Chapter 4 - Energy and Potential

# Chapter 4 - Energy and Potential - Chapter 4 Energy and...

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1 Chapter 4 Energy and Potential

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2 Consider a charge Q in a region in space where an electric field E exists. The force experienced by the charge due to the field is F E = Q E If we want to move the charge along d L , then we should apply a force opposite the component of F E along d L . The component of F E along d L is F EL = F E a L = Q E a L a L is the unit vector in the direction of d L . Energy Expended in Moving a Point Charge in an Electric Field
3 E F E F EL F appl d L Then the force that should be applied is –F EL or F appl = -F EL = -Q E a L The differential work done is equal to the product of the force applied and the displacement: dW = -Q E a L dL F E = Q E F EL = F E a L = Q E a L F appl = - Q E a L

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4 Example: Determine the work done when moving a 20 nC charge from B(0,1.5,10) to A(0,1.49,9.99). An infinite line charge density = 10 nC/m is at the z-axis. A B y dW = -Q E a L dL E (B) = ρ L /2 πε 0 ρ a y = 10n / 2 πε 0 1.5 a y = 119.84 a y V/m R AB = -0.01 a y – 0.01 a z dL = 0.01414 m
5 A B y y x y x L a a a a a 707 . 0 707 . 0 01 . 0 01 . 0 01 . 0 01 . 0 2 2 - - = + - - = Thus, dW = -Q E a L dL = -20n ( 119.84 a y ) • ( -0.707 a x - 0.707 a y )(0.01414) = 23.96 nJ

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6 The work required to move the charge a finite distance: dW = -Q E a L dL - = final init d Q W L E On what conditions is the differential work equal to zero? 1. E a L = 0, that is, d L is normal to the electric field. 2. Trivial condition: E , d L , or Q = 0.
7 The expression above is an example of a line integral. Integration is performed over a specified path. - = final init d Q W L E The Line Integral

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8 The work done in moving a charge Q from B to A is the sum of the work done in moving the charge along each segment. The electric field throughout the region is assumed constant. W = -Q(E L1 Δ L 1 + E L2 Δ L 2 + … + E L6 Δ L 6 ) = -Q( E • Δ L 1 + E • Δ L 2 + … + E • Δ L 6 ) = -Q E ( Δ L 1 + Δ L 2 + … + Δ L 6 ) = -Q E L BA
9 The result above can also be obtained using the line integral: W = -Q(E L1 Δ L 1 + E L2 Δ L 2 + … + E L6 Δ L 6 ) = -Q( E • Δ L 1 + E • Δ L 2 + … + E • Δ L 6 ) = -Q E ( Δ L 1 + Δ L 2 + … + Δ L 6 ) = -Q E L BA BA L E L E L E - = - = - = Q d Q d Q W A B A B The last expression is valid only if the electric field is constant.

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10 Example: Determine the work done in moving a 2 C charge from B(1,0,1) to A(0.8,0.6,1) along the shorter arc of the circle x 2 + y 2 = 1, z = 1 in the presence of the non- uniform field E = y a x + x a y + 2 a z . J 96 . 0 ) dz 2 dy y 1 dx x 1 ( 2 ) dz 2 xdy ydx ( 2 ) dz dy dx ( ) 2 x y ( 2 d Q W 1 1 6 . 0 0 2 8 . 0 1 2 A B A B A B A B z y x z y x A B - = + - + - - = + + - = + + + + - = - = a a a a a a L E
11 Now, determine the work done in moving the same charge from B to A, but using a straight line as a path. Equation of line passing through A(1,0,1) and

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## Chapter 4 - Energy and Potential - Chapter 4 Energy and...

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