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Chapter 22: The Electric Field II: Continuous Charge Distributions . Section 22-1: Calculating E from Coulomb’s Law.
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Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law
A conducting circular disk has a uniform positive surface charge density. Which of the following diagrams best represents the electric field lines from the disk? (The disk is drawn as a cross–section.) • 1 • 2 • 3 • 4 • None of the diagrams.
A conducting circular disk has a uniform positive surface charge density. Which of the following diagrams best represents the electric field lines from the disk? (The disk is drawn as a cross–section.) • 1 • 2 • 3 • 4 • None of the diagrams.
An infinite plane lies in the yzplane and it has a uniform surface charge density. The electric field at a distance x from the plane • decreases linearly with x. • decreases as 1/x2. • is constant and does not depend on x. • increases linearly with x. • is undetermined.
An infinite plane lies in the yzplane and it has a uniform surface charge density. The electric field at a distance x from the plane • decreases linearly with x. • decreases as 1/x2. • is constant and does not depend on x. • increases linearly with x. • is undetermined.
A uniform circular ring has charge Q and radius r. A uniformly charged disk also has charge Q and radius r. Calculate the electric field due to the ring at a distance of r along the axis of the ring divided by the electric field due to the disk at a distance of r along the axis of the disk. • 1.0 • 0.60 • 1.7 • 0.50 • 0.85
A uniform circular ring has charge Q and radius r. A uniformly charged disk also has charge Q and radius r. Calculate the electric field due to the ring at a distance of r along the axis of the ring divided by the electric field due to the disk at a distance of r along the axis of the disk. • 1.0 • 0.60 • 1.7 • 0.50 • 0.85
Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-2: Gauss’s Law
A cubical surface with no charge enclosed and with sides 2.0 m long is oriented with right and left faces perpendicular to a uniform electric field E of (1.6 ´ 105 N/C)in the +x direction. The net electric flux fE through this surface is approximately • zero • 6.4 ´ 105 N · m2/C • 13 ´ 105 N · m2/C • 25 ´ 105 N · m2/C • 38 ´ 105 N · m2/C
A cubical surface with no charge enclosed and with sides 2.0 m long is oriented with right and left faces perpendicular to a uniform electric field E of (1.6 ´ 105 N/C)in the +x direction. The net electric flux fE through this surface is approximately • zero • 6.4 ´ 105 N · m2/C • 13 ´ 105 N · m2/C • 25 ´ 105 N · m2/C • 38 ´ 105 N · m2/C
A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that • the surface encloses a net positive charge. • the surface encloses a net negative charge. • the surface encloses no net charge.
A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that • the surface encloses a net positive charge. • the surface encloses a net negative charge. • the surface encloses no net charge.
A surface is so constructed that, at all points on the surface, the E vector points outward. Therefore, it can be said that • the surface encloses a net positive charge. • the surface encloses a net negative charge. • the surface encloses no net charge.
A surface is so constructed that, at all points on the surface, the E vector points outward. Therefore, it can be said that • the surface encloses a net positive charge. • the surface encloses a net negative charge. • the surface encloses no net charge.
The figure shows a surface enclosing the charges q and –q. The net flux through the surface surrounding the two charges is • q/0 • 2q/0 • –q/0 • zero • –2q/0
The figure shows a surface enclosing the charges q and –q. The net flux through the surface surrounding the two charges is • q/0 • 2q/0 • –q/0 • zero • –2q/0
The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is
The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is
The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is
The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is
A hollow spherical shell of radius 5.36 cm has a charge of 1.91 mC placed at its center. Calculate the electric flux through a portion of the shell with an area of 1.20 ´ 10–2 m2. • 6.48 ´ 105 N.m2/C • 2.16 ´ 105 N.m2/C • 7.20 ´ 104 N.m2/C • 2.16 ´ 101 N.m2/C • None of the above.
A hollow spherical shell of radius 5.36 cm has a charge of 1.91 mC placed at its center. Calculate the electric flux through a portion of the shell with an area of 1.20 ´ 10–2 m2. • 6.48 ´ 105 N.m2/C • 2.16 ´ 105 N.m2/C • 7.20 ´ 104 N.m2/C • 2.16 ´ 101 N.m2/C • None of the above.
A horizontal surface of area 0.321 m2 has an electric flux of 123 N.m2/C passing through it at an angle of 25° to the horizontal. If the flux is due to a uniform electric field, calculate the magnitude of the electric field. • 907 N/C • 423 N/C • 1.10 ´ 10–3 N/C • 2.36 ´ 10–3 N/C • 383 N/C
A horizontal surface of area 0.321 m2 has an electric flux of 123 N.m2/C passing through it at an angle of 25° to the horizontal. If the flux is due to a uniform electric field, calculate the magnitude of the electric field. • 907 N/C • 423 N/C • 1.10 ´ 10–3 N/C • 2.36 ´ 10–3 N/C • 383 N/C
Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-3: Using Symmetry to Calculate E with Guass’s Law, and Concept Check 22-1
The electric field E in Gauss’s Law is • only that part of the electric field due to the charges inside the surface. • only that part of the electric field due to the charges outside the surface. • the total electric field due to all the charges both inside and outside the surface.
The electric field E in Gauss’s Law is • only that part of the electric field due to the charges inside the surface. • only that part of the electric field due to the charges outside the surface. • the total electric field due to all the charges both inside and outside the surface.
A rod of infinite length has a charge per unit length of l (= q/l). Gauss's Law makes it easy to determine that the electric field strength at a perpendicular distance r from the rod is, in terms of k = (40)–1,
A rod of infinite length has a charge per unit length of l (= q/l). Gauss's Law makes it easy to determine that the electric field strength at a perpendicular distance r from the rod is, in terms of k = (40)–1,
A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a uniform charge distribution totaling +Q and the hollow sphere a charge of –Q, the electric field magnitude at radius r, where r < a, is which of the following, in terms of k = (40)–1?
A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a uniform charge distribution totaling +Q and the hollow sphere a charge of –Q, the electric field magnitude at radius r, where r < a, is which of the following, in terms of k = (40)–1?
A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a uniform charge distribution totaling +Q and the hollow sphere a charge of –Q, the electric field magnitude at radius r, where a < r < b, is which of the following, in terms of k = (40)–1?
A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a uniform charge distribution totaling +Q and the hollow sphere a charge of –Q, the electric field magnitude at radius r, where a < r < b, is which of the following, in terms of k = (40)–1?
A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a uniform charge distribution totaling +Q and the hollow sphere a charge of –Q, the electric field magnitude at radius r, where r > b, is which of the following, in terms of k = (40)–1?
A solid sphere of radius a is concentric with a hollow sphere of radius b, where b > a. If the solid sphere has a uniform charge distribution totaling +Q and the hollow sphere a charge of –Q, the electric field magnitude at radius r, where r > b, is which of the following, in terms of k = (40)–1?
A sphere of radius 8.0 cm carries a uniform volume charge density r = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm? • 0.12 kN/C • 1.5 kN/C • 0.74 kN/C • 2.3 kN/C • 12 kN/C
A sphere of radius 8.0 cm carries a uniform volume charge density r = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm? • 0.12 kN/C • 1.5 kN/C • 0.74 kN/C • 2.3 kN/C • 12 kN/C
A spherical shell of radius 9.0 cm carries a uniform surface charge density s = 9.0 nC/m2. The electric field magnitude at r = 4.0 cm is approximately • 0.13 kN/C • 1.0 kN/C • 0.32 kN/C • 0.75 kN/C • zero
A spherical shell of radius 9.0 cm carries a uniform surface charge density s = 9.0 nC/m2. The electric field magnitude at r = 4.0 cm is approximately • 0.13 kN/C • 1.0 kN/C • 0.32 kN/C • 0.75 kN/C • zero
A spherical shell of radius 9.0 cm carries a uniform surface charge density s = 9.0 nC/m2. The electric field magnitude at r = 9.1 cm is approximately • zero • 1.0 kN/C • 0.65 kN/C • 0.32 kN/C • 0.13 kN/C
A spherical shell of radius 9.0 cm carries a uniform surface charge density s = 9.0 nC/m2. The electric field magnitude at r = 9.1 cm is approximately • zero • 1.0 kN/C • 0.65 kN/C • 0.32 kN/C • 0.13 kN/C
An infinite plane of surface charge density s = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density s = –8.00 nC/m2 lies in a plane parallel to the yz plane at x = 4.00 m. The electric field magnitude at x = 3.50 m is approximately • 226 N/C • 339 N/C • 904 N/C • 452 N/C • zero
An infinite plane of surface charge density s = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density s = –8.00 nC/m2 lies in a plane parallel to the yz plane at x = 4.00 m. The electric field magnitude at x = 3.50 m is approximately • 226 N/C • 339 N/C • 904 N/C • 452 N/C • zero
An infinite plane of surface charge density s = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density s = –8.00 nC/m2 lies in a plane parallel to the yz plane at x =4.00 m. The electric field magnitude at x = 5.00 m is approximately • 226 N/C • 339 N/C • 904 N/C • 452 N/C • zero
An infinite plane of surface charge density s = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density s = –8.00 nC/m2 lies in a plane parallel to the yz plane at x =4.00 m. The electric field magnitude at x = 5.00 m is approximately • 226 N/C • 339 N/C • 904 N/C • 452 N/C • zero
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density r. The electric field magnitude at y = b where 0 < b < d is
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density r. The electric field magnitude at y = b where 0 < b < d is
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density r. The electric field magnitude at y = b where b > d is
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density r. The electric field magnitude at y = b where b > d is
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density r. Which diagram best represents the electric field along the y–axis? E. None of the diagrams.