PART 4
ELECTROMAGNETISM ELECTR OMAGNETISM
CHAPTER 23
ELECTRIC CHARGE, FORCE, AND FIELD
ActivPhysics can help with these problems: Activities 11.1–11.8 Section 23-2:
Electric Charge
Problem 1.
Suppose the electron and and proton charges differed differed by one part in one one billion. Estimate the net charge you you would carry. carry.
Solution Nearly all of the mass of an atom is in its nucleus, and about one half of the nuclear mass of the light elements in living matter (H, O, N, and C) is protons. Thus, the number of protons in a 65 kg average-sized person is approximately 1 (65 kg )=(1.67 × 10 1 0 −27 kg ) ¼ 2 × 10 28 , which is also the number of electrons, since an average person is electrically 2 neutral. If there were a charge imbalance of qproton − qelectron ±2
19 × 10 28 × 10 −9 × 1.6 × 10 10 − C
= ± 3. 2
=
10 −9 e, a person’s net charge would be about about
C, or several coulombs (huge by ordinary standards).
Problem 2.
A typical lightning flash delivers delivers about 25 25 C of negative negative charge charge from cloud to ground. ground. How many many electrons are involved?
Solution The number is Q=e
=
25 C=1.6 × 10 −19 C
=
1.56 × 10 20 .
Problem 3.
Protons Protons and neutr neutrons ons are made made from combina combination tionss of the two most most common common quarks, quarks, the the u quark and the d quark. The 2 1 u quark’s charge is + 3 e while the d quark carries − 3 e . How could three of these quarks combine to make (a) a proton and (b) a neutron?
Solution (a) The proton’s charge is 1e 2 3
−
1 3
−
1 3
2
2
= 3e + 3e
− 13 e, corresponding corresponding to a combination of uud quarks; (b) for neutrons, 0
=
corresponds to udd . (See Chapter 39, or Chapter 45 in the extended version of the text.)
Problem 4.
A 2-g ping-pong ping-pong ball rubbed against a wool wool jacket acquires a net net positive positive charge of 1 µC . Estimate the fraction of the ball’s electrons that have been removed.
Solution If half the ball’s mass is protons, their number (equal to the original number of electrons) is 1 g=m p . The number of electrons removed is 1 µ C=e, so the fraction removed removed is (1 µ C=e) (1 g=m p ) (a hundred billionth).
=
10 −6 C × 1.67 × 10 −24 g 1.6 × 10 10
−19
C ×1g
=
1.04 × 10 −11
CHAPTER 23 541
Section 23-3:
Coulomb’s Law
Problem 5.
If the charge charge imbalance imbalance of Problem Problem 1 existed, existed, what would would be the approximate approximate force between you you and another another person person 10 m away? Treat the people as point charges, and compare the answer with your weight.
Solution The magnitude of the Coulomb force between two point charges of 3.2 C (see solution to Problem 1), at a distance of 10 m, is k q 2=r 2 = (9 × 10 9 N ⋅ m 2/C 2 )(3.2 C=10 m)2 = 9 .22 × 10 8 N. This is approximately 1.45 million times the weight of an average-sized 65 kg person.
Problem 6.
Find the the ratio of the the electrical force force between a proton and and an electron electron to the gravitation gravitational al force between between the two. Why doesn’t it matter that you aren’t told the distance between them?
Solution At all distances (for which the particles can be regarded as classical point charges), the Coulomb force is stronger than the gravitational force by a factor of: F elec
=
F grav
2
2
Gr K JH GGm m K J H ke
r
2
p
=
e
(9 × 10 9 N ⋅ m 2/C 2 )(1.6 × 10 −19 C)2 (6.67 × 10 −11 N ⋅ m 2/ kg 2 )(1.67 × 10 −27 kg )(9.11 × 10 −31 kg )
¼
2.3 × 10 39.
The spacial dependence of both forces is the same, and cancels out.
Problem 7.
The electron electron and proton proton in in a hydrogen hydrogen atom are are 52.9 pm apart. apart. What What is the magnitude magnitude of the the electric electric force between between them? them?
Solution 2
2
separated by a Bohr Bohr radius, FCoulomb = ke =a0 ' a0 = 52.9 pm is called the Bohr radius. For a proton and electron separated (9 × 10 10 9 N ⋅ m 2/C 2 )(1.6 × 10 10 −19 C=5.29 × 10 10 −11 m ) 2
=
8.23 × 10 10 −8 N.
Problem 8.
How far apart should an electron and proton be so the force of Earth’s gravity gravity on the electron is equal to the electric force arising from the proton? Your answer shows why gravity is unimportant unimportant on the molecular scale!
Solution The electric force between a proton and an electron has magnitude ke 2=r 2 , while the weight of an electron is me g. These are equal when r =
2
ke =me g =
(almost fifty billion atomic diameters).
(9 × 10 9 N ⋅ m 2/C 2 ) (1.6 × 10 −19 C) 2 (9.11 × 10 −31 kg)(9.8 m/s 2 )
=
5.08 m
542 CHAPTER 23
Problem 9.
Two charges, one twice as large as the other, are located 15 cm apart and experience a repulsive force of 95 N. What is the magnitude of the larger charge?
Solution The product of the charges is q1q2 is twice the other, q1
=
2q2 , then
= 1 2
2 2 9 2 2 −10 C 2 . If one charge r FCoulomb=k = (0.15 m) (95 N )=(9 × 10 N ⋅ m /C ) = 2.38 × 10
− q1 = 2.38 × 10 2
10
C and q1
= ±218 . µ C.
Problem 10. Earth carries a net charge of −4.3 × 10 5 C. The force due to this charge is the same as if it were concentrated at Earth’s center. How much charge would you have to place on a 1.0-g mass in order for the electrical and gravitational forces on it to balance?
Solution The mass could be suspended at the Earth’s surface if the electric force were repulsive ( q same sign as q E ) and 2 kqq E= RE = mg. Thus, q =
(10 −3 kg)(9.8 m/s 2 )(6.37 × 10 6 m ) 2 (9 × 10 9 N ⋅ m 2/C 2 )(−4.3 × 10 5 C)
=
−103 µ C.
Problem 11. A proton is on the x-axis at x = 1.6 nm. An electron is on the y-axis at y a helium nucleus (charge +2e ) at the origin.
=
0.85 nm. Find the net force the two exert on
Solution A unit vector from the proton’s position to the origin is − î , so the Coulomb force of the proton on the helium nucleus is FP,He
=
k(e)(2e)( − î )=(1.6 nm )2 = −0.180 î nN. (Use Equation 23-1, with q1 for the proton, q2 for the helium nucleus, and
the approximate values of k and e given.) A unit vector from the electron’s position to the origin is − $j , so its force on the helium nucleus is Fe,He
=
2 k (−e)(2e)( − $j )=(0.85 nm ) = 0.638 $j nN. The net Coulomb force on the helium nucleus is the sum
of these. (The vector form of Coulomb’s law and superposition, as explained in the solution to Problems 15 and 19, provides a more general approach.)
Problem 12. Let V
=
V x î + V y $j be an arbitrary vector, with V its magnitude. Show that V=V is a unit vector—i.e., that its
magnitude is 1.
Solution The magnitude of V=V is always V=V
=
V =V = 1, since the magnitude of V is V = V . (The components of V are not
needed here.)
Problem 13. A charge q is at the point x = 1 m, y = 0. Write expressions for the unit vectors you would use in Coulomb’s law if you were finding the force that q exerts on other charges located at (a) x = 1 m, y = 1 m; (b) the origin; (c) x = 2 m, y = 3 m. Note that you don’t know the sign of q. Why doesn’t this matter?
CHAPTER 23 543
Solution =
A unit vector from rq
( x− 1 m, y)= ( x− 1 m) 2
( x, y) is n$
=
(1 m, 0) , the position of charge q, to any other point r
(r − rq )= r − rq
=
=
2
+
y . The sign of q doesn’t affect this unit vector, but the signs of both charges do
$ (a) When the other determine whether the force exerted by q is repulsive or attractive, i.e., in the direction of + n$ or −n.
charge is at position r ( −1 m )
2
+
0
=
=
( −1, 0 )
(1 m, 1 m), n$ =
(0, 1 m)= 0
=
− î . (c) Finally, when r
+
(1 m) 2
=
(0 , 1)
(2 m, 3 m ), n$
=
=
= $j .
(b) When r
(1 m, 3 m)= (1 m)
= 2
$ (0, 0), n
+
(3 m )
2
=
=
(−1 m, 0 ) ÷
(1, 3)= 10
=
0. 316 î + 0. 949$j .
Problem 14. A proton is at the origin and an electron is at the point x
=
0.41 nm, y
=
0.36 nm. Find the electric force on the proton.
Solution The magnitude of the force is F p
=
ke r
2
=
2
(9 × 10 9 N ⋅ m 2 /C 2 )(1.6 × 10 −19 C)2
and its direction is from the proton (at r p angle θ
=
tan −1 ( 0.36=0.41)
=
(0.412 =
+
0.36 2 ) × 10 −18 m 2
0) to the electron (at re
=
=
7.74 × 10 −10 N,
(0.41î + 0.36 $j ) nm), for an attractive force, at an
41.3° to the x-axis. The vector form of Coulomb’s law, F p
=
3
− ke 2 (r p − r e)= r p − r e (see
solution of next problem) gives the same result: F p
=
−(9 × 10 9 N ⋅ m 2/C 2 )(1.6 × 10 −19 C) 2 ( −0.41î − 0.36 $j )=(0 .412
=
(5.82 î + 511 . $j ) × 10 −10 N.
+ 0 .36
2 3=2
)
(10 −9 m) 2
Problem 16 cm, y
=
5.0 cm, and a −3.2 - µ C charge is at x
Denote the positions of the charges by r1
=
(16 î + 5$j ) cm for q1
15. A 9.5 - µ C charge is at x the negative charge.
=
=
4.4 cm, y
=
11 cm. Find the force on
Solution The vector from q1 to q2 is r
=
=
9.5 µ C, and r2
r2 − r1 , and a unit vector in this direction is r$
Coulomb’s law for the electric force of q1 on q2 is F12
=
=
=
(4.4 î + 11$j ) cm for q2
=
−3.2 µ C.
(r2 − r1 )= r2 − r1 . The vector form of
3
kq1q 2 (r2 − r1 )= r2 − r1 . (This gives the Coulomb force between
two point charges, as a function of their positions, and is a convenient form to memorize because of its direct applicability.) Substituting the given values for this problem, we find: F12
=
=
9 × 10 9 N ⋅ m 2
G H
C
2
JK(9.5
µ C)( −3.2 µ C)
×
(4.4 î
+ 11$j
[(4.4 − 16) 2
+
− 16 î − 5 $j ) cm 3=2
(11 − 5 ) 2 ]
cm 3
(14.2 î − 7.37 $j ) N,
with magnitude 16.0 N and direction θ
=
−27.3° to the x-axis (negative angle measured CW).
Problem 16. A charge 3q is at the origin, and a charge −2q is on the positive x-axis at x = a. Where would you place a third charge so it would experience no net electric force?
544 CHAPTER 23
Solution The reasoning of Example 23-3 implies that for the force on a third charge Q to be zero, it must be placed on the x-axis to the right of the (smaller) negative charge, i.e., at x > a. The net Coulomb force on a third charge so placed is 2 2 −2 −2 2 2 F x = kQ[3qx − 2q( x − a ) ], so F x = 0 implies that 3( x − a) = 2 x , or x − 6 xa+ 3 a = 0. Thus, x = 3 a±
9 a2 − 3 a2
=
(3 ±
6 )a. Only the solution (3 +
6 )a
=
5.45a is to the right of x = a.
Problem 17. A 60- µ C charge is at the origin, and a second charge is on the positive x-axis at x x = 50 cm experiences no net force, what is the second charge?
=
75 cm. If a third charge placed at
Solution In order for the net force to be zero at a position between the first two charges, they must both have the same sign, i.e., q1 = 60 µ C at x1 = 0 and q2 > 0 at x 2 = 75 cm. (Then the separate forces of the first two charges on the third are in opposite directions.) Therefore, for the third charge q3 at x3 kq3 [60
µ C(50
cm ) −2 − q2 (25 cm ) −2 ]
=
0 implies q2
=
60
=
50 cm, F3 x
µ C(25=50 )
2
=
=
− − kq3 [ q1 ( x3 − x1 ) − q2 ( x2 − x3 ) ] = 2
2
15 µ C.
Problem 18. You have two charges +4q and one charge −q . (a) How would you place them along a line so there’s no net force on any of the three? (b) Is this equilibrium stable or unstable?
Solution By symmetry, the negative charge must be at the midpoint between the two positive charges (the force on it is zero there) such that its attractive force on one positive charge cancels the repulsive force of the other. Thus, 2 2 2 k(4 q) =(2 a) = k (4 q) − q =a ,
which holds for any a. The equilibrium is unstable, since if −q is displaced slightly toward one charge, the net force on it will be in the direction of that charge.
Problem 18 Solution.
Problem 19. In Fig. 23-39 take q1
=
68 µ C, q2
=
−34 µ C, and q3
=
15 µ C. Find the electric force on q3.
Solution Denote the positions of the charges by r1 = $j , r2 = 2î , and r3 = 2 î + 2 $j (distances in meters). The vector form of Coulomb’s law (in the solution to Problem 15) and the superposition principle give the net electric force on q3 as: F3
=
=
F13
+
F23
=
kq1q3 (r3 − r1 )
r3 − r1
(1.64 î − 0.326 $j ) N,
3
+
kq2 q3 (r3 − r2 )
r3 − r2
3
=
(9 × 10 9 N )(15 × 10 −6 )[(68 × 10 −6 )(2 î + $j )=5 5
+
( −34 × 10 −6 )2 $j=8 ]
CHAPTER 23 545
or F3
=
− F3 x + F 3y = 1.67 N at an angle of θ = tan ( F3 y =F 3 x ) = −11.2° to the x-axis. 2
2
1
FIGURE 23-39
Problem 19 Solution.
Problem 20. In Fig. 23-39 take q1 = 25 µ C and q2 = 20 µ C . If the force on q1 points in the − x direction, (a) what is q3 and (b) what is the magnitude of the force on q1?
Solution The positions of the charges are the same as in the previous problem, so the net force on q1 is =
F1
kq1
q (r − r ) M M Nr − r 2
1
2 3
1
(a) If F 1 y
=
0, then q2 − q3
=
0, or q3
+
r1 − r3
2
=
q3 (r1 − r3 ) 3
Lq (−2 î P kq M P N5 m Q M
20 µ C. (b) Then F1
=
+ $j )
2
=
1
3=2
2
+
q3 (−2 î − $j ) 5
(9 × 10 9 N )(25 × 20
3=2
µC
2
m2
. P P Q
)( −4 î )5
−3=2
=
−1.61î N.
Problem 21. Four identical charges q form a square of side a. Find the magnitude of the electric force on any of the charges.
Solution By symmetry, the magnitude of the force on any charge is the same. Let’s find this for the charge at the lower left corner, which we take as the origin, as shown. Then r1 = 0, r2 = a$j , r3 = a( î + $j ), r4 = aî , and F1
=
kq 2
r −r M M Nr − r 1
1
2 3
+
r1 − r3 r1 − r3
2
3
+
r1 − r4 r1 − r4
3
L−aj a( î j ) − aî O kq P − kq M − P P Q Na 2 2a a Q a =
2
$
3
+ $
3
3
=
2
2
F G H
( î + $j ) 1 +
1 2
IJ, 2K
(Use the vector form of Coulomb’s law in the solution to Problem 15, and the superposition principle.) Since î F1
=
(kq 2=a 2 ) 2 (1 + 1=2 2 )
=
( kq 2=a2 )( 2
1
+ 2) =
1.91kq 2=a2 .
Problem 21 Solution.
+ $j =
2,
546 CHAPTER 23
Problem 22. Three identical charges + q and a fourth charge −q form a square of side a. (a) Find the magnitude of the electric force on a charge Q placed at the center of the square. (b) Describe the direction of this force.
Solution The magnitudes of the forces on Q from each of the four charges are equal to kqQ=( 2 a=2) 2 = 2 kqQ=a 2 . But the forces from the two positive charges on the same diagonal are in opposite directions, and cancel, while the forces from the positive and negative charges on the other diagonal are in the same direction (depending on the sign of Q) and add. Thus, the net force on Q has magnitude 2(2 kqQ=a 2 ) and is directed toward (or away from) the negative charge for Q > 0 (or Q < 0).
Problem 23. Three charges lie in the x-y plane: q1 y = 3.0 m. If the force on q3 is
=
55 µ C at x= 0,
8.0 î + 15$j
y= 2.0 m;
q 2 at x = 3.0 m, y = 0; and q3 at x = 4.0 m,
N , find q2 and q3.
Solution Using the vector form of Coulomb’s law explained in the solution to Problem 15, and superposition, we can write the force on q3 as F3 = kq3 [q1 (r3 − r1 ) r3 − r1
r2
=
3î m, and r3
=
−3
+
q2 (r3 − r2 ) r3 − r2
( 4 î + 3$j ) m, we find (8
î $j ) + 15 2
Equating x and y components, we get 8 N ⋅ m =kq3
−3
=
=
−3=2
4 q117 =
Substituting this into either component equation, we get q3 [q117
+
3q2 10
−3=2 −1
]
=
116
kq3 [q1 (4 + $jî)(4 2 + 12 )
N ⋅ m2
Dividing these equations and solving for q2 in terms of q1 −3=2
]. Substituting the given values, F3 +
q210
−3=2
and 15 N ⋅ m =kq3
=
=
(8 N ⋅ m /k )[4 q117
(10=17)
−3=2
+
+
(8î
15$j ) N, r1
q2 ( + 3î$j )(12 + 3 2 )
+
2
55 µ C, we find q2 2
−3=2
=
3=2
=
q117
(52=9 )q1
q2 10
−3=2 −1
]
−3=2
=
=
+
143
=
−3=2
3q2 10
2 $j m,
].
−3=2
.
µ C.
(15 N ⋅ m 2 /k )
µ C.
Problem 24. Two identical small metal spheres initially carry charges q1 and q2, respectively. When they’re 1.0 m apart they experience a 2.5-N attractive force. Then they’re brought together so charge moves from one to the other until they have the same net charge. They’re again placed 1.0 m apart, and now they repel with a 2.5-N force. What were the original values of q1 and q2?
Solution The charges initially attract, so q1 and q2 have opposite signs, and 2 .5 N together, they share the total charge equally, each acquiring 2.5 N
=
k
1 4
( q1
+
2
( q1
+
− kq1q 2 =1 m 2 . When the spheres are brought
q2 ). The magnitude of their repulsion is
2
q2 ) =1 m . Equating these two forces, we find a quadratic equation
2 2 q1 + 6q1q2 + q2 = 0, with solutions q1
=
( −3 ±
8 (16.7
µ C) = ±40 .2 µ C
and q 2
1 4
(q1
+
2 q2 ) = − q1q2 , or
8 ) q2 . Both solutions are possible, but since 3 +
merely represent a relabeling of the charges. Since − q1q 2 q1 = ± 3 +
1 2
=
=
2.5 N ⋅ m 2=(9 × 10 9 N ⋅ m 2/ C2 )
= m40.2 µC=(3 +
8) =
m 6.90 µ C,
=
(16.7
8 = (3 − µ C)
2
−1 8 ) , they
, the solutions are
or the same values with q1 and q2
interchanged.
Section 23-4:
The Electric Field
Problem 25. An electron placed in an electric field experiences a 61 . × 10 −10 N electric force. What is the field strength?
CHAPTER 23 547
Solution − F=e = 6.1 × 10
=
From Equation 23-3a, E field.)
10
N=1.6 × 10 −19 C
=
3.81 × 10 9 N/C. (The field strength is the magnitude of the
Problem 26. What is the magnitude of the force on a 2 .0 - µ C charge in a 100 N /C electric field?
Solution =
From Equation 23-3b, F
qE = (2
µ C )(100
N/C)
=
2 × 10 −4 N.
Problem 27. A 68-nC charge experiences a 150-mN force in a certain electric field. Find (a) the field strength and (b) the force that a 35- µC charge would experience in the same field.
Solution Equations 23-3a and b give (a) E = 150 mN=68 nC
=
2.21 MN/C, and (b) F = (35 µ C)(2.21 MN/C)
=
77.2 N.
Problem 28. A −1.0 - µC charge experiences a 10 î -N electric force in a certain electric field. What force would a proton experience in the same field?
Solution The electric field is E −1.6 î pN.
=
10 î N=(−1 µ C)
=
−10 î MN/C. The force on a proton is eE
(1.6 × 10 −19 C)( −10 î MN/C)
=
=
Problem 29. The electron in a hydrogen atom is 0.0529 nm from the proton. What is the proton’s electric field strength at this distance?
Solution The proton in a hydrogen atom behaves like a point charge, for an electron one Bohr radius away (see solution to Problem 7), so Equation 23-4 gives E = ke= a02 = (9 × 10 9 N ⋅ m 2 /C 2 )(1.6 × 10 −19 C)=(5.29 × 10 −11 m) 2 = 515 . × 10 11 N/C.
Problem 30. A 65- µ C point charge is at the origin. Find the electric field at the points (a) x = y 50 cm; (c) x= −25 cm, = y 75 cm.
=
50 cm, y
=
0; (b) x
=
50 cm,
Solution The electric field from a point charge at the origin is E(r) and q E
=
=
65 µ C, E
=
(9 × 10 9 N ⋅ m 2/C 2 )(65 µ C)î=(0 .5 m)2
(9 × 65 × 10 3 N ⋅ m 2/C)(0.5 m)( î
to the x axis.) (c) When r ( −296 î
+ 888 $j )
=
=
kN/C ( E
(−0.25î =
+
+ $j )=(0.5
2 m )3
0.75$j ) m, E
936 kN/C, θ x
=
=
=
kqr$=r 2 = kqr=r 3 , since r$ = r=r. (a) For r = 0.5î m =
2 .34 î MN/C. (b) At r
0.5 m ( î + $j ),
(827 kN/C)(î + $j ). (The field strength is 117 . MN/C at 45°
(5.85 × 10 5 N ⋅ m 2/C)( −0 .25 î
108° ).
=
+
0 .75 $j ) m=[( −0 .25 )2
+
(0 .75 ) 2 ]
3=2
m3
=
548 CHAPTER 23
Section 23-5:
Electric Fields of Charge Distributions
Problem 31. In Fig. 23-40, point P is midway between the two charges. Find the electric field in the plane of the page (a) 5.0 cm directly above P, (b) 5.0 cm directly to the right of P, and (c) at P.
Solution Take the origin of x-y coordinates at the midpoint, as indicated, and use Equation 23-5. Let r±
= ±(2.5
cm )$j denote the
positions of the charges, and r that of the field point. A unit vector from one charge to the field point is ( r − r±) = r − r± , so the spacial factors in Coulomb’s law are r$i=r 2i
=
3
ri =r i
=
3
(r − r± )= r − r± . (a) For r
=
(5.0 cm )$j , r1
=
r − r+
=
( 5.0 cm )$j − ( 2.5 cm ) $j = ( 2.5 cm ) $j , and r2 = r2 = r − r− = (7.5 cm) $j . Then
E (b) For r
=
=
q1r1
G Hr
k
3 1
+
q2 r2 3 r2
JK G H =
9 × 10 9
N ⋅ m2 C
2
$j
JK M N(2.5 cm) (2 µ C)
−
2
$j
(7.5 cm)
2
(25.6 MN/C) j . P Q $
=
(5.0 cm )î , E
(c) For r
=
=
G H
9 × 10 9
N ⋅ m2 C
2
2 CI (5.0 î − 2.5 j ) M JKF Hcm KN(5.0 (−2.5) ) $
µ
2
2
+
2 3=2
−
(5.0 î + 2.5$j ) (5.0
2
+
2 3=2
2.5 )
−(515 . MN/C) j. P Q $
=
0, E
=
G H
9 × 10 9
N ⋅ m2 C
2
− $j
JKHcm KM N(2.5) 2
µC 2
FIGURE 23-40
2
−
$j
(2.5)
2
−(57.6 MN/C) j. P Q =
$
Problem 31 Solution.
Problem 32. A 1.0 - µ C charge and a 2.0 - µ C charge are 10 cm apart, as shown in Fig. 23-41. Find a point where the electric field is zero.
FIGURE
23-41 Problem 32 Solution.
Solution The field can be zero only along the line joining the charges (the x-axis). To the left or right of both charges, the fields due to each are in the same direction, and cannot add to zero. Between the two, a distance x > 0 from the 1 µ C charge, the
CHAPTER 23 549
electric field is E
=
2 2 2 k[q1 î=x + q2 (− î )=(10 cm − x ) ], which vanishes when 1 µ C= x = 2
µ C=(10
cm − x ) 2 , or
x = 10 cm=( 2 + 1) = 4.14 cm.
Problem 33. A proton is at the origin and an ion is at x the ion?
=
5.0 nm. If the electric field is zero at x
=
−5 nm, what is the charge on
Solution The proton, charge e, is at r p
=
0, and the ion, charge q, is at r I
=
5 î nm. The field at point r
=
−5î nm is given by
Equation 23-5, with spacial factors written as in the solutions to Problems 15 or 31: E(r)
=
∑ kq
i
i
(r − ri ) r − ri
3
=
ke
î ) (−5 nm (5 nm )
3
+
kq
î − 5 nm î ) ( −5 nm . 3 (10 nm)
Therefore, E = 0 implies 2 q=(10 )3 = −e=(5)3 , or q = −4e . (Note how we used the general expression for the electric field, at position r, due to a distribution of static point charges at positions ri .)
Problem 34. For the situation of Example 23-3, (a) write an expression for the electric field as a function of x for points to the right of the charge −q shown in Fig. 23-13. (b) Taking q = 1.0 µ C and a = 1.0 m, plot the field as a function of position for x = 5 m to x = 25 m.
Solution (a) For points on the x-axis with x
>
a, the electric field from the charge +2q is in the x direction ( r$ = î ) , and that of
the charge −q is in the negative x direction ( r$ Example 23-3). (b) For q
=
1 µ C and a
=
=
− î ) . Thus, E( x > a)
1 m, E(
> x 1) =
=
(9î kN/C)[2(
kqî [2( x + a) − − ( x − a) − ] (which is F=Q in 2
+x 1)
−2
2
− ( −x 1) −2 ], with x in meters. A graph of
E for 5 ≤ x ≤ 25 is shown.
Problem 34 Solution.
Problem 35. (a) Find an expression for the electric field on the y-axis due to the two charges q in Fig. 23-11. (b) At what point is the field on the y-axis a maximum?
550 CHAPTER 23
Solution −3=2
(a) The electric field is the force per unit charge, so Example 23-2 shows that E( y) = 2 kqy$j ( a2 + y2 ) . (b) The magnitude of the field, a positive function, is zero for y = 0 and y = ∞, hence it has a maximum in between. Setting the derivative equal to zero, we find 0 maxima are at y
= ± a=
=
(a2
+
2 −3=2
y )
− 23 y( a 2
+
2 −5=2
y )
(2 y), or a 2
+
2
y − 3y
2
=
0. Thus, the field strength
2 (the directions at these points, of course, are opposite, by symmetry).
Problem 36. Write an expression for the dipole moment vector of the dipole shown in Fig. 23-18.
Solution The vector separation of the positive charge from the negative charge is r+ − r− vector is q times this, p = 2qaî .
=
the dipole moment a î − ( − aî ) = 2aî , and
Problem 37. A dipole lies on the y axis, and consists of an electron at y = 0.60 nm and a proton at y = −0.60 nm. Find the electric field (a) midway between the two charges, (b) at the point x = 2.0 nm, y = 0, and (c) at the point x = −20 nm, y = 0.
Solution We can use the result of Example 23-6, with y replaced by x, and x by − y (or equivalently, $j by î , and î by − $j ) . Then E( x)
=
2 kqa$j ( a2
constant 2 kq
=
+
=
−3=2
, where q
=
e = 1.6 × 10 −19 C and a = 0.6 nm. (Look at Fig. 23-18 rotated 90 ° CW.) The
2(9 × 10 9 N ⋅ m 2/C 2 )(1.6 × 10 −19 C)
(2.88 GN/C) $j=(0.6 ) 2 (c) At x
x2 )
20 nm, E
= =
(2 .88 GN/C)( nm) 2 . (a) At
=
(8.00 GN/C)$j . (b) For x
=
2 nm, E
(2.88 GN/C) $j (0.6 )(0.6 2
+
20 2 )
−3=2
$j= 2a = x= 0, E(0) = 2 kq
=
(2.88 GN/C) $j (0.6 )(0.6 2
=
(216 kN/C)$j .
+
22 )
−3=2
=
(190 MN/C)$j .
Problem 38. What is the electric field strength 10 cm from a point dipole with dipole moment 3.8 perpendicular bisector and (b) on its axis?
µC
⋅ m (a) on the dipole’s
Solution (a) Equation 23-7a gives E = kp= y3
=
(9 × 10 9 Nm 2/C 2 )(3.8 µC ⋅ m)=(0.1 m)3 3
same distance, gives a field strength twice as large, E = 2 kp= x
=
=
34.2 MN/C. (b) Equation 23-7b, for the
68.4 MN/C.
Problem 39. The dipole moment of the water molecule is 6 .2 × 10 −30 C ⋅ m. What would be the separation distance if the molecule consisted of charges ±e ? (The effective charge is actually less because electrons are shared by the oxygen and hydrogen atoms.)
Solution The distance separating the charges of a dipole is d
=
−30 C ⋅ m=1.6 × 10 −19 C = 38.8 pm. p=q = 6.2 × 10
Problem 40. You’re 1.5 m from a charge distribution whose size is much less than 1 m. You measure an electric field strength of 282 N/C. You move to a distance of 2.0 m and the field strength becomes 119 N /C . What is the net charge of the distribution? Hint: Don’t try to calculate the charge. Determine instead how the field decreases with distance, and from that infer the charge.
CHAPTER 23 551
Solution n
Taking the hint, we suppose that the field strength varies with a power of the distance, E ' r . Then 282=119 or n
=
ln(282=119 )=ln(0.75)
=
=
n
(1.5=2 ) ,
−3.00. A dipole field falls off like r −3, hence the net charge is zero.
Problem 41. Three charges form an equilateral triangle of side a. At one vertex is a charge +2 q; at the other two vertices are charges − q. The triangle is oriented with the charge 2 q on the positive x axis and both charges −q on the y-axis. (a) Find an expression for the electric field on the x-axis, in the approximation x a. (b) Compare with Equation 23-7b to show that your result in (a) is a dipole field, and give an expression for the magnitude of the triangle’s dipole moment.
Solution (a) With the charges positioned as shown, the electric field on the positive x-axis, due to the two negative charges at (0, ± a=2) , matches the field found in Example 23-2 (replace q with − q, a with a=2, y with x, and $j with î ): F=Q ( x >
=
2k ( − q ) × ( x 2
+
2
a =4 )
−3=2
3a=2) −2 î (the unit vector in Equation 23-5 is î and the distance is x −
3a=2 ) is just k(2q)( x −
field is the sum of these, E( x)
î . For the positive charge at ( 3a=2, 0 ), the electric field on the x-axis to the right
=
2 kqî [( x −
3 a=2 )
−2
2
− x( x
+
2
a =4 )
3 a=2 ). The total
−3=2
]. For x a, one can use the binomial −3=2 −2 = x (1 + . . . ), where approximation (see Appendix A): ( x − 3 a=2) −2 = x−2 (1 + 3 a= x + . . . ) and x( x2 + a2=4) −3 . . . indicates terms of order a 2=x 2 or higher. Therefore, E( )x= 2 kq x . . . − (1 + . . . )] ' 2 3 kqax î −x2 [1 + 3 =a + î . (b) This field is the same as Equation 23-7b, with
p=
3 qa . (In general, the dipole moment for a distribution of point
charges is ∑ ri qi . Also note that the field in part (a) can also be found from Equation 23-5, with r$i=ri2
=
(r − ri )= r − ri
3
as
in the solution to Problem 15.)
Problem 41 Solution.
Problem 42. Three identical charges q form an equilateral triangle of side a, with two charges on the x-axis and one on the positive y-axis. (a) Find an expression for the electric field at points on the y-axis above the uppermost charge. (b) Show that your result reduces to the field of a point charge 3 q for y a.
552 CHAPTER 23
Solution (a) The electric field on the y-axis ( y 2kqy$j ( y E( y)
=
2
1 2 −3=2 + 4a ) . 2 $
4 kqj [4 y( a
4 y(2 y) −3
+
(2 y) −2
+
=
>
3 a=2) due to the two charges on the x-axis, follows from Example 23-2: F=Q =
That from the charge on the y axis is just kq$j ( y − 4 y2 )
−3=2
+
(2 y − 3 a) −2 ], for y
2
2 −3=2
3 a=2. (b) For y À a, [ 4 y( a + 4 y )
>
3(2 y) −2 , so E( y) → 4 kq$j 3(2 y) −2
3a=2) −2 (see Equation 23-4). The total field is
=
+
(2 y −
3 a) −2 ] →
2 k(3 q)$j= y , like the field from a point charge of magnitude 3 q.
Problem 42 Solution.
Problem 43. A 30-cm-long rod carries a charge of 80 axis, 45 cm from the end of the rod.
µC
spread uniformly over its length. Find the electric field strength on the rod
Solution Applying the result of Example 23-7, at a distance a (9 × 10 9 N ⋅ m 2 /C 2 )(80
µ C)=(0 .45
m)(0.45 m
+
=
0.45 m from the near end of the rod, we get E
0.30 m)
=
=
kQ=a( a + l) =
2.13 MN/ C.
Problem 44. A thin rod of length l carries charge Q distributed evenly over its length. A point charge with the same charge Q lies a distance b from the end of the rod, as shown in Fig. 23-42. Find a point where the electric field is zero.
FIGURE
23-42 Problem 44.
Solution At a point between the right end of the rod and the point charge, the fields from each oppose one another, and can cancel. At such a point, a distance x from the right end of the rod, the electric field is zero when kQ=x( x + l) = kQ=(b − x) 2 , or 2
2
2
2
x + xl = b − 2 bx + x , or x = b =( l + 2 b).
Problem 45. A thin rod of length l has its left end at the origin and its right end at the x = l . It carries a line charge density given by λ = λ 0 ( x 2=l 2 ) sin(π x=l) , where λ 0 is a constant. Find the electric field strength at the origin.
CHAPTER 23 553
Solution The electric field at the origin, due to a small element of charge, dq Using λ= x 2
=
(λ 0=l 2 )sin(π x=l) and integrating from x
z l
E
=
0
=
0 to x
Fk λ I Fπ x I dx − î G Jsin H KHK l
0 2
l
= î (kλ 0=lπ )( −1 − 1) =
=
λ dx, located at position x, is d E
= l, =
î
=
− î kλ dx=x 2.
we find
Fπ x I cos HK π
k λ 0 l
l
l
2
l
0
−2 î k λ 0 =π l.
Problem 45 Solution.
Problem 46. Two identical rods of length l lie on the x-axis and carry uniform charges ± Q, as shown in Fig. 23-43. (a) Find an expression for the electric field strength as a function of position x for points to the right of the right-hand rod. (b) Show that your result has the 1= x 3 dependence of a dipole field for x l . (c) What is the dipole moment of this configuration? Hint: See Equation 23-7b.
FIGURE 23-43
Problem 46 Solution.
Solution (a) The field due to each rod, for a point on their common axis, can be obtained from Example 23-7: E+ the right, and E− = kq= x( x + l), to the left. The resultant field (positive right) is E = E+ − E− =
(b) For x p = Ql.
l,
kQ x
1
Hx−
l
−
1 x+
2 kQl
K x( x =
l
2
− l2 )
=
kQ=x( x − l), to
.
E¼ 2 kQl= x3 . (c) Comparison with Equation 23-7b shows that the rods appear like a dipole with moment
Problem 47. A uniformly charged ring is 1.0 cm in radius. The electric field on the axis 2.0 cm from the center of the ring has magnitude 2.2 MN/C and points toward the ring center. Find the charge on the ring.
Solution From Example 23-8, the electric field on the axis of a uniformly charged ring is kQx( x 2 from the center of the ring. For the given ring in this problem, −2.2 MN/C ( −2.2 MN/C)( 5.59 cm 2 )=(9 × 10 9 N ⋅ m 2 /C)
=
−0.137
µ C.
=
+
2 −3=2
a ) 2
, where x is positive away 2 −3=2
kQ(2 cm)( 4 cm + 1 cm )
, or Q
=
554 CHAPTER 23
Problem 48. Figure 23-44 shows a thin, uniformly charged disk of radius R. Imagine the disk divided into rings of varying radii r, as suggested in the figure. (a) Show that the area of such a ring is very nearly 2 π r dr. (b) If the surface charge density on the disk is σ C/m 2 , use the result of (a) to write an expression for the charge dq on an infinitesimal ring. (c) Use the result of (b) along with the result of Example 23-8 to write the infinitesimal electric field dE of this ring at a point on the disk axis, taken to be the positive x axis. (d) Integrate over all such rings (that is, from r = 0 to r = R ), to show that the net electric field on the disk axis is
G H
E = 2π k σ 1 −
FIGURE
x 2
x + R
2
JK.
23-44 Problem 48.
Solution (a) The area of an anulus of radii R1
2 2 R2 is just π ( R2 − R1 ). For a thin ring, R1 = r and R2 = r+ dr, so the area is
<
π [( r + dr ) 2 − r 2 ] = π (2 r dr + dr 2 ). When dr is very small, the square term is negligible, and dA = 2π r dr. (This is equal to the circumference of the ring times its thickness.) (b) For surface charge density σ , dq = σ dA = 2πσ r dr. (c) From Example 23-8, dE x
=
2 2 k( dq) x( x + r )
center. (d) Integrating from r
=
−3=2
2π kσ xr( x 2
0 to R, one finds
z R
E x = 2π kσ x
(Note: For x > 0,
=
0
( x
r dr 2
+
2 3=2
r )
xE=
=
+
z R 0
2π kσ x
2 r )
−3=2
dr, which holds for x positive away from the ring’s
dE x , or R
−1 x 2 + r 2 2
2 −1=2
E x = 2π kσ [ −1 + x ( x + R )
0
2 −1=2
x = x and the field is E x = 2π kσ [1 − x( x + R ) 2
=
2π k σ
x − M x N ( x
x 2
+
2 1=2
R )
]. However, for x < 0,
] . This is consistent with symmetry on the axis, since E x ( x)
=
P.
x = − x and
− Ex (− x). )
Problem 49. Use the result of the preceding problem to show that the field of an infinite, uniformly charged flat sheet is 2π k σ , where σ is the surface charge density. Note that this result is independent of distance from the sheet.
Solution An infinite flat sheet is the sa me as an infinite flat disk (as long as the dimensions are infinite in all directions, the shape is irrelevant). Thus, we can find the magnitude of the electric field from a uniformly changed infinite flat sheet by letting R → ∞ in the result of the previous problem. Then, the limit of the second term is zero, and the magnitude is constant, E = 2π k σ . (The direction is perpendicularly away from (towards) the sheet for positive (negative) σ .)
CHAPTER 23 555
Problem 50. A semicircular loop of radius a carries positive charge Q distributed uniformly over its length. Find the electric field at the center of the loop (point P in Fig. 23-45). Hint: Divide the loop into charge elements dq as shown in Fig. 23-45, and write dq in terms of the angle d θ . Then integrate over θ to get the net field at P.
FIGURE
23-45 Problem 50 Solution.
Solution This problem is the same as Problem 73, with θ 0
=
0. Thus, E( P)
=
2kQî =π a 2 .
Problem 51. The electric field 22 cm from a long wire carrying a uniform line charge density is 1.9 kN/C. What will be the field strength 38 cm from the wire?
Solution For a very long wire ( l 38 cm) , Example 23-9 shows that the magnitude of the radial electric field falls off like 1=r. Therefore, E(38 cm)=E (22 cm) = 22 cm=38 cm; or E (38 cm) = (22=38)1.9 kN/C = 110 . kN/C.
Problem 52. What is the line charge density on a long wire if the electric field 45 cm from the wire has magnitude 260 kN/C and points toward the wire?
Solution If the electric field points radially toward the long wire ( l 45 cm) , the charge on the wire must be negative. The magnitude of the field is given by the result of Example 23-9, Er = 2 kλ= r, so λ = (−260 kN/C)(0.45 m) ÷ (2 × 9 × 10 9 N ⋅ m 2/C)
=
−6.50
µ C/ m.
Problem 53. A straight wire 10 m long carries 25 µC distributed uniformly over its length. (a) What is the line charge density on the wire? Find the electric field strength (b) 15 cm from the wire axis, not near either end and (c) 350 m from the wire. Make suitable approximations in both cases.
Solution (a) For a uniformly charged wire, λ = Q=l = 2.5 µC/m. (b) Since r = 15 cm ¿ 10 m = l and the field point is far from either end, we may regard the wire as approximately infinite. Then Example 23-9 gives Er = 2 kλ= r =
556 CHAPTER 23
(2 × 9 × 10 9 N ⋅ m 2 /C 2 )(2.5
µ C/ m)=(0.15
m)
=
300 kN/ C. (c) At r
like a point charge, so the field strength is kQ=r
2
=
=
9
350 m
(9 × 10 × 25 × 10
−6
10 m
= l,
the wire behaves approximately
2
N ⋅ m /C)=(350 m)2
=
184 . N/C.
Problem 54. Figure 23-46 shows a thin rod of length l carrying charge Q distributed uniformly over its length. (a) What is the line charge density on the rod? (b) What must be the electric field direction on the rod’s perpendicular bisector (taken to be the y axis)? (c) Modify the calculation of Example 23-9 to find an expression for the electric field at a point P a distance y along the perpendicular bisector. (d) Show that your result for (c) reduces to the field of a point charge Q for l. y
FIGURE
23-46 Problem 54 Solution.
Solution (a) λ = Q=l . (b) The x components of the fields from symmetrically placed elements of charge, dq = λ dx at ± x, cancel, so the net field is along the y axis (see Fig. 23-23). (c) Proceed exactly as in Example 23-9, except that the limits of integration are from −l=2 to + l=2 . Thus, E y = kλ y
(d) For y l
l,
z
l=2
+ l=2
− l=2
dx
( x
2
+
2 3=2
y )
=
x
kλy
2
y
we can neglect l in the square root, so we obtain E y
2
2
x + y =
k λ l
= − l=2 2
2
y y +
l
2
=4
.
2
kλ l= y = k Q= y as for a point charge. (Of course, for
→ ∞ , the result of Example 23-9 is recaptured.)
Section 23-6:
Matter in Electric Fields
Problem 55. In this famous 1909 experiment that demonstrated quantization of electric charge, R. A. Millikan suspended small oil drops in an electric field. With a field strength of 20 MN/C, what mass drop can be suspended when the drop carries a net charge of 10 elementary charges?
Solution In equilibrium under the gravitational and electrostatic forces, mg 7
2
(2 × 10 N/C)=(9.8 m/s ) in terms of its radius, R
=
=
3.27 × 10
−12 1=3
=
qE, or m
=
(10 × 1.6 × 10 −
19
C) ×
kg. (Because this is so small, the size of such a drop may be better appreciated
(3m=4πρ oil ) . Millikan used oil of density 0 .9199 g/cm 3 , so R
=
9.46
µm
for this drop.)
Problem 56. How strong an electric field is needed to accelerate electrons in a TV tube from rest to one-tenth the speed of light in a distance of 5.0 cm?
CHAPTER 23 557
Solution For uniform acceleration, a Therefore, ( c=10)
2
=
=
eE=m, electrons, starting from rest, reach speed v 2
=
2 ax , traversing a region of length x.
2(eE=m) x, or E =
mc
2
=
200ex
(9.11 × 10 −31 kg) (3 × 10 8 m/s) 2 200(1.6 × 10
−19
C )( 0.05 m )
=
512 . × 10 4 N/C.
Problem 57. A proton moving to the right at 3 .8 × 10 5 m/s enters a region where a 56 kN/C electric field points to the left. (a) How far will the proton get before its speed reaches zero? (b) Describe its subsequent motion.
Solution (a) Choose the x axis to the right, in the direction of the proton, so that the electric field is negative to the left. If the Coulomb force on the proton is the only important one, the acceleration is a x = e( − E )=m. Equation 2-11, with vox = 3.8 × 10 5 m/s a nd v x
=
0, gives a maximum penetration into the field region of x− x0 (1.67 × 10 −27 kg)(3.8 × 10 5 m/s) 2 2(1.6 × 10 −19 C)(56 × 10 3 N/C)
=
=
2 =2 ax −v ox
=
2
mv ox =2 eE=
1.35 cm.
(b) The proton then moves to the left, with the same constant acceleration in the field region, until it exits with the initial velocity reversed.
Problem 58. An oscilloscope display requires that a beam of electrons moving at 8.2 Mm/s be deflected through an angle of 22 ° by a uniform electric field that occupies a region 5.0 cm long. What should be the field strength?
Solution Referring to Example 23-10, one sees that v =v y 6
2
(8.2 × 10 m/s) tan 22°=(1.6 × 10
−19
x=
C)(0 .05 m )
=
tan θ
=
− qE y∆ x=mv x. Therefore, in this case, E y = (9.11 × 10 2
31
kg) ×
3.09 kN/C.
Problem 59. An ink-jet printer works by “steering” charged ink drops to the right place on the page by passing moving drops through a uniform electric field that deflects them by the appropriate amount. Figure 23-47 shows an ink drop approaching the field region, which has length l and width d between the charged plates that establish the field. Find an expression for the minimum speed a drop with mass m and charge q must have if it is to get through the region without hitting either plate.
FIGURE
23-47 Problem 59.
Solution If they enter the field region midway, moving horizontally, the maximum vertical deflection, during the transit time t = l=v , can be d =2, for the ink drops to pass through. Thus, y
1
= 2
at
2
1
= 2
(qE=m)(l=v) 2
1
< 2
d , or v >
l
qE=md .
558 CHAPTER 23
Problem 60. An electrostatic analyzer like that of Example 23-11 has b select protons moving at 84 km/s?
=
7.5 cm. What should be the value of E 0 if the device is to
Solution From the analysis in Example 23-11, E0
=
− mv = eb = (1.67 × 10 2
27
kg)(84 × 10 3 m/s)2=(1.6 × 10 −19 C)(0 .075 m )
=
982 N/C.
Problem 61. An electron is moving in a circular path around a long, uniformly charged wire carrying 2.5 nC/m. What is the electron’s speed?
Solution The electric field of the wire is radial and falls off like 1=r (see Example 23-9). For an attractive force (negative electron encircling a positively charged wire), this is the same dependance as the centripetal acceleration. For circular motion around the wire, the Coulomb force provides the electron’s centripetal acceleration, or −eE=m = −2 keλ=mr = − v 2=r. Thus,
v
=
2 keλ=m
=
1=2
[2(9 × 10 9 N ⋅ m 2/C 2 )(1.6 × 10 −19 C)(2 .5 × 10 −9 C/m)=(9 .11 × 10 −31 kg)]
=
2 .81 Mm/s.
Problem 62. Figure 23-48 shows a device its inventor claims will separate isotopes of a particular element. (Isotopes of the same element have nuclei with the same charge but different masses.) Atoms of the element are first stripped completely of their electrons, then accelerated from rest through an electric field chosen to give the desired isotope exactly the right speed to pass through the electrostatic analyzer (see Example 23-11). Prove that the device won’t work—that is, that it won’t separate different isotopes.
FIGURE 23-48
Problem 62.
Solution When the device is in operation, an isotope, of nuclear charge q and mass m, is accelerated from rest to a speed v, in a distance d , by the field E 1, where v 2 = 2a1 d = 2( qE1=m)d . This will be the proper speed to pass through the analyzer if 2 a2 = qE2 =m = v =r = 2(qE1=m) d=r, or E2 = 2 E1 d=r. This condition depends on the fields and the geometry, but not on q=m, so different isotopes cannot be separated. (Essentially, the device compares two accelerations, both of which are
proportional to q=m. )
Problem 63. What is the line charge density on a long wire if a 6 .8 - µ g particle carrying 2.1 nC describes a circular orbit about the wire with speed 280 m/s?
CHAPTER 23 559
Solution The solution to Problem 61 reveals that λ = − mv 2=2kq = −(6.8 × 10 −9 kg)(280 m/s) 2=2 (9 × 10 9 N ⋅ m 2/C 2 )(2 .1 × 10 −9 C) −14.1 µ C/m. (In this case, the force on a positively charged orbiting particle is attractive for a wire with negative linear charge density.)
=
Problem 64. The electron in a hydrogen atom has kinetic energy 2 .18 × 10 −18 J. Assuming the electron is in a circular orbit around the central proton, estimate the size of the atom. (Although this problem gives a reasonable answer, the simple model of an electron orbiting a proton is not consonant with the quantum mechanical description of the atom.)
Solution In a circular orbit, the electron’s acceleration ( −v 2=r) is provided by the Coulomb force of the proton ( F=m Thus, r
=
2
2
ke =mv =
1 2
2
ke =Ek =
1 2
(9 × 10 9 N ⋅ m 2/C 2 )(1.6 × 10 −19 C) 2=(2 .18 × 10 −18 J )
=
=
− ke 2=mr 2).
5.28 × 10 −11 m (essentially a
Bohr radius, see Problem 7).
Problem 65. A dipole with dipole moment 1.5 nC ⋅ m is oriented at 30 ° to a 4.0-MN/C electric field. (a) What is the magnitude of the torque on the dipole? (b) How much work is required to rotate the dipole until it’s antiparallel to the field?
Solution (a) The torque on an electric dipole in an external electric field is given by Equation 23-11; τ
=
p×E
=
pE sin θ =
(1.5 nC ⋅ m)(4.0 MN/C) sin 30° = 3.0 mN ⋅ m. (b) The work done against just the electric force is equal to the change in the dipole’s potential energy (Equation 23-12); W = ∆U = ( −p ⋅ E) f − ( −p ⋅ E )i = pE(cos 30 ° − cos 180 ° ) = (1.5 nC ⋅ m ) × ( 4.0 MN/C)(1866 . ) = 11.2 mJ.
Problem 66. A molecule has its dipole moment aligned with a 1.2-kN/C electric field. If it takes 31 . × 10 −27 J to reverse the molecule’s orientation, what is its dipole moment?
Solution From Equation 23-12, the energy required to reverse the orientation of such a dipole is ∆U 1 2
∆U=E =
1 2
(3.1 × 10
−27
3
J )=(1.2 × 10 N/C)
=
1.29 × 10
−30
=
2 pE , therefore p
=
C ⋅ m.
Problem 67. Two identical dipoles, each of charge q and separation a, are a distance x apart as shown in Fig. 23-49. By c onsidering forces between pairs of charges in the different dipoles, calculate the net force between the dipoles. (a) Show that, in the limit a ¿ x , the force has magnitude 6 kp2=x 4 , where p= qa is the dipole moment. (b) Is the force attractive or repulsive?
FIGURE
23-49 Problem 67 Solution.
560 CHAPTER 23
Solution All the forces are along the same line, so take the origin at the center of the left-hand dipole and the positive x axis in the direction of the right-hand dipole in Fig. 23-49. The right-hand dipole has charges + q at x + a=2, −q at x − a=2 , each of which experiences a force from both charges of the left-hand dipole, which are + q at a=2 and −q at − a=2. (There are forces between four pairs of changes.) The Coulomb force on a charge in the right-hand dipole, due to one in the left-hand one, is kqr ql ( x r − xl ) î = xr − xl
3
(see solution to Problem 15), so the total force on the right-hand dipole is 2
F x = kq î
− M Nx ( x 1
2
1 +
a)
2
−
1 ( x − a)2
+
1 2
x
− P Q =
2 kq 2 a 2 (3 x 2 − a2 ) 2 2 2 2 x (x − a )
î .
(a) In the limit a ¿ x, F x → −2kq 2 a 2 (3 x 2 ) =xî 6 = −6 kq 2 a 2 =xî4 = −6 kp 2 =xî 4 , where p= qa is the dipole moment of both dipoles. (b) The force on the right-hand dipole is in the negative x direction, indicating an attractive force.
Problem 68. A dipole with charges ± q and separation 2 a is located a distance x from a point charge +Q, with its dipole moment vector perpendicular to the x axis, as shown in Fig. 23-50. Find expressions for the magnitude of (a) the net torque and (b) the net force on the dipole, both in the limit x a. (c) What is the direction of the net force?
FIGURE
23-50 Problem 68 Solution.
Solution (a) In the limit x
a, the torque on the dipole is p × E, where E is the field from the point charge Q, at the position of the
dipole. With x axis in the direction from Q to p and y axis parallel to the dipole in Figure 23-50, p
=
2qa$j and E
=
kQî =x 2 .
Then τ = 2 qa$j × kQî=x 2 = −(2 kQqa=x 2 )k$ (i.e., into the page, or CW, to align p with E). (b) and (c) The Coulomb force obeys Newton’s third law. The field of the dipole at the position of Q is (Example 23-6 adapted to new axes) E dip = −(2kqa=x 3 )$j , so the force on Q due to the dipole is QE dip = −(2kQqa=x 3 )$j . The force on the dipole due to Q is the opposite of this,
+( 2 kQqa=x
3 $ )j
(magnitude 2 kQqa=x 3 , parallel to the dipole moment).
Paired Problems Problem 69. An electron is at the origin and an ion with charge
+5e
is at x
=
10 nm. Find a point where the electric field is zero.
Solution The electron’s field is directed toward the electron (a negative charge) and the ion’s field is directed away from the ion (a positive charge). Therefore, the fields can cancel only at points on the negative x axis ( x < 0), since the directions are opposite there and the smaller charge is closer. The field from one point charge is E q ( x)
=
3
kqî ( x − xq )= x − xq , where
CHAPTER 23 561
q = − e, x q = 0 for the electron, and q = 5e, x q = 10 nm for the ion. The total field is zero when 0 = k[( −e) x x
5e( x − 10 nm) x − 10 nm so this implies x
−2
−3
]. (See note to solution of Problem 33.) Since x< 0,
− 5(10 nm − x ) x =
−2
=
[−10 nm −
2
−3
+
x = − x and x − 10 nm = 10 nm − x,
2
0, or 4 x + 2(10 nm ) x − (10 nm ) = 0. The negative solution to this quadratic is
(10 nm )2
+
4(10 nm )2 ]
=
4
−2.5 nm(1 + 5 )
=
−8.09 nm.
Problem =
70. A proton is at the origin and an ion is at x the ion?
5.0 nm. If the electric field is zero at x
=
−6.83 nm, what is the charge on
Solution If the field is zero on the negative x axis (the other side of the proton from the ion), the ion must have a negative charge of greater magnitude than the proton’s. (See the previous solution.) Thus, e(6.83 nm) −2 = − q I (6.83 nm + 5 nm ) −2 , q I =
−(1183 . =6.83) 2 e
=
−3.00 e
=
−3e (since ionic charges are multiples of e).
Problem 71. A thin rod of length
l
has its left end at x
=
−l and its right end at the origin. It carries a line charge density given by λ
=
λ0
x
2
l
2
,
where λ 0 is a constant. Find the electric field at the origin.
Solution The electric field at the origin, due to an element of charge dq d E = kλ î dx=x
2
=
=
λ dx, located at x, where −l ≤ x ≤ 0, is
2
k(λ 0 =l )î dx . The total field is the integral of this over the rod,
E(0)
=
z 0
−l
( k λ 0 =l 2 )î dx
=
(kλ 0 =l 2 )î[0 − ( − l)]
=
( kλ 0 =l )î .
Problem 72. Repeat the preceding problem for the case when λ
=
λ0
x
4
l
4
.
Solution With λ= x 2
=
λ 0 x 2=l 4 in the previous solution, E(0)
=
zGH 0
−l
k λ 0 î l
4
JKx
2
dx =
kλ î 1 G H JK3 [0 − (− ) ] l
0 4
l
3
=
k λ 0
3l
î .
Problem 73. A thin, flexible rod carrying charge Q spread uniformly over its length is bent into a quarter circle of radius a, as shown in Fig. 23-51 a. Find the electric field strength at the point P, which is the center of the circle. Hint: Consult Problem 50.
562 CHAPTER 23
Solution It should be clear from the symmetry that the electric field is along the radius bisecting the arc, so take this as the x-axis, with P at the origin, and θ as defined in Fig. 23-45. The electric field at P, from each charge element, dq = λ d l =
λa d θ , at θ , has the same magnitude, dE = k dq=a 2 = kλ dθ=a, but direction r$ = î sinθ − $j cos θ , as sketched. λ = Q=l is constant, and the arc extends from θ 0 = 45° to π − θ 0 = 135° , so the total field at P is the integral of dE r$ from θ 0 to π − θ 0: E(0)
=
(kλ=a)
z
π −θ 0
( î sin θ − $j cos θ ) dθ
=
( kλ=a) − î cos θ − $j sin θ
θ0
=
Here, we used sin θ 0 Q=( 12
π a), so E(0)
=
=
(kλ=a)2 î cos θ 0
sin(π − θ 0 ), cos θ 0
=
2
=
θ0
2 kλ î =a.
− cos(π − θ 0 ), and θ 0
2 2 kQ î =πa . [Note: in general,
FIGURE
π −θ 0
l =
=
45°. In terms of the total charge, λ
=
Q=l =
(π − 2θ 0 )a.]
23-51 Problems 73 and 74 Solution.
Problem 74. A thin, flexible rod carrying charge Q spread uniformly over its length is bent into a circular arc of radius a, as shown in Fig. 23-50 b. Find the electric field strength at the point P, which is the center of the circular arc.
Solution This problem is the same as the previous one, except θ 0
=
60°. Thus, E(0)
=
(k=a)(Q=13 πa)2 î cos 60°
=
3kQ î =πa 2.
Problem 75. Ink-jet printers work by deflecting moving ink droplets with an electric field so they hit the right place on the paper. Droplets in a particular printer have mass 11 . × 10 −10 kg, charge 2.1 pC, speed 12 m/s, a nd pass through a uniform 97-kN/C electric field in order to be deflected through a 10 ° angle. What is the length of the field region?
Solution Suppose the ink droplets enter the field region perpendicular to the field, as in the geometry of Example 23-10. Then the 2 2 v x2 tan θ= qE analysis of that example shows that v =v m y x = tan θ = qE y∆x=mv x, so ∆ x= y = ( 0.11 µ g)(12 m/s) × tan 10°=(2.1 pC)(97 kN/C) = 1.37 cm.
Problem 76. If the drop speed in the printer of Problem 75 is doubled, what should be done to the electric field to have the drops hit the same point on the paper?
CHAPTER 23 563
Solution Since tan θ » E y=v x2 (for droplets of given q=m and fixed printer field region), droplets will hit the same point if E y′ when v x′
=
=
4 E y
2v x .
Supplementary Problems Problem 77. A spring of spring constant 100 N/m is stretched 10 cm beyond its 90-cm equilibrium length. If you want to keep it stretched by attaching equal electric charges to the opposite ends, what magnitude of charge should you use?
Solution The repulsive force between like charges, kq 2=r 2 ( r = 90 cm 10 cm is the stretch and k s is the spring constant). Thus, 2
+ 10
cm
=
1 m), must balance the spring force, ks x ( x
2
9
2
2
1=2
q = ± r ks x=k = ±[(100 N/m)(1 m ) (0.1 m)=(9 × 10 N ⋅ m / C )]
=
= ±33.3 µ C.
Problem 78. Two small spheres with the same mass m and charge q are suspended from massless strings of length l, as shown in Fig. 23-52. Each string makes an angle θ with the vertical. Show that the charge on each sphere is q = ±2l
sin θ mg tan θ=k .
Solution The magnitudes of the forces on either sphere (drawn acting on the righthand one in Fig. 23-52) are Fgrav 2 2 kq =(2l sin θ ) , and T (the unknown string tension). In equilibrium, ∑ F x
0
=
=
∑ F y
2
=
0, or 0
=
=
mg, Felec =
Felec − T sin θ and
2
T cos θ − mg. Eliminating T , we obtain tan θ = Felec=mg = kq =mg(2l sin θ ) , or q = ±2 l sin θ mg tan θ=k .
FIGURE
23-52 Problem 78 Solution.
Problem 79. A charge −q and a charge
4 9
q are located a distance a apart, as shown in Fig. 23-53. Where would you place a third
charge so that all three are in static equilibrium? What should be the sign and magnitude of the third charge?
564 CHAPTER 23
Solution Because of the vector nature of the forces, the third charge, Q, must be placed along the line joining the other two, as in Example 23-3. Q cannot go to the left of −q , since the magnitude of the force on it due to −q would always be greater than that due to
4 9
q . It cannot go between −q and
must go to the right of for Q: 0
=
c q =x
kQ
4 9
2
4 9
+
4 9
q , since the forces on it would always be in the same direction. Thus, Q
q , as shown ( x-axis to the right with origin at 2
kQ( −q )=( x + a) , or
4 9
( x + a) 2
=
2
x . For
4 9
4 9
q ). The net force on each charge must be zero, so,
q: 0 = k
c q (−q)=a 4 9
2
−k
c q Q=x , or −qx 4 9
2
2
=
2
Qa .
(The equation for the force on the third charge follows from the equations for the other two plus Newton’s third law.) The solution of these two equations (for x > 0) is x = 2 a and Q = −4 q. (The equilibrium is unstable. A slight displacement of the positive charge to the right, for example, would cause it to be attracted more strongly to the right.)
FIGURE
23-53 Problem 79 Solution.
Problem 80. Two 34-µC charges are attached to the opposite ends of a spring of spring constant 150 N/m and equilibrium length 50 cm. By how much does the spring stretch?
Solution Suppose that the Coulomb repulsion is the only force stretching the spring. When balanced with the spring force, 2 2 2 9 2 2 2 3 −2 kq =(l 0 + x ) = ks x, or x (0.5 m + x ) = (9 × 10 N ⋅ m /C )(34 µC) =(150 N/m) = 6.94 × 10 m . This cubic equation can be solved by iteration or by Newton’s method to yield x = 1595 . . . . cm.
Problem 81. A 3.8-g particle with a 4.0- µC charge experiences a downward force of 0.24 N in a uniform electric field. Find the electric field, assuming that the gravitational force is not negligible.
Solution If gravity and Coulomb forces both act, then Fnet ( mg − 0.24 N)$j=q
=
=
− mg$j
+
qE = −(0.24 N) $j , where $j is upward. Thus, E =
[(3.8 × 10 −3 kg)(9.8 m/s 2 ) − 0 .24 N]$j=(4 .0 µC)
=
−50 .7 $j kN/C.
Problem 82. A rod of length 2 l lies on the x-axis, centered on the origin. It carries a line charge density given by λ = λ 0 ( x=l ), where λ 0 is a constant. (a) What is the net charge on the rod? (b) Find an expression for electric field strength at all points x
> l.
(c) Show that your result has the 1= x 3 dependence of dipole field when x 3
3
l. Hint:
For
l
¿ x,
ln( x − l)=( x + l) becomes approximately −2l= x − 2l =3 x . (d) By comparing with Equation 23-7b, determine the dipole moment of the rod.
CHAPTER 23 565
Solution (a) Q
=
z λ dx z (λ = )x dx l
−l
=
l
(λ 0 =l)
=
0 l
−l
2 dE = k dq î =( x − x ′) at position x >
l
=
z
l
2
= −l
=
0. (b) An element of charge, dq
λ dx ′, produces an electric field
z
λ k λ 1 x ′ dx ′ x − î î J dx ′ M k J G G H K( x − x ′) H K N( x − x ′) ( x − x ′) P k λ 2x 1 x x x − x − − λ î M î ln ln k G J H x KP H x KP. H KN x − x x − Q M N 0
0
l
0
= 2
l
−l
=
x
on the x-axis, where we are using x ′ for the variable of integration. Thus,
l
E( x )
1 2
l
l
+
l
+l
l
2
−l
=
0
2
+ 2
. . . ,
+ l
l
+ 2
l
+ l
l
(c) For l= x ¿ 1, 2 x 2
x −
l
F1 − I G J xH x K
2
= 2
l
−1
2
=
2
F1 G xH
2
+
l
x
I JK
2
while 1 l
1 1 F x − I 1 L F I − lnF1 + IO − − ln ln 1 − M M P Hx K NH K H K x xQ M N x 2x l
l
=
+ l
l
l
=
l
l
2
l
=
−
2 x
−
2 l2 3x 3
+
2
−
1
l
3
3 x3
. . .−
+
G Hx − 2 x l
1
l
2 2
+
1
l
3
3 x3
+
JKP P
. . .
. . .
Thus, E( x ) = k λ 0 î
M x N 2
+
2l2 x
3
+
(d) On the axis of a point dipole, Equation 23-7b gives E( x) Explicit use of the formula p=
z l
−l
x dq =
zλ
=
x
−
2l 2 3x 3
+
P Q
. . . '
4 k λ 0 l 2 î 3 x 3
2 kp î = x3, so comparison reveals that p
=
2λ 0 l 2=3. (Note:
2
l
−l
2
. . .−
0
x dx=l gives the same result.)
Problem 83. The electric field on the axis of a uniformly charged ring has magnitude 380 kN/C at a point 5.0 cm from the ring center. The magnitude 15 cm from the center is 160 kN/C; in both cases the field points away from the ring. Find the radius and charge of the ring.
Solution The electric field on the axis of a uniformly charged ring is calculated in Example 23-8, so the data given in the question −3=2 −3=2 imply 380 kN/C = kQ(5 cm )[(5 cm )2 + a 2 ] , and 160 kN/C = kQ(15 cm )[(15 cm )2 + a 2 ] . Dividing these two equations and taking the
2 3
root we get
F380 × 15 I H160 × 5 K
2=3
(15 cm) 2
+
a
(5 cm )2
+
a
[(15 cm) 2 − (3.70 )(5 cm)2 ]=2.70
=
7.00 cm.
=
3.70
=
2
2
,
which when solved for the radius, gives a=
566 CHAPTER 23
Substituting for a in either of the field equations allows us to find (380 kN/C)[(5 cm )2
Q =
+
3=2
(7 cm) 2 ]
=
(9 × 10 9 N ⋅ m2/C 3 )(5 cm)
538 nC.
Problem 84. Use the binomial theorem to show that, for x R, the result of Problem 48 reduces to the field of a point charge whose total charge is the charge density σ times the disk area.
Solution The result of Problem 48 for the field on the axis of a uniformly charged disk, of radius R, at a distance x axis (away from the disk’s center) was E x
=
2
2π kσ (1 − x= x
+
2
2 −1=2
2
R ) = 2π kσ [1 − (1 + R = x )
>
0 along the
]. The binomial
expansion in Appendix A, for R 2=x 2 ¿ 1, gives
M H M NG
E x = 2π k σ 1 − 1 −
which is the field from a point charge Q
=
1 R2 2 x
+ 2
JKP P Q
. . .
'
2π kσ R2 2 x
=
2
kQ x
2
,
π R 2σ at a distance x.
Problem 85. A molecule with dipole moment p is located a distance r from a proton, oriented with its dipole moment vector p as shown in Fig. 23-54. (a) Use Equation 23-7b to find the force the molecule exerts on the proton. (b) Now find the net force on the molecule in the proton’s nonuniform electric field by considering that the molecule consists of two opposite charges ± q, separated by a distance d such that qd = p. Take the limit as d becomes very small compared with r, and show that the resulting force has the same magnitude as that of part (a), as required by Newton’s third law.
Solution Take the origin at the molecule, with the x-axis parallel to its dipole moment, as shown on Fig. 23-54. (a) Equation 23-7b gives the dipole’s electric field at the position of the proton, so the force on the latter is F p = eE dip = e(2 kpr −3 î ) = 2kepr −3 î . (b) The electric field of the proton, for points on the x-axis to its left ( x Equation 23-4). If the molecular dipole is considered to consist of charges the force on it is Fdip
=
qE p ( d=2) − qE p ( − d=2 ) = − keq [ r î−(d=2 )
−2
− (r
±q +
<
−2 r) is E p ( x ) = − ke(r − x) î (see
located at x
d=2)
−2
]
=
= ± d =2,
where p= qd, then
−2 kepr( r − d 2=4 )−2 . Inî the limit 2
−3 d ¿ r, this becomes Fdip = −2 kepr î = − F p , in agreement with the result of part (a) and Newton’s third law.
FIGURE
23-54 Problem 85.
Problem 86. An electric quadrupole consists of two oppositely directed dipoles in close proximity. (a) Calculate the field of the quadrupole shown in Fig. 23-55 for points to the right of x = a, and (b) show that for x a the quadrupole field falls off as 1= x 4.
CHAPTER 23 567
FIGURE
23-55 Problem 86.
Solution (a) The electric field, from the three point charges shown, at points on the x-axis with x E( x ) (b) For x À
=
k î
M N( x − a) q
2
−
2q x
2
q
+
( x
+
a)
2
P Q =
2 kqa2 î
>
a is:
(3x 2 − a 2 ) 2 2 2 x ( x − a)
2 4 2 a, E( x) → 6 kqa î = x . (The quadrupole moment of this “linear quadrupole” is Q xx = 4qa .)
Problem 87. Derive Equation 23-9 in Example 23-9 by making θ the integration variable, then evaluating the resulting integral.
Solution In Fig. 23-23, r
=
(x 2
+
2 1=2
y )
=
2
y=sin θ , and cot θ = x=y. Differentiating the latter with respect to x, we get −csc θ d θ =
−1 dx=y , which relates dx to d θ . (Extend the results in Appendix B, and recall that cosecant = sine .) The limits x = −∞ to ∞ correspond to θ = π to 0 (since when x is negative so is cot θ ), thus the integral in Example 23-9 le ading to
Equation 23-9 becomes E (kλ=y )[−cos π
+
cos 0]
=
=
z ∞
kλ y
dx
( x 2 + y 2 ) 2 kλ=y as before. −∞
3=2
=
z 0
kλy
π
− y csc 2θ d θ ( y=sin θ )3
=
k λ y
z π
sin θ d θ
0
=
