Hwk2Solns

# Hwk2Solns - Assignment 2 Solutions Chpt 23 r r 2 We use = E...

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Assignment 2 Solutions Chpt 23 2. We use A d E r r = Φ and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m. (a) On the top face of the cube y = 2.0 m and ( ) ˆ j dA dA = r . Therefore, we have ( ) ( ) 2 ˆ ˆ 4i 3 2.0 2 j 4i 18j E = + = r ˆ ˆ . Thus the flux is ( ) ( ) ( )( ) 2 2 2 top top top ˆ ˆ ˆ 4i 18j j 18 18 2.0 N m C 72 N m C. E dA dA dA Φ = = = − = − = − r r (b) On the bottom face of the cube y = 0 and ) ˆ )( ( j dA A d = r . Therefore, we have j i j i E ˆ 6 ˆ 4 ˆ ) 2 0 ( 3 ˆ 4 2 = + = r . Thus, the flux is ( ) ( ) ( ) ( ) 2 2 2 bottom bottom bottom ˆ ˆ ˆ 4i 6j j 6 6 2.0 N m C 24 N m C. E dA dA dA Φ = = = = = + r r (c) On the left face of the cube ( ) ( ) ˆ i dA dA = r . So ( ) ( ) ( ) ( ) 2 2 2 left left bottom ˆ ˆ ˆ ˆ 4i j i 4 4 2.0 N m C 16 N m C. y E dA E dA dA Φ = = + = − = − = − r (d) On the back face of the cube ( ) ( ) ˆ k dA dA = r . But since E r has no z component . Thus, Φ = 0. 0 E dA = r r (e) We now have to add the flux through all six faces. One can easily verify that the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or +16 N·m 2 /C. Thus the net flux through the cube is Φ = (–72 + 24 – 16 + 0 + 0 + 16) N·m 2 /C = – 48 N·m 2 /C. 20. Using Eq. 23-11, the surface charge density is ( )( ) 5 12 2 2 0 2.3 10 N C 8.85 10 C / N m 2.0 10 C/m . E σ ε = = × × = × 6 2 27. We assume the charge density of both the conducting cylinder and the shell are uniform, and we neglect fringing effect. Symmetry can be used to show that the electric 1

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