Solution 2 2 e ds n r 0 0 4 2 z0 r10er

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Unformatted text preview: gion enclosed by r = 2, z = 0, and z = 4. Solution: 2π 2 E · ds = n r =0 φ =0 4 2π + z=0 ˆ ˆ ˆ (r10e−r − z3z) · (rr d φ dz) r =2 ˆ ˆ ˆ (r10e−r − z3z) · (zr dr d φ ) z=4 φ =0 z=0 2 2π + r =0 φ =0 4 2π = 0+ φ =0 z=0 −2 = 160π e ∇·E d V = ˆ ˆ ˆ (r10e−r − z3z) · (−zr dr d φ ) 4 2 10e−2 2 d φ dz + r =0 2π r =0 φ =0 −12r dr d φ − 48π ≈ −82.77, 2π z=0 r=0 φ =0 2 −r = 8π 2 10e−r (1 − r) − 3 r d φ dr dz r (10e (1 − r) − 3r) dr = 8π −10e−r + 10e−r (1 + r) − = 160π e−2 − 48π ≈ −82.77. 3r2 2 2 r =0 ˆ ˆ Problem 3.52 Verify Stokes’s theorem for the vector field B = (rr cos φ + φ sin φ ) by evaluating: (a) B · d l over the semicirc...
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This homework help was uploaded on 02/20/2014 for the course EEC 130 at UC Davis.

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