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![Page 1: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/1.jpg)
Chapter 22: The Electric Field II: Continuous Charge Distributions
Section 22-1: Calculating E from Coulomb’s Law
![Page 2: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/2.jpg)
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.)
A. 1
B. 2
C. 3
D. 4
E. None of the diagrams.
1 2 3 4
![Page 3: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/3.jpg)
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.)
A. 1
B. 2
C. 3
D. 4
E. None of the diagrams.
1 2 3 4
![Page 4: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/4.jpg)
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
A. decreases linearly with x.
B. decreases as 1/x2.
C. is constant and does not depend on x.
D. increases linearly with x.
E. is undetermined.
![Page 5: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/5.jpg)
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
A. decreases linearly with x.
B. decreases as 1/x2.
C. is constant and does not depend on x.
D. increases linearly with x.
E. is undetermined.
![Page 6: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/6.jpg)
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.
A. 1.0
B. 0.60
C. 1.7
D. 0.50
E. 0.85
![Page 7: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/7.jpg)
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.
A. 1.0
B. 0.60
C. 1.7
D. 0.50
E. 0.85
![Page 8: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/8.jpg)
Chapter 22: The Electric Field II: Continuous Charge Distributions
Section 22-2: Gauss’s Law
![Page 9: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/9.jpg)
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 E through this
surface is approximately
A. zero
B. 6.4 105 N · m2/C
C. 13 105 N · m2/C
D. 25 105 N · m2/C
E. 38 105 N · m2/C
![Page 10: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/10.jpg)
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 E through this
surface is approximately
A. zero
B. 6.4 105 N · m2/C
C. 13 105 N · m2/C
D. 25 105 N · m2/C
E. 38 105 N · m2/C
![Page 11: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/11.jpg)
A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that
A. the surface encloses a net positive charge.
B. the surface encloses a net negative charge.
C. the surface encloses no net charge.
![Page 12: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/12.jpg)
A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that
A. the surface encloses a net positive charge.
B. the surface encloses a net negative charge.
C. the surface encloses no net charge.
![Page 13: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/13.jpg)
A surface is so constructed that, at all points on the surface, the E vector points outward. Therefore, it can be said that
A. the surface encloses a net positive charge.
B. the surface encloses a net negative charge.
C. the surface encloses no net charge.
![Page 14: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/14.jpg)
A surface is so constructed that, at all points on the surface, the E vector points outward. Therefore, it can be said that
A. the surface encloses a net positive charge.
B. the surface encloses a net negative charge.
C. the surface encloses no net charge.
![Page 15: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/15.jpg)
The figure shows a surface enclosing the charges q and –q. The net flux through the surface surrounding the two charges is
A. q/0
B. 2q/0
C. –q/0
D. zero
E. –2q/0
![Page 16: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/16.jpg)
The figure shows a surface enclosing the charges q and –q. The net flux through the surface surrounding the two charges is
A. q/0
B. 2q/0
C. –q/0
D. zero
E. –2q/0
![Page 17: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/17.jpg)
The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is
3
E.
zero D.
C.
2 B.
A.
0
0
0
0
q
q
ε
q
q
![Page 18: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/18.jpg)
The figure shows a surface enclosing the charges 2q and –q. The net flux through the surface surrounding the two charges is
3
E.
zero D.
C.
2 B.
A.
0
0
0
0
q
q
ε
q
q
![Page 19: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/19.jpg)
+q 2q
s
The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is
0
0
0
0
3 E.
zero D.
C.
2 B.
A.
q
q
ε
q
q
![Page 20: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/20.jpg)
+q 2q
s
The figure shows a surface, S, with two charges q and –2q. The net flux through the surface is
0
0
0
0
3 E.
zero D.
C.
2 B.
A.
q
q
ε
q
q
![Page 21: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/21.jpg)
A hollow spherical shell of radius 5.36 cm has a charge of 1.91 C placed at its center. Calculate the electric flux through a portion of the shell with an area of 1.20 10–2 m2.
A. 6.48 105 N.m2/C
B. 2.16 105 N.m2/C
C. 7.20 104 N.m2/C
D. 2.16 101 N.m2/C
E. None of the above.
![Page 22: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/22.jpg)
A hollow spherical shell of radius 5.36 cm has a charge of 1.91 C placed at its center. Calculate the electric flux through a portion of the shell with an area of 1.20 10–2 m2.
A. 6.48 105 N.m2/C
B. 2.16 105 N.m2/C
C. 7.20 104 N.m2/C
D. 2.16 101 N.m2/C
E. None of the above.
![Page 23: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/23.jpg)
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.
A. 907 N/C
B. 423 N/C
C. 1.10 10–3 N/C
D. 2.36 10–3 N/C
E. 383 N/C
![Page 24: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/24.jpg)
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.
A. 907 N/C
B. 423 N/C
C. 1.10 10–3 N/C
D. 2.36 10–3 N/C
E. 383 N/C
![Page 25: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/25.jpg)
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
![Page 26: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/26.jpg)
The electric field E in Gauss’s Law is
a. only that part of the electric field due to the charges inside the surface.
b. only that part of the electric field due to the charges outside the surface.
c. the total electric field due to all the charges both inside and outside the surface.
![Page 27: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/27.jpg)
The electric field E in Gauss’s Law is
A. only that part of the electric field due to the charges inside the surface.
B. only that part of the electric field due to the charges outside the surface.
C. the total electric field due to all the charges both inside and outside the surface.
![Page 28: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/28.jpg)
zero E.
2 D.
4 C.
B.
A.2
r
kr
kr
kr
k
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,
![Page 29: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/29.jpg)
zero E.
2 D.
4 C.
B.
A.2
r
kr
kr
kr
k
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,
![Page 30: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/30.jpg)
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?
zero E.
D.
C.
B.
A.
2
2
3
2
b
kQa
kQa
kQrr
kQ
![Page 31: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/31.jpg)
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?
zero E.
D.
C.
B.
A.
2
2
3
2
b
kQa
kQa
kQrr
kQ
![Page 32: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/32.jpg)
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?
2
2
2
2
2
)( E.
D.
C.
2 B.
A.
ab
kQb
kQa
kQr
kQr
kQ
![Page 33: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/33.jpg)
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?
2
2
2
2
2
)( E.
D.
C.
2 B.
A.
ab
kQb
kQa
kQr
kQr
kQ
![Page 34: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/34.jpg)
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?
zero E.
D.
C.
2 B.
A.
2
2
2
2
b
kQa
kQr
kQr
kQ
![Page 35: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/35.jpg)
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?
zero E.
D.
C.
2 B.
A.
2
2
2
2
b
kQa
kQr
kQr
kQ
![Page 36: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/36.jpg)
A sphere of radius 8.0 cm carries a uniform volume charge density = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm?
A. 0.12 kN/C
B. 1.5 kN/C
C. 0.74 kN/C
D. 2.3 kN/C
E. 12 kN/C
![Page 37: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/37.jpg)
A sphere of radius 8.0 cm carries a uniform volume charge density = 500 nC/m3. What is the electric field magnitude at r = 8.1 cm?
A. 0.12 kN/C
B. 1.5 kN/C
C. 0.74 kN/C
D. 2.3 kN/C
E. 12 kN/C
![Page 38: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/38.jpg)
A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 4.0 cm is approximately
A. 0.13 kN/C
B. 1.0 kN/C
C. 0.32 kN/C
D. 0.75 kN/C
E. zero
![Page 39: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/39.jpg)
A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 4.0 cm is approximately
A. 0.13 kN/C
B. 1.0 kN/C
C. 0.32 kN/C
D. 0.75 kN/C
E. zero
![Page 40: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/40.jpg)
A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 9.1 cm is approximately
A. zero
B. 1.0 kN/C
C. 0.65 kN/C
D. 0.32 kN/C
E. 0.13 kN/C
![Page 41: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/41.jpg)
A spherical shell of radius 9.0 cm carries a uniform surface charge density = 9.0 nC/m2. The electric field magnitude at r = 9.1 cm is approximately
A. zero
B. 1.0 kN/C
C. 0.65 kN/C
D. 0.32 kN/C
E. 0.13 kN/C
![Page 42: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/42.jpg)
An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –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
A. 226 N/C
B. 339 N/C
C. 904 N/C
D. 452 N/C
E. zero
![Page 43: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/43.jpg)
An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –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
A. 226 N/C
B. 339 N/C
C. 904 N/C
D. 452 N/C
E. zero
![Page 44: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/44.jpg)
An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –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
A. 226 N/C
B. 339 N/C
C. 904 N/C
D. 452 N/C
E. zero
![Page 45: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/45.jpg)
An infinite plane of surface charge density = +8.00 nC/m2 lies in the yz plane at the origin, and a second infinite plane of surface charge density = –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
A. 226 N/C
B. 339 N/C
C. 904 N/C
D. 452 N/C
E. zero
![Page 46: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/46.jpg)
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
x
y
z
d d
2
4 E.
2 D.
4 C.
2 B.
4 A.
b
πkρb
πkρb
πkρ
πkρb
πkρb
![Page 47: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/47.jpg)
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
x
y
z
d d
2
4 E.
2 D.
4 C.
2 B.
4 A.
b
πkρb
πkρb
πkρ
πkρb
πkρb
![Page 48: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/48.jpg)
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
x
y
z
d d
d
k
dk
dkd
kb
k
4 E.
2 D.
4 C.
4 B.
4 A.
2
![Page 49: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/49.jpg)
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
x
y
z
d d
d
k
dk
dkd
kb
k
4 E.
2 D.
4 C.
4 B.
4 A.
2
![Page 50: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/50.jpg)
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density . Which diagram best represents the electric field along the y–axis?
E. None of the diagrams.
x
y
z
d d
![Page 51: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/51.jpg)
An infinite slab of thickness 2d lies in the xz–plane. The slab has a uniform volume charge density . Which diagram best represents the electric field along the y–axis?
E. None of the diagrams.
x
y
z
d d
![Page 52: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/52.jpg)
Chapter 22: The Electric Field II: Continuous Charge Distributions
Section 22-4: Discontinuity of En
![Page 53: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/53.jpg)
A thin conducting plane with surface charge density is exposed to an external electric Eext. The difference in the electric field between one surface of the plane to the other surface is
A. /0
B. /0 Eext
C. /0 Eext
D. 2/0 Eext
E. /20 Eext
Eext
![Page 54: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/54.jpg)
A thin conducting plane with surface charge density is exposed to an external electric Eext. The difference in the electric field between one surface of the plane to the other surface is
A. /0
B. /0 Eext
C. /0 Eext
D. 2/0 Eext
E. /20 Eext
Eext
![Page 55: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/55.jpg)
Chapter 22: The Electric Field II: Continuous Charge Distributions
Section 22-5: Charge and Field at Conductor Surfaces
![Page 56: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/56.jpg)
Electrical conductors contain
A. only free electrons.
B. only bound electrons.
C. both free and bound electrons.
D. neither bound nor free electrons.
E. only protons and neutrons.
![Page 57: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/57.jpg)
Electrical conductors contain
A. only free electrons.
B. only bound electrons.
C. both free and bound electrons.
D. neither bound nor free electrons.
E. only protons and neutrons.
![Page 58: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/58.jpg)
The electric field at the surface of a conductor
A. is parallel to the surface.
B. depends only on the total charge on the conductor.
C. depends only on the area of the conductor.
D. depends only on the curvature of the surface.
E. depends on the area and curvature of the conductor and on its total charge.
![Page 59: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/59.jpg)
The electric field at the surface of a conductor
A. is parallel to the surface.
B. depends only on the total charge on the conductor.
C. depends only on the area of the conductor.
D. depends only on the curvature of the surface.
E. depends on the area and curvature of the conductor and on its total charge.
![Page 60: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/60.jpg)
Qra
rb1
rb2
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for ra r rb1 is
ero E.
ˆ2
D.
ˆ2
C.
ˆ B.
ˆ A.
2
2
z
rr
kQ
rr
kQ
rr
kQ
rr
kQ
![Page 61: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/61.jpg)
Qra
rb1
rb2
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for ra r rb1 is
ero E.
ˆ2
D.
ˆ2
C.
ˆ B.
ˆ A.
2
2
z
rr
kQ
rr
kQ
rr
kQ
rr
kQ
![Page 62: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/62.jpg)
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for rb1 r rb2 is
Qra
rb1
rb2ero E.
ˆ2
D.
ˆ2
C.
ˆ B.
ˆ A.
2
2
z
rr
kQ
rr
kQ
rr
kQ
rr
kQ
![Page 63: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/63.jpg)
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for rb1 r rb2 is
Qra
rb1
rb2ero E.
ˆ2
D.
ˆ2
C.
ˆ B.
ˆ A.
2
2
z
rr
kQ
rr
kQ
rr
kQ
rr
kQ
![Page 64: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/64.jpg)
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for r rb1 is
Qra
rb1
rb2
ero E.
ˆ2
D.
ˆ2
C.
ˆ B.
ˆ A.
2
2
z
rr
kQ
rr
kQ
rr
kQ
rr
kQ
![Page 65: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/65.jpg)
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The electric field for r rb1 is
Qra
rb1
rb2
ero E.
ˆ2
D.
ˆ2
C.
ˆ B.
ˆ A.
2
2
z
rr
kQ
rr
kQ
rr
kQ
rr
kQ
![Page 66: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/66.jpg)
The charge on an originally uncharged insulated conductor is separated by induction from a positively charged rod brought near the conductor. For which of the various Gaussian surfaces represented by the dashed lines
does ?
A. S1
B. S2
C. S3
D. S4
E. S5
0sdE
![Page 67: Chapter 22: The Electric Field II: Continuous Charge Distributions Section 22-1: Calculating E from Coulomb’s Law.](https://reader035.fdocuments.us/reader035/viewer/2022081501/56649c895503460f94941c83/html5/thumbnails/67.jpg)
The charge on an originally uncharged insulated conductor is separated by induction from a positively charged rod brought near the conductor. For which of the various Gaussian surfaces represented by the dashed lines
does ?
A. S1
B. S2
C. S3
D. S4
E. S5
0sdE