Physics Solution Manual for 1100 and 2101

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Unformatted text preview: 2 C 2 /(N ⋅ m 2 ) −2.3 × 10 – 6 C 8.85 × 10 −12 C /(N ⋅ m ) 2 2 = 4.0 × 10 5 N ⋅ m 2 /C = –2.6 × 10 5 N ⋅ m 2 /C (3.5 × 10 – 6 C ) + ( −2.3 × 10 – 6 C) −12 = 1.4 × 10 5 N ⋅ m 2 /C 8.85 × 10 C /(N ⋅ m ) ______________________________________________________________________________ 2 2 990 55. ELECTRIC FORCES AND ELECTRIC FIELDS SSM REASONING As discussed in Section 18.9, the electric flux Φ E through a surface is equal to the component of the electric field that is normal to the surface multiplied by the area of the surface, Φ E = E ⊥ A , where E⊥ is the component of E that is normal to the surface of area A. We can use this expression and the figure in the text to determine the flux through the two surfaces. SOLUTION a. The flux through surface 1 is (Φ E )1 = ( E cos 35° ) A1 = (250 N/C)(cos 35° )(1.7 m 2 ) = 350 N ⋅ m 2 /C b. Similarly, the flux through surface 2 is (Φ E )2 = ( E cos 55° ) A2 = (250 N/C)(cos 55° )(3.2 m2 ) = 460 N ⋅ m 2 /C ______________________________________________________________________________ 56. REASONING The electric flux ΦE through the circular surface is determined by the angle φ between the electric field and the normal to the surface, as well as the magnitude E of the electric field and the area A of the surface: Φ E = ( E cos φ ) A (18.6) We will use Equation 18.6 to determine the angle φ. SOLUTION Solving Equation 18.6 for the angle φ, we obtain cos φ = ΦE EA ΦE EA φ = cos −1 or (1) The surface is circular, with a radius r, so its area is A = π r 2 . Making this substitution in Equation (1) yields 78 N ⋅ m 2 /C −1 Φ E −1 = 58o φ = cos = cos = cos 2 4 2 EA 1.44 × 10 N/C π ( 0.057 m ) Eπ r −1 Φ E ( ) 57. REASONING AND SOLUTION Since the electric field is uniform, its magnitude and direction are the same at each point on the wall. The angle φ between the electric field and the normal to the wall is 35°. Therefore, the electric flux is ΦE = (E cos φ) A = (150 N/C)(cos 35°)[(5.9 m)(2.5 m)] = 1.8 × 10 3 N ⋅ m 2 /C Chapter 18 Problems 991 ______________________________________________________________________________ 58. REASONING The charge Q inside the rectangular box is related to the electric flux ΦE that passes through the surfaces of the box by Gauss’ law, Q = ε 0Φ E (Equation 18.7), where ε0 is the permittivity of free space. The electric flux is the algebraic sum of the flux through each of the six surfaces. SOLUTION The charge inside the box is Q = ε 0 Φ E = ε 0 ( Φ1 + Φ 2 + Φ3 + Φ 4 + Φ5 + Φ 6 ) C2 N ⋅ m2 N ⋅ m2 N ⋅ m2 = 8.85 × 10−12 + 2200 + 4600 +1500 C C C N ⋅ m2 − 1800 N ⋅ m2 N ⋅ m2 N ⋅ m2 − 3500 − 5400 C C C = −2.1 × 10−8 C ______________________________________________________________________________ 59. SSM REASONING The electric flux through each face of the cube is given by Φ E = ( E cos φ ) A (see Section 18.9) where E is the magnitude of the electric field at the face, A is the area of the face, and φ is the angle between the electric field and the outward normal of that face. We can use this expression to calculate the electric flux Φ E through each of the six faces of the cube. y E x z SOLUTION a. On the bottom face of the cube, the outward normal points parallel to the –y axis, in the opposite direction to the electric field, and φ = 180°. Therefore, ( Φ E )bottom = (1500 N/C )( cos 180° )( 0.20 m )2 = −6.0 × 101 N ⋅ m 2 /C On the top face of the cube, the outward normal points parallel to the +y axis, and φ = 0.0°. The electric flux is, therefore, (Φ E ) top = (1500 N/C)(cos 0.0°)(0.20 m) 2 = +6.0 × 101N.m 2 /C 992 ELECTRIC FORCES AND ELECTRIC FIELDS On each of the other four faces, the outward normals are perpendicular to the direction of the electric field, so φ = 90°. So for each of the four side faces, (Φ E )sides = (1500 N/C)(cos 90°)(0.20 m) 2 = 0 N ⋅ m 2 / C b. The total flux through the cube is (Φ E ) total = (Φ E ) top + (Φ E )bottom + (Φ E )side 1 + (Φ E )side 2 + (Φ E )side 3 + (Φ E )side 4 Therefore, (Φ E ) total = ( + 6.0 × 101N.m 2 /C) + (–6.0 × 101N.m 2 /C) + 0 + 0 + 0 + 0 = 0 N ⋅ m 2 / C ______________________________________________________________________________ 60. REASONING Gauss’ Law, Σ ( E cos φ ) ∆A = Q (Equation 18.7), relates the electric field ε0 magnitude E on a Gaussian surface to the net charge Q enclosed by that surface; Φ E = Σ ( E cos φ ) ∆A (Equation 18.6) is the electric flux through the Gaussian surface (divided into many tiny sections of area ∆A) and ε0 is the permittivity of free space. We are to determine the magnitude E of the electric field due to electric charges that are spread uniformly over the surfaces of two concentric spherical shells. The electric field due to these charges possesses spherical symmetry, so we will choose Gaussian surfaces in the shape of spheres concentric with the shells. The radius r of each Gaussian surface will be equal to...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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