Explanation Given: Charge of particle q = 0.01 C Intital point = ( 1 , − 5 , 0 ) final point = 3,4,4 Dipole moment p = 4 × 10 − 5 c − m f ( x , y , z ) = k p r 5 〈 3 x z , 3 y z , 2 z 2 − x 2 − y 2 〉 k = 8.99 × 10 9 N − m 2 / c 2 r = x 2 + y 2 + z 2 Formula used: The force on q is qf Newtons ∫ d d t f ( t ) d t = f ( t ) Calculation: Point p ( 1 , − 5 , 0 ) to Q ( 3 , 4 , 4 ) can be parameterized as follows: c ( t ) = 〈 2 t + 1 , 9 t − 5 , 4 t 〉 0 ≤ t ≤ 1 Since, f ( x , y , z ) = k p r 5 〈 3 x z , 2 y z , 2 z 2 − x 2 − y 2 〉 Substitute x = 2 t + 1 , y = 9 t − 5 and z = 4t f ( c ( t ) ) = k p r 5 〈 3 ( 2 t + 1 ) 4 t , 3 × ( 9 t − 5 ) 4 t , 2 ( 4 t ) 2 − ( 2 t + 1 ) 2 − ( 9 t − 5 ) 2 〉 = k p r 5 〈 12 × ( 2 t 2 + t ) , 12 t ( 9 t − 5 ) , 32 t 2 − 4 t 2 − 4 t − 1 − ( 81 t 2 + 25 − 90 t ) 〉 = k p r 5 〈 24 t 2 + 12 t , 108 t 2 − 60 t , 0.28 t 2 − 4 t − 1 − 81 t 2 − 25 + 90 t 〉 = k p r 5 〈 24 t 2 + 12 t , 108 t 2 − 60 t , − 53 t 3 + 86 t − 26 〉 f ( c ( t ) ) = k p r 5 〈 24 t 2 + 12 t , 108 t 2 − 60 t , − 5 t 2 + 86 − 26 〉 Work done for parametric function can be calculated as w = − ∫ 0 1 q f ( c ( t ) ) c ' ( t ) d t q f ( c ( t ) ) c ' ( t ) d t = k p q r 5 〈 24 t 2 + 12 t , 108 t 2 − 60 t , − 53 t 2 + 86 t − 26 〉 〈 2 , 9 , 4 〉 d t = k p q r 5 [ 48 t 2 + 24 t + 972 t 2 − 540

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ISBN: 9781319055844

Author: FRANZOSA

Publisher: VST

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Chapter 17.3, Problem 28E

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The work against $f$ required to move a particle of charge $q = 0.01 C$ from $(1, - 5, 0)$ to $(3, 4, 4)$ .

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Chapter 17.3, Problem 27E

Chapter 17.3, Problem 29E

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Chapter 17 Solutions

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The work against f required to move a particle of charge q = 0.01 C from ( 1 , − 5 , 0 ) to ( 3 , 4 , 4 ) .

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To calculate: The work against f required to move a particle of charge q=0.01C from (1,−5,0) to (3,4,4).

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Chapter 17 Solutions

To calculate:

The work against $f$ required to move a particle of charge $q = 0.01 C$ from $(1, - 5, 0)$ to $(3, 4, 4)$ .