fundamental of physics halliday ed. 9 chapter 23 homework solution

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.)


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=


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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)


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.)


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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.

fundamental of physics halliday ed. 9 chapter 23 homework solution

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