) cylindrical coordinates. Now, RP Q (5ax 4ay 6az ) 5 cos 4 sin 6.20 RP Q (5ax 4ay 6az ) sin 4 cos 1.60 Thus and Q RP Q 6az 6.202 1.602 62 8.8 c) spherical coordinates. At P, r 9 16 25 50 7.07,, and 3). RP Q ar (5ax 4ay 6az ) ar 5 sin cos 4 sin sin 6 cos 0.14 RP Q (5ax 4ay 6az ) 5 cos cos 4 cos sin sin 8.62 RP Q (5ax 4ay 6az ) sin 4 cos 1.60 10 1.24. Thus and Q RP Q 0.14ar 0.142 8.622 1.602 8.8 d) Show that each of these vectors has the same magnitude. Each does, as shown above. Given point P (r 0.8,, ), and sin 1 E 2 cos ar r sin a) Find E at P: E. Again, converting C to cartesian, obtain xC 20 ) ) 14.14, yC 20 ) ) 14.14, and zC 20 ) 0. So RAC RC RA (7.07, 7.07, and 20.0. A to C on a great circle path: Note that A and C share the same r and thus moving from A to C involves only a change in of. The requested arc length is then distance 20 60 360 13 20.9 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: Arrange the charges in the xy plane at locations (4,4), and 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. To solve this problem, the z coordinate of the third charge is immaterial, so we can place it in the xy plane at coordinates (x, y, 0). The force will be: (50 )2 RCA RDA RBA 3 3 where RCA ax ay, RDA ax ay, and RBA 2ax. Substituting these leads to (50 )2 1 1 2 ax 21.5ax 2 2 2 2 8 where distances are in meters. Point charges of 120 nC are located at A(0, 0, 1) and B(0, 0, in free space. The field will take the general form: R1 2R2 R3 EP where R1, R2, R3 are the vectors to P from each of the charges in their original listed order. So Er EP ar 89.9(ay ar ) 179.8(az ar ) 89.9 sin sin 179.8 cos 160.9 EP 89.9(ay ) 179.8(az ) 89.9 cos sin sin EP 89.9(ay ) 179.8(az ) 89.9 cos 0 16 2.9. A 100 nC point charge is located at 1, 3) in free space. Substituting these into the expression for EP produces 20 EP from which (9 y 2 )1.5 1079 (9 y 2 )1.5 2.11. A charge Q0 located at the origin in free space produces a field for which Ez 1 at point P 1, a) Find Q0: The field at P will be Q0 EP ay az 61.5 Since the z component is of value 1 we find Q0 61.5 103 17 2.14. Let 1 10 in the region 0 10, all z, and 0 elsewhere. 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 3.35 1.24 106 (0.003)3 2.16. The region in which 4 r 5, 0, and contains the volume charge density of 10(r 4)(r 5) sin Outside the region, 0. Find the charge within the region: The integral that gives the charge will be Q 10 0 5 (r 4)(r 5) sin r 2 sin dr 4 19 2.16. Carrying out the integral, we obtain r5 r4 r3 Q 10 9 20 5 4 3 4 1 1 2 4 cos 2 0 0.57 C 2.17. A uniform line charge of 16 is located along the line defined y z 5.
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