Chapter%204 - Chapter 4 Energy and Potential Dr. Ray Chen...

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Dr. Ray Chen© Chapter 4 Energy and Potential
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Dr. Ray Chen© 4.1 Energy expended in moving a point charge in an electric field dL initial final L a ˆ Q L appl L L E EL E a ˆ E Q F a ˆ E Q a ˆ F F E Q F = = = = v v v v v v v dL a ˆ E Q L = v Q moving source external by done work al Differenti ∫∫ = = = final Initial final initial L d E Q dw W L d E Q dw v v v v
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Dr. Ray Chen© 4.1 Energy expended in moving a point charge in an electric field Example 1 . E of direction the along (d) , a (c) , a (b) , a (a) direction the in um. 0.8 of distance a at charge C 10 a moving in done work l incrementa the Find . 60 30 20 as 60 30 5, = (r at coordinate spherical in given is field A r 0 0 v v φ θ µ ˆ ˆ ˆ a ˆ a ˆ a ˆ E ) , Q r + = = = Solution: ) , , . ( L d ), , , ( E Q ) p ( pJ . L d E Q dw 0 0 10 8 0 60 30 20 10 10 160 10 8 0 20 10 10 ) a 6 5 12 6 6 × = = = = = = v v v v ( ) pJ . dw 240 10 8 0 30 10 10 b) 6 6 = × × × × =
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Dr. Ray Chen© 4.1 Energy expended in moving a point charge in an electric field Solution to Example 1 (continued): pJ . dw 480 10 8 0 60 10 10 c) 6 6 = × × × × = pJ . cos L d E Q L d E Q dw 560 10 8 0 60 30 20 10 10 d) 6 2 2 2 6 = × × + + × × = = = θ v v v v
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Dr. Ray Chen© 4.2 The line Integral Fig. 4.1 () field. electric uniform a for notation vector using E E E E E E If 6 2 1 6 5 4 3 2 1 6 6 2 2 1 1 BA L L L L E Q W L ...... L L E Q W , L E ...... L E L E Q W v v v v v v = + + + = = = = = = + + + =
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Dr. Ray Chen© 4.2 The line Integral = A B L d E Q W v v : form General ) E ( L E Q L d E Q W dz E d dy E d dx E d A B BA v v v v v v v v uniform have we , 0 If = = = = =
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Dr. Ray Chen© 4.2 The line Integral Conclusion: ( ) BA L , E , Q f W v v * * Due to energy conservation, the value of W is independent of the route or routes been taken as long as the initial & final positions are the same.
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Dr. Ray Chen© 4.2 The line Integral Example 2: () integral. line Direct (b (2,1,3). to (2,1,0) to (2,0,0) to (0,0,0) segment line straight the (a path the use 8 2 4 4 if A(2,1,3), to B(0,0,0) from E of integral line the Compute 2 ) ) , a ˆ z a ˆ x y a ˆ xy E z y x + + = v v B(0,0,0) A(2,1,3) x y z
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