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북스힐 일반물리학 연습문제 솔루션 & 해설
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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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