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

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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 ˆˆ 4i 3 2.0 2 j 4i 18j E =− + =− r ˆ ˆ . Thus the flux is () ( ) ( ) 2 22 top top top ˆ 4i 18j j 18 18 2.0 N m C 72 N m C. EdA 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 bottom bottom bottom ˆ 4 i 6 j j 6 62 . 0 Nm C 2 4 Nm C . = − = = =+ r r (c) On the left face of the cube ( ) ˆ i dA dA = r . So ( ) ( ) 2 left left bottom ˆ ˆ 4i j i 4 4 2.0 N m C 16 N m C. y E dA E dA dA = + − =− 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 ⋅= 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 ( ) 51 2 2 2 0 2.3 10 N C 8.85 10 C / N m 2.0 10 C/m . E σε −− ==× × × 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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field is radial, both between the cylinder and the shell and outside the shell. It is zero, of course, inside the cylinder and inside the shell. (a) We take the Gaussian surface to be a cylinder of length L , coaxial with the given cylinders and of larger radius r than either of them. The flux through this surface is where E is the magnitude of the field at the Gaussian surface. We may ignore any flux through the ends. Now, the charge enclosed by the Gaussian surface is q enc = Q
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Hwk2Solns - Assignment 2 Solutions Chpt 23 r r 2. We use =...

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