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7/22/2019 Fundamental of Physics Halliday Ed. 9 Chapter 23 Homework Solution
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Fundamentals of Physics, 8thEdPrinciple of Physics, 9thEdHalliday & Resnic
8thEdCH23Finding the Electric Field II9thEdCH23Gausss Law
8th
EdHomework of Chapter 231, 3, 5, 7, 15, 19, 21, 23, 29, 33, 35, 37, 41, 45, 49, 51
8thEdSample Problem 23-1Figure 23-4 shows a Gaussian surface in the form of a cylinder of radius R immersed in a uniform
electric field E
, with the cylinder axis parallel to the field. What is the flux of the electric fieldthrough this closed surface?
23-4 E dA =
a b c
E dA E dA E dA= + +
0EA EA= + + 0=
0(cos180 )a
E dA E dA = E dA= EA= 0(cos0 )
c
E dA E dA =
EA=
0(cos90 )b
E dA E dA =
0=
8thEdSample Problem 23-2A nonuniform electric field given by 3 4E xi j= +
pierces the Gaussian cube shown in Fig. 23-5.
(E is in newtons per columb and x is in meters.) What is the electric flux through the right face, theleft face, and the top face? (we consider the other faces in Sample Problem 23-4.)
Fundamentals of Physics, 8thEdPrinciple of Physics, 9thEdHalliday & Resnic
23-5Right face : dA dAi=
rE dA =
(3 4 ) ( )xi j dAi= + (3 0)xdA= + 3 xdA=
3x m= 3 (3)dA= 9 dA=
24A m= 2 2(9 / )(4 ) 36 /r N C m N m C = =
Left face :
dA dAi=
1x m= 3l xdA = 212 /N m C=
Top face : dA dA j=
(3 4 ) ( )t xi j dA j = + (0 4 )dA= + 4 dA=
216 /N m C=
8th
EdProblem 23-19th
EdProblem 23-1The square surface shown in Fig. 23-26 measures 3.2 mm on each side. It is immersed in a uniform
electric field with magnitude 1800 /E N C= and with field lines at an angle of 035= with anormal to the surface, as shown. Take that normal to be directed outward, as though the surface
were one face of a box. Calculate the electric flux through the surface.
23-26 3.2 mm 1800 /E N C= 035=
7/22/2019 Fundamental of Physics Halliday Ed. 9 Chapter 23 Homework Solution
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Fundamentals of Physics, 8thEdPrinciple of Physics, 9thEdHalliday & Resnic
23-26The vector area
A and the electric field
E are shown on the diagram below. The angle
between them is 180 35 = 145, so the electric flux through the area is
( ) ( )23 2 2cos 1800 N C 3.2 10 m cos145 1.5 10 N m C.E A EA = = = =
8thEdProblem 23-39thEdProblem 23-3The cube in Fig. 23-27 has edge length 1.40 m and is oriented as shown in a region of uniform
electric field. Find the electric flux through the right face if the electric field, in newtons per
coulomb, is given by (a) 6i , (b) 2j , and (c) 3 4i k + . (d) What is x the total flux through the
cube for each field?
Fig. 23-27 1.40 m
(a)6iNC (b) 2j NC (c) 3 4i k + N
C (d)
Fundamentals of Physics, 8thEdPrinciple of Physics, 9thEdHalliday & Resnic
23-27
We use =
E A , where
A A= = . j m j2
140b g .
(a)( ) ( )
2
6.00 N C i 1.40 m j 0. = =
(b) ( ) ( )2 22.00 N C j 1.40 m j 3.92 N m C. = =
(c) ( ) ( ) ( )2
3.00 N C i 400 N C k 1.40 m j 0 = + = .
(d) The total flux of a uniform field through a closed surface is always zero.
8thEdProblem 23-59thEdProblem 23-7
A point charge of 1.8 C is at the center of a cubical Gaussian surface 55 cm on edge. What is thenet electric flux through the surface?
1.8 C 55 cm
We use Gauss law: 0 q = , where is the total flux through the cube surface and qis thenet charge inside the cube. Thus,
6
5 212 2 2
0
1.8 10 C 2.0 10 N m C.8.85 10 C N m
q
= = =
8thEdProblem 23-79thEdProblem 23-5
In Fig. 23-29, a proton is a distance2
d directly above the center of a square of side d. What is
the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a
cube with edge d.)
7/22/2019 Fundamental of Physics Halliday Ed. 9 Chapter 23 Homework Solution
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7/22/2019 Fundamental of Physics Halliday Ed. 9 Chapter 23 Homework Solution
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7/22/2019 Fundamental of Physics Halliday Ed. 9 Chapter 23 Homework Solution
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Fundamentals of Physics, 8thEdPrinciple of Physics, 9thEdHalliday & Resnic
For r< a, the charge enclosed by the Gaussian surface is q1(r/a)3. Gauss law yields
3
2 1 13
0 0
4 .
4
q q rrr E E
a a
= =
(a) For r = 0, the above equation impliesE= 0.
(b) For r = a/2, we have9 2 2 15
213 2 2
0
( / 2) (8.99 10 N m /C )(5.00 10 C)5.62 10 N/C.
4 2(2.00 10 m)
q aE
a
= = =
(c) For r = a, we have9 2 2 15
12 2 2
0
(8.99 10 N m /C )(5.00 10 C)0.112 N/C.
4 (2.00 10 m)
qE
a
= = =
In the case where a< r< b, the charge enclosed by the Gaussian surface is q1, so Gausslaw leads to
2 1 12
0 0
4 .4
q qr E E
r
= =
(d) For r = 1.50a, we have9 2 2 15
12 2 2
0
(8.99 10 N m /C )(5.00 10 C)0.0499 N/C.
4 (1.50 2.00 10 m)
qE
r
= = =
(e) In the region b< r< c, since the shell is conducting, the electric field is zero. Thus, for r
= 2.30a, we haveE= 0.(f) For r > c, the charge enclosed by the Gaussian surface is zero. Gauss law yields
4 0 02r E E= = . Thus,E= 0 at r= 3.50a.(g) Consider a Gaussian surface that lies completely within the conducting shell. Since the
electric field is everywhere zero on the surface,
E dA =z 0 and, according to Gausslaw, the net charge enclosed by the surface is zero. If Qiis the charge on the inner surface
of the shell, then q1+ Qi= 0 and Qi= q1=5.00 fC.
(h) Let Qobe the charge on the outer surface of the shell. Since the net charge on the shell
is q, Qi+ Qo= q1. This means
Qo= q1 Qi= q1(q1) = 0.