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Unformatted text preview: CHAPTER 4 4.1. The value of E at P( = 2 , = 40 , z = 3 ) is given as E = 100 a 200 a + 300 a z V/m. Determine the incremental work required to move a 20 C charge a distance of 6 m: a) in the direction of a : The incremental work is given by dW = q E d L , where in this case, d L = d a = 6 10 6 a . Thus dW = ( 20 10 6 C )( 100 V / m )( 6 10 6 m ) = 12 10 9 J = 12 nJ b) in the direction of a : In this case d L = 2 d a = 6 10 6 a , 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 200 a + 300 a z [100 2 + 200 2 + 300 2 ] 1 / 2 = . 267 a . 535 a + . 802 a z Thus dW = ( 20 10 6 ) [100 a 200 a + 300 a z ] [0 . 267 a . 535 a + . 802 a z ] ( 6 10 6 ) = 44 . 9 nJ 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 = . 371 a x . 557 a y + . 743 a z So now dW = ( 20 10 6 ) [100 a 200 a + 300 a z ] [0 . 371 a x . 557 a y + . 743 a z ] ( 6 10 6 ) = ( 20 10 6 ) 37 . 1 ( a a x ) 55 . 7 ( a a y ) 74 . 2 ( a a x ) + 111 . 4 ( a a y ) + 222 . 9] ( 6 10 6 ) where, at P , ( a a x ) = ( a a y ) = cos ( 40 ) = . 766, ( a a y ) = sin ( 40 ) = . 643, and ( a 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 . 8 nJ 42 4.2. Let E = 400 a x 300 a y + 500 a z in the neighborhood of point P( 6 , 2 , 3 ) . Find the incremental work done in moving a 4C charge a distance of 1 mm in the direction specified by: a) a x + a y + a z : We write dW = q E d L = 4 ( 400 a x 300 a y + 500 a z ) ( a x + a y + a z ) 3 ( 10 3 ) = ( 4 10 3 ) 3 ( 400 300 + 500 ) = 1 . 39 J b) 2 a x + 3 a y a z : The computation is similar to that of part a , but we change the direction: dW = q E d L = 4 ( 400 a x 300 a y + 500 a z ) ( 2 a x + 3 a y a z ) 14 ( 10 3 ) = ( 4 10 3 ) 14 ( 800 900 500 ) = 2 . 35 J 4.3. If E = 120 a V / m, find the incremental amount of work done in moving a 50 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 ( a x a y + a z 3 ( 2 10 3 ) = ( 50 10 6 )( 120 ) ( a a x ) ( a a y ) 1 3 ( 2 10 3 ) At P , = tan 1 ( 2 / 1 ) = 63 . 4 . Thus ( a a x ) = cos ( 63 . 4 ) = . 447 and ( a a y ) = sin ( 63 . 4 ) = . 894. Substituting these, we obtain dW = 3 . 1 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 or z . Note also that P and Q are at the same radius ( 5) from the z axis, but have different and z coordinates. We could just as well position the two points at the same z location and the problem would not change. If this were so, then moving along a straight line between P and Q would thus involve moving along a chord of a circle whose radius is 5. Halfway along this line is a point of symmetry in the field (make a sketch to see 5....
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas.
 Spring '08
 Brown
 Electromagnet

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