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 )
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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
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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.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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