chap4 - CHAPTER 4 4.1 The value of E at P = 2 = 40 z = 3 is...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 4-C 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....
View Full Document

{[ snackBarMessage ]}

Page1 / 18

chap4 - CHAPTER 4 4.1 The value of E at P = 2 = 40 z = 3 is...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online