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일반물리학2 개정10판 22장 솔루션

191

Chapter 22

THINK Our system consists of two point charges of opposite signs fixed to the x axis.

Since the net electric field at a point is the vector sum of the electric fields of individual

charges, there exists a location where the net field is zero.

EXPRESS At points between the charges, the individual electric fields are in the same

direction and do not cancel. Since charge q 2 = – 4 q 1 located at x 2 = 70 cm has a greater

magnitude than q 1 = 2 ¥ 10

8 C located at x 1 = 20 cm, a point of zero field must be closer

to q 1 than to q 2. It must be to the left of q 1.

Let x be the coordinate of P , the point where the field vanishes. Then, the total electric

field at P is given by

####### ( )

2 1 2 2 021

1 | | | |

4 ( )

q q E pe x x x x

Ê ˆ = Á – ̃ Á – – ̃ Ë ̄

.

ANALYZE (a) If the field is to vanish, then

####### ( ) ( )

2 2 1 2 2 2 2 2 2111

| | | | | | ( ) . ( ) | |

q q q x x

x x x x q x x

= fi =

Taking the square root of both sides, noting that | q 2 |/| q 1 | = 4, we obtain

70 cm 2. 20 cm

x

x

= ±

.

Choosing –2 for consistency, the value of x is found to be x = -30 cm.

(b) If the particles are interchanged, the condition becomes, choosing +1/2 for

consistency,

70 cm 1

20 cm 2

x

x

= +

and the answer is x = +1 m.

192 CHAPTER 22

Referring to Eq. 22-6, we use the binomial expansion (see Appendix E) but keeping

higher order terms than are shown in Eq. 22-7:

E =

q

4 peo z

2 Ë

Á

Ê ̄

̃

ˆ Ë

Á

Ê ̄

̃

ˆ 1 +

d

z

3

4

d

2

z

2 +

1

2

d

3

z

3 + … – Ë

Á

Ê ̄

̃

ˆ 1 –

d

z

3

4

d

2

z

2 –

1

2

d

3

z

3 + …

=

q d

2 peo z

3 +

q d

3

4 peo z

5 + …

Therefore, in the terminology of the problem, E next = q d

3 / 4 pe 0 z

5 .

194 CHAPTER 22

(a) We use the usual notation for the linear charge density: l = q/L. The arc length is L

= r q with q is expressed in radians. Thus,

L = (0 m)(0 rad) = 0 m.

With q = -300(1 ¥ 10

19 C), we obtain l = -1 ¥ 10

15 C/m.

(b) We consider the same charge distributed over an area A = p r

2 = p(0 m)

2 and

obtain s = q/A = -3 ¥ 10

14 C/m².

(c) Now the area is A sphere = 4p r

2 and thus obtain:

s = q/A sphere = -2 ¥ 10

15 C/m².

(d) Finally, we consider that same charge spread throughout a volume of V = 4p r

3 /3 and

obtain the charge density r = q V / = -1 ¥ 10

12 C/m

3 .

195

We take the charge Q =45 pC of the bee to be concentrated as a particle at the

center of the sphere. The magnitude of the induced charges on the sides of the grain is

| | 1 pC. q =

(a) The electrostatic force on the grain by the bee is

2 2 2 2

( ) 1 1 | | ( / 2) ( / 2) ( / 2) ( / 2)

kQq kQ q F kQ q d D D D d D

È ̆ = + = – Í – ̇

Î + ̊

where D =1 cmis the diameter of the sphere representing the honeybee, and

d =40 mm is the diameter of the grain. Substituting the values, we obtain

( )

9 2 2 12 12 3 2 3 2

10

1 1 8 10 N m C (60 10 C)(1 10 C) (5 10 m) (5 10 m)

3 10 N.

F

È ̆ = – ¥ ◊ ¥ ¥ Í – ̇ Î ¥ ¥ ̊

= – ¥

The negative sign implies that the force between the bee and the grain is attractive. The

magnitude of the force is

10 | F | 3 10 N

= ¥.

(b) Let | Q ¢| 60 pC= be the magnitude of the charge on the tip of the stigma. The force

on the grain due to the stigma is

2 2 2 2

| | | | ( ) 1 1 | || | ( ) ( ) ( ) ( )

k Q q k Q q F k Q q d D D D d D

¢ ¢ – È ̆ ¢= + = – ¢ Í – ̇

¢ ¢ Î ¢ + ¢ ̊

where D ¢=1 mmis the distance between the grain and the tip of the stigma.

Substituting the values given, we have

( )

9 2 2 12 12 3 2 3 2

8

1 1 8 10 N m C (60 10 C)(1 10 C) (1 10 m) (1 10 m)

4 10 N.

F

È ̆ ¢= – ¥ ◊ ¥ ¥ Í – ̇ Î ¥ ¥ ̊ = – ¥

The negative sign implies that the force between the grain and the stigma is attractive.

The magnitude of the force is

8 | F | 4 10 N

¢ = ¥.

(c) Since | F ¢| |> F | , the grain will move to the stigma.

197

We take the positive direction to be to the right in the figure. The acceleration of the

proton is ap = eE / mp and the acceleration of the electron is ae = – eE / me , where E is the

magnitude of the electric field, mp is the mass of the proton, and me is the mass of the

electron. We take the origin to be at the initial position of the proton. Then, the

coordinate of the proton at time t is x = 12 a tp

2 and the coordinate of the electron is

x = L + 21 a te

2 . They pass each other when their coordinates are the same, or

1212 . 2 2

a tp = + L a te

This means t

2 = 2 L /( ap – ae ) and

( ) ( )

####### ( )

31

31 27

5

9 10 kg 0 m 9 10 kg 1 10 kg

4 10 m

p p e

p e p e e p

a eE m m x L L L a a eE m eE m m m

Ê ˆ = = =Á ̃

Á + ̃ Ë ̄

Ê ¥ ˆ =Á ̃ Ë ¥ + ¥ ̄

= ¥

or about 44 mm.

198 CHAPTER 22

The field of each charge has magnitude

( )

19 9 2 2 6 2 2 2

1 10 C (8 10 N m C ) 2 10 N C. 0 m (0 m)

kq e E k r

¥ – = = = ¥ ◊ = ¥

The directions are indicated in standard format below. We use the magnitude-angle

notation (convenient if one is using a vector-capable calculator in polar mode) and write

(starting with the proton on the left and moving around clockwise) the contributions to r E net as follows:

b E – – 20 ∞ +g b E – 130 ∞ +g b E – – 100 ∞ +g b E – – 150 ∞ +g b E – ∞ 0 g.

This yields ( )

6 2 10 76.

¥ – – ∞ , with the N/C unit understood. Alternatively, one can

use the Catesian coordinates, and add up the x and the y components:

0

1 1 1

2 1 2 1 2

3 3 4 3 4

ˆr ˆi

ˆr cos( )i sin(ˆ ) j cos(210 )i sin(210 ) jˆ ˆ ˆ 0 0 jˆ ˆ

ˆr cos( )i sin(ˆ ) j cos(260 )i sin(260 ) jˆ ˆ ˆ 0 0ˆ 85 jˆ

ˆr cos( )i sin(ˆ ) j cos(130 )i sin(13ˆ ˆ

q p q p

q q p q q p

p q q p q q

=

= + + + = ∞ + ∞ = – –

= + + + + + = ∞ + ∞ = – –

= – – + – – = ∞ +

4 4 4

0 1 2 3 4

0 ) jˆ 0 0 jˆ ˆ

ˆr cos( )i sin(ˆ ) j cos( 20 )i sin( 20 ) j 0 0 jˆ ˆ ˆ ˆ ˆ

r ˆ ˆ ˆ ˆ ˆr r r r r 0 1 jˆ ˆ

q q

∞ = – +

= – + – = – ∞ + – ∞ = –

= + + + + = –

r

(a) The result above shows that the magnitude of the net electric field is

6 | E net| 2 10 N/C.

= ¥

r

(b) Similarly, the direction of

r E net is –76∞ from the x axis.

200 CHAPTER 22

We assume q > 0. Using the notation l = q / L we note that the (infinitesimal) charge

on an element dx of the rod contains charge dq = l dx. By symmetry, we conclude that all

horizontal field components (due to the dq ’s) cancel and we need only “ sum ” (integrate)

the vertical components. Symmetry also allows us to integrate these contributions over

only half the rod (0 £ x £ L /2) and then simply double the result. In that regard we note

that sin q = R / r where

2 2 r = x + R.

(a) Using Eq. 22-3 (with the 2 and sin q factors just discussed) the magnitude is

( )

####### ( )

####### ( )

2 2

0 2 0 2 2 2 2 0 0 / 2 2

0 2 2 3 2 2 2 2 000

2 2 2 2 0 0

2 2 sin 4 4

2 2

2 1

2224

L L

L L

dq dx y E r x R x R

R dx q L R x

x R R x R

q L q

LR L R R L R

q pe pe

pe pe

pe pe

Ê ˆ Ê l ˆÊ ˆ = Á ̃ = Á ̃Á ̃ Ë ̄ Ë + ̄Ë + ̄

l = = ◊

= = + +

Ú Ú

Ú

r

where the integral may be evaluated by elementary means or looked up in Appendix

E (item #19 in the list of integrals). With

12 q 9 10 C

= ¥ , L =0 m,and R =

0 m, we have | | 13 N/C E =

r .

(b) As noted above, the electric field E

r points in the + y direction, or

90 ∞counterclockwise from the + x axis.

201

From symmetry, we see that the net field at P is twice the field caused by the upper

semicircular charge + = ( q l p R )(and that it points downward). Adapting the steps leading

to Eq. 22-21, we find

( )

90

net 2 2 0 90 0

2 ˆj sin ˆj. 4

q E R R

q e e

∞

∞

Ê ˆ = – = -Á ̃ Ë p ̄

r l

p

(a) With R = 4¥ 10

2 m and q = 1¥ 10

11 C, we obtain

11

net 2 2 12 2 2 2 2 2 0

1 10 C | | 95 N/C. (8 10 C /N m ) (4 10 m)

q E e p R p

¥ = = = ¥ ◊ ¥

r

(b) The net electric field E net

r points in the -ˆjdirection, or – 90 ∞counterclockwise from

the + x axis.

203

THINK Our system is a non-conducting rod with uniform charge density. Since the

rod is an extended object and not a point charge, the calculation of electric field requires

an integration.

EXPRESS The linear charge density l is the charge per unit length of rod. Since the total

charge – q is uniformly distributed on the rod of length L , we have l= – q L /. To

calculate the electric at the point P shown in the figure, we position the x- axis along the

rod with the origin at the left end of the rod, as shown in the diagram below.

Let dx be an infinitesimal length of rod at x. The charge in this segment is dq = l dx. The

charge dq may be considered to be a point charge. The electric field it produces at point P

has only an x component and this component is given by

dE

dx

L a x

x = + –

1

40

2 pe

l

b g

.

The total electric field produced at P by the whole rod is the integral

( )

( ) ( )

0 2 0 0 0 0

0 0

1 1 1

4 4 4

, 4 4

L L x

dx E L a x L a x a L a

L q

a L a a L a

e e e

e e

l l l Ê ˆ = = = Á – ̃ p + – p + – p Ë + ̄

l 1 = = – p + p +

Ú

upon substituting – = q l L.

ANALYZE (a) With q = 4 ¥ 10

15 C, and L = 0 m, the linear charge density of

the rod is

15 4 10 C 14 5 10 C/m. 0 m

q

L

l

¥ – = = = – ¥

(b) With a = 0 m, we obtain

( )

9 2 2 15 3

0

(8 10 N m C )(4 10 C) 4 10 N/C 4 (0 m)(0 m 0 m)

x

q E e a L a

1 ¥ ◊ ¥ – = – = – = – ¥ p + +

,

or

3 | Ex | 4 10 N/C

= ¥.

204 CHAPTER 22

(c) The negative sign in Ex indicates that the field points in the – x direction, or – 180 ∞

counterclockwise from the + x axis.

(d) If a is much larger than L , the quantity L + a in the denominator can be approximated

by a , and the expression for the electric field becomes

E

q

a

x = – 40

2 pe

.

Since a =50 mis much greater than L =0 m, the above approximation applies and

we have

8 Ex 1 10 N/C

= – ¥ , or

8 | Ex | 1 10 N/C

= ¥.

(e) For a particle of charge

15 q 4 10 C,

= – ¥ the electric field at a distance a = 50 m

away has a magnitude

8 | Ex | 1 10 N/C

= ¥.

LEARN At a distance much greater than the length of the rod ( a? L ), the rod can be

effectively regarded as a point charge – q , and the electric field can be approximated as

2 0

. 4

x

q E pe a

ª

206 CHAPTER 22

THINK The electric quadrupole is composed of two dipoles, each with a dipole

moment of magnitude p = qd. The dipole moments point in the opposite directions and

produce fields in the opposite directions at points on the quadrupole axis.

EXPRESS Consider the point P on the axis, a distance z to the right of the quadrupole

center and take a rightward pointing field to be positive. Then the field produced by the

right dipole of the pair is given by qd /2pe 0 ( z – d /2)

3 while the field produced by the left

dipole is – qd /2pe 0 ( z + d /2)

3 .

ANALYZE Use the binomial expansions

( z – d /2)

3 ª z

3 3 z

4 (– d /2)

( z + d /2)

3 ª z

3 3 z

4 ( d /2)

we obtain

2

3 3 3 4 3 4 4 0 0 0 0

1 3 1 3 6 . 2 ( / 2) 2 ( / 2) 2 2 2 4

qd qd qd d d qd E pe z d pe z d pe z z z z pe z

È ̆ = – ª + – + =

Í ̇ Î ̊

Since the quadrupole moment is

2 Q = 2 qd , we have

LEARN For a quadrupole moment Q , the electric field varies with z as

4 E : Q z /. For a

point charge q , the dependence is

2 E : q z / , and for a dipole p , we have

3 E : p z /.

207

With x 1 = –5 cm and x 2 = 10 cm, the point midway between the two charges is

located at x = 2 cm. The values of the charge are

q 1 = – q 2 = – 4 ¥ 10

7 C,

and the magnitudes and directions of the individual fields are given by:

####### ( )

####### ( )

9 2 2 7 1 5 1 2 2 0 1

9 2 2 7 2 5 222 0 2

| | ˆ (8 10 N m C ) | 4 10 C|ˆ ˆ i i (6 10 N C)i 4 ( ) 0 m 0 m

ˆ (8 10 N m C ) (4 10 C)ˆ ˆ i i (6 10 N C)i 4 ( ) 0 m 0 m

q E x x

q E x x

pe

pe

¥ ◊ – ¥ = – = – = – ¥

¥ ◊ ¥ = – = – = – ¥

r

r

Thus, the net electric field is

6 net 1 2 E = E + E = -(1 10 N C)i¥ ˆ

r r r .

209

Examining the lowest value on the graph, we have (using Eq. 22-38)

U = – p

Æ · E

Æ = – 100 ¥ 10

28 J.

If E = 50 N/C, we find p = 2 ¥ 10

28 C· m.

210 CHAPTER 22

Our system consists of four point charges that are placed at the corner of a square.

The total electric field at a point is the vector sum of the electric fields of individual

charges. Applying the superposition principle, the net electric field at the center of the

square is

4 4

2 110

1 ˆr 4

i i i i i i

q E E = = pe r

=Â =Â

r r .

With q 1 = +30 nC, q 2 = -15 nC, q 3 = +15 nC,and q 4 = -30 nC, the x component of the

electric field at the center of the square is given by, taking the signs of the charges into

consideration,

####### ( )

1 2 3 4 2 2 2 2 0

2 1 2 3 4 0

1 | | | | | | | | cos 45 4 ( / 2 ) ( / 2 ) ( / 2 ) ( / 2 )

1 1 1 | | | | | | | |. 4 / 2 2

x

q q q q E a a a a

q q q q a

e

e

È ̆ = Í + – – ̇ ∞ Î ̊

= + – –

p

p

Similarly, the y component of the electric field is

####### ( )

1 2 3 4 2 2 2 2 0

2 1 2 3 4 0

1 | | | | | | | | cos 45 4 ( / 2 ) ( / 2 ) ( / 2 ) ( / 2 )

1 1 1 | | | | | | | |. 4 / 2 2

y

q q q q E a a a a

q q q q a

pe

pe

È ̆ = Í- + + – ̇ ∞ Î ̊

= – + + –

The magnitude of the net electric field is

2 2 E = Ex + Ey.

Substituting the values given, we obtain

####### 2 ( 1234 ) 2 ( )

0 0

1 2 1 2 | | | | | | | | 30 nC 15 nC 15 nC 30 nC 0 4 4

Ex q q q q e a e a

= + – – = + – – = p p

and

####### ( ) ( )

( )

2 1 2 3 4 2 0 0 9 2 2 8

2

5

1 2 1 2 | | | | | | | | 30 nC 15 nC 15 nC 30 nC 4 4

8 10 N m / C (3 10 C) 2

(0 m)

1 10 N/C.

Ey q q q q pe a pe a

= – + + – = – + + –

¥ ◊ ¥

= – ¥

Thus, the electric field at the center of the square is ˆ 5 ˆ E E = y j ( 1 10 N/C)j.= – ¥

r

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북스힐 대학 물리학 10 판 솔루션 pdf

북스힐 일반물리학 연습문제 솔루션 & 해설

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일반물리학2 개정10판 22장 솔루션

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