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Unformatted text preview: Chapter 4 – OddNumbered 4.1. The value of E at P ( ρ = 2 , φ = 40 ◦ , z = 3) is given as E = 100 a ρ − 200 a φ + 300 a z V/m. Determine the incremental work required to move a 20 µ C charge a distance of 6 µ m: a) in the direction of a ρ : The incremental work is given by dW = − q E · d L , where in this case, d L = dρ a ρ = 6 × 10 − 6 a ρ . Thus dW = − (20 × 10 − 6 C)(100V / m)(6 × 10 − 6 m) = − 12 × 10 − 9 J = − 12nJ b) in the direction of a φ : In this case d L = 2 dφ a φ = 6 × 10 − 6 a φ , and so dW = − (20 × 10 − 6 )( − 200)(6 × 10 − 6 ) = 2 . 4 × 10 − 8 J = 24 nJ c) in the direction of a z : Here, d L = dz a z = 6 × 10 − 6 a z , and so dW = − (20 × 10 − 6 )(300)(6 × 10 − 6 ) = − 3 . 6 × 10 − 8 J = − 36 nJ d) in the direction of E : Here, d L = 6 × 10 − 6 a E , where a E = 100 a ρ − 200 a φ + 300 a z [100 2 + 200 2 + 300 2 ] 1 / 2 = 0 . 267 a ρ − . 535 a φ + 0 . 802 a z Thus dW = − (20 × 10 − 6 )[100 a ρ − 200 a φ + 300 a z ] · [0 . 267 a ρ − . 535 a φ + 0 . 802 a z ](6 × 10 − 6 ) = − 44 . 9nJ e) In the direction of G = 2 a x − 3 a y + 4 a z : In this case, d L = 6 × 10 − 6 a G , where a G = 2 a x − 3 a y + 4 a z [2 2 + 3 2 + 4 2 ] 1 / 2 = 0 . 371 a x − . 557 a y + 0 . 743 a z So now dW = − (20 × 10 − 6 )[100 a ρ − 200 a φ + 300 a z ] · [0 . 371 a x − . 557 a y + 0 . 743 a z ](6 × 10 − 6 ) = − (20 × 10 − 6 )[37 . 1( a ρ · a x ) − 55 . 7( a ρ · a y ) − 74 . 2( a φ · a x ) + 111 . 4( a φ · a y ) + 222 . 9](6 × 10 − 6 ) where, at P , ( a ρ · a x ) = ( a φ · a y ) = cos(40 ◦ ) = 0 . 766, ( a ρ · a y ) = sin(40 ◦ ) = 0 . 643, and ( a φ · a x ) = − sin(40 ◦ ) = − . 643. Substituting these results in dW = − (20 × 10 − 6 )[28 . 4 − 35 . 8 + 47 . 7 + 85 . 3 + 222 . 9](6 × 10 − 6 ) = − 41 . 8nJ 45 4.3. If E = 120 a ρ V / m, find the incremental amount of work done in moving a 50 µ m charge a distance of 2 mm from: a) P (1 , 2 , 3) toward Q (2 , 1 , 4): The vector along this direction will be Q − P = (1 , − 1 , 1) from which a PQ = [ a x − a y + a z ] / √ 3. We now write dW = − q E · d L = − (50 × 10 − 6 ) · 120 a ρ · ( a x − a y + a z √ 3 ¸ (2 × 10 − 3 ) = − (50 × 10 − 6 )(120)[( a ρ · a x ) − ( a ρ · a y )] 1 √ 3 (2 × 10 − 3 ) At P , φ = tan − 1 (2 / 1) = 63 . 4 ◦ . Thus ( a ρ · a x ) = cos(63 . 4) = 0 . 447 and ( a ρ · a y ) = sin(63 . 4) = 0 . 894. Substituting these, we obtain dW = 3 . 1 µ J . b) Q (2 , 1 , 4) toward P (1 , 2 , 3): A little thought is in order here: Note that the field has only a radial component and does not depend on φ or z . Note also that P and Q are at the same radius ( √ 5) from the z axis, but have different φ and z coordinates. We could just as well position the two points at the same z location and the problem would not change....
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This note was uploaded on 10/02/2011 for the course EE 1 taught by Professor Joshi during the Fall '08 term at UCLA.
 Fall '08
 Joshi
 Electrical Engineering

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