D E F I N I T I O N S The electric field at some point in space is defined as the electric force that acts on a small positive test charge placed at that point divided by the magnitude q
Trang 1664 Chapter 23 Electric Fields
Substitute numerical values:
We have neglected the gravitational force acting on the electron, which represents a good approximation when
deal-ing with atomic particles For an electric field of 200 N/C, the ratio of the magnitude of the electric force eE to the magnitude of the gravitational force mg is on the order of 1012for an electron and on the order of 109for a proton.
Summary
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D E F I N I T I O N S
The electric field at some point in space is defined as the electric force that acts on a small positive test charge
placed at that point divided by the magnitude q0of the test charge:
Electric charges have the following important properties:
■ Charges of opposite sign attract one another, and
charges of the same sign repel one another.
■ The total charge in an isolated system is conserved.
■ Charge is quantized.
Conductors are materials in which electrons
move freely Insulators are materials in which
electrons do not move freely.
Coulomb’s law states that the electric force exerted by a point
charge q1on a second point charge q2is
(23.6)
where r is the distance between the two charges and is a unit
vector directed from q1toward q2 The constant ke, which is called
the Coulomb constant, has the value ke 8.99 109N m2/C2.
The electric force on a charge q placed in an electric field is
(23.8) F
At a distance r from a point charge q, the
electric field due to the charge is
(23.9)
where is a unit vector directed from the charge toward the point in question The electric field is directed radially outward from a positive charge and radially inward toward a negative charge.
r^
ES keq
r2 r
^
The electric field due to a group of point
charges can be obtained by using the
superpo-sition principle That is, the total electric field
at some point equals the vector sum of the
electric fields of all the charges:
Trang 2Questions 665
Questions
denotes answer available in Student Solutions Manual/Study Guide; O denotes objective question
1. Explain what is meant by the term “a neutral atom.”
Explain what “a negatively charged atom” means
Does its mass (a) increase measurably, (b) increase by an
amount too small to measure directly, (c) remain
unchanged, (d) decrease by an amount too small to
mea-sure directly, or (e) decrease measurably? (ii) Now the
coin is given a negative electric charge What happens to
its mass? Choose from the same possibilities
3. A student who grew up in a tropical country and is
study-ing in the United States may have no experience with
static electricity sparks and shocks until their first
Ameri-can winter Explain
4. Explain the similarities and differences between Newton’s
law of universal gravitation and Coulomb’s law
5. A balloon is negatively charged by rubbing, and then it
clings to a wall Does that mean the wall is positively
charged? Why does the balloon eventually fall?
6 OIn Figure 23.8, assume the objects with charges q2and
q3 are fixed Notice that there is no sightline from the
location of object 2 to the location of object 1 We could
say that a bug standing on q1is unable to see q2because it
is behind q3 How would you calculate the force exerted
on the object with charge q1? (a) Find only the force
exerted by q2 on charge q1 (b) Find only the force
exerted by q3 on charge q1 (c) Add the force that q2
would exert by itself on charge q1 to the force that q3
would exert by itself on charge q1 (d) Add the force that
q3would exert by itself to a certain fraction of the force
that q2would exert by itself (e) There is no definite way
to find the force on charge q1
7 OA charged particle is at the origin of coordinates The
particle produces an electric field of kN/C at the
point with position vector cm (i) At what location
does the field have the value kN/C? (a) cm
(b) cm (c) cm (d) cm (e) nowhere (ii) At
what location is the value kN/C? Choose from the
same possibilities
8. Is it possible for an electric field to exist in empty space?
Explain Consider point A in Figure 23.21a Does charge
exist at this point? Does a force exist at this point? Does a
field exist at this point?
A exerts on charged particle B, located at distance r away
from A, from the largest to the smallest in the following
cases In your ranking, note any cases of equality (a) qA
20 nC, qB 20 nC, r 2 cm (b) qA 30 nC, qB 10 nC,
r 2 cm (c) qA 10 nC, qB 30 nC, r 2 cm (d) qA
30 nC, qB 20 nC, r 3 cm (e) qA 45 nC, qB 20 nC,
r 3 cm (ii) Rank the magnitudes of the electric fields
charged particle A creates at the location of charged
par-ticle B, a distance r away from A, from the largest to the
smallest in the same cases In your ranking, note any cases
of equality
square as shown in Figure Q23.10, with charge Q on
both the particle at the upper left corner and the particle
at the lower right corner, and charge 2Q on the particle
at the lower left corner (i) What is the direction of the
electric field at the upper right corner, which is a point inempty space? (a) It is upward and to the right (b) It isstraight to the right (c) It is straight downward (d) It isdownward and to the left (e) It is perpendicular to theplane of the picture and outward (f) There is no direc-tion; no field exists at that corner because no charge isthere (g) There is no direction; the total field there is
zero (ii) Suppose the 2Q charge at the lower left corner
is removed Then does the magnitude of the field at theupper right corner (a) become larger, (b) becomesmaller, (c) stay the same, or (d) change unpredictably?
(a)(b)(c)(d)
–Q
Figure Q23.10
Uni-verse, 8 cm apart The charge of A is 40 nC The net tric field at one certain point 4 cm from A is zero Whatcan you conclude about the charge of B? Choose everycorrect answer (a) It can be 40 nC (b) It can be 120 nC.(c) It can be 360 nC (d) It can be 40 nC (e) It can be
elec-120 nC (f) It can be 360 nC (g) It can have any of
an infinite number of values (h) It can have any of eral values (i) It must have one of three values (j) Itmust have one of two values (k) It must have one certain
sev-value (l) No possible value for qBexists; the situation isimpossible
12. Explain why electric field lines never cross Suggestion:
Begin by explaining why the electric field at a particularpoint must have only one direction
13. Figures 23.12 and 23.13 show three electric field vectors
at the same point With a little extrapolation, Figure 23.19would show many electric field lines at the same point Is
it really true that “no two field lines can cross”? Are thediagrams drawn correctly? Explain your answers
uniformly distributed around it What is the magnitude of
the electric field at the center of the ring? (a) 0 (b) k e q/b2
(c) k e q2/b2(d) k e q2/b (c) none of these answers
charge Q produces an electric field Eringat a point P on its axis, at distance x away from the center of the ring Now the charge Q is spread uniformly over the circular area
the ring encloses, forming a flat disk of charge with the
same radius How does the field Ediskproduced by the disk
at P compare to the field produced by the ring at the same point? (a) Edisk Ering (b) Edisk Ering (c) Edisk
Ering(d) impossible to determine
Trang 316 OA free electron and a free proton are released in
iden-tical electric fields (i) How do the magnitudes of the
elec-tric force exerted on the two particles compare? (a) It is
millions of times greater for the electron (b) It is
thou-sands of times greater for the electron (c) They are
equal (d) It is thousands of times smaller for the
elec-tron (e) It is millions of times smaller for the elecelec-tron
(f) It is zero for the electron (g) It is zero for the proton
(ii) Compare the magnitudes of their accelerations
Choose from the same possibilities
space where the electric field is directed vertically
upward What is the direction of the electric force exerted
on this charge? (a) It is up (b) It is down (c) There is no
force (d) The force can be in any direction
666 Chapter 23 Electric Fields
18. Explain the differences between linear, surface, and ume charge densities and give examples of when eachwould be used
vol-19. Would life be different if the electron were positivelycharged and the proton were negatively charged? Doesthe choice of signs have any bearing on physical andchemical interactions? Explain
20. Consider two electric dipoles in empty space Each dipolehas zero net charge Does an electric force exist betweenthe dipoles; that is, can two objects with zero net chargeexert electric forces on each other? If so, is the force one
of attraction or of repulsion?
2= intermediate; 3= challenging; = SSM/SG; = ThomsonNOW; = symbolic reasoning; = qualitative reasoning
Problems
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Section 23.1 Properties of Electric Charges
1. (a) Find to three significant digits the charge and the
mass of an ionized hydrogen atom, represented as H
Suggestion: Begin by looking up the mass of a neutral atom
on the periodic table of the elements in Appendix C
(b) Find the charge and the mass of Na, a singly ionized
sodium atom (c) Find the charge and the average mass
of a chloride ion Clthat joins with the Nato make one
molecule of table salt (d) Find the charge and the mass
of Ca Ca2, a doubly ionized calcium atom (e) You
can model the center of an ammonia molecule as an N3
ion Find its charge and mass (f) The plasma in a hot star
contains quadruply ionized nitrogen atoms, N4 Find
their charge and mass (g) Find the charge and the mass
of the nucleus of a nitrogen atom (h) Find the charge
and the mass of the molecular ion H2O
2. (a) Calculate the number of electrons in a small,
electri-cally neutral silver pin that has a mass of 10.0 g Silver has
47 electrons per atom, and its molar mass is 107.87 g/mol
(b) Imagine adding electrons to the pin until the negative
charge has the very large value 1.00 mC How many
elec-trons are added for every 109electrons already present?
Section 23.2 Charging Objects by Induction
Section 23.3 Coulomb’s Law
3. Nobel laureate Richard Feynman (1918–1988) once
said that if two persons stood at arm’s length from each
other and each person had 1% more electrons than
pro-tons, the force of repulsion between them would be
enough to lift a “weight” equal to that of the entire Earth
Carry out an order-of-magnitude calculation to
substanti-ate this assertion
4. A charged particle A exerts a force of 2.62 mN to the right
on charged particle B when the particles are 13.7 mm
apart Particle B moves straight away from A to make the
distance between them 17.7 mm What vector force does
it then exert on A?
5. (a) Two protons in a molecule are 3.80 1010 mapart Find the electrical force exerted by one proton onthe other (b) State how the magnitude of this force com-pares with the magnitude of the gravitational force
exerted by one proton on the other (c) What If? What
must be a particle’s charge-to-mass ratio if the magnitude
of the gravitational force between two of these particles isequal to the magnitude of electrical force between them?
6. Two small silver spheres, each with a mass of 10.0 g, areseparated by 1.00 m Calculate the fraction of the elec-trons in one sphere that must be transferred to the other
to produce an attractive force of 1.00 104 N (about
1 ton) between the spheres (The number of electronsper atom of silver is 47, and the number of atoms pergram is Avogadro’s number divided by the molar mass ofsilver, 107.87 g/mol.)
7. Three charged particles are located at the corners of anequilateral triangle as shown in Figure P23.7 Calculatethe total electric force on the 7.00-mC charge
0.500 m7.00mC
Trang 48. Two small beads having positive charges 3q and q are
fixed at the opposite ends of a horizontal insulating rod,
extending from the origin to the point x d As shown in
Figure P23.8, a third small charged bead is free to slide
on the rod At what position is the third bead in
equilib-rium? Explain whether it can be in stable equilibrium
Problems 667
13. What are the magnitude and direction of the electric fieldthat balances the weight of (a) an electron and (b) a pro-ton? You may use the data in Table 23.1
14. Three charged particles are at the corners of an eral triangle as shown in Figure P23.7 (a) Calculate theelectric field at the position of the 2.00-mC charge due tothe 7.00-mC and 4.00-mC charges (b) Use your answer
equilat-to part (a) equilat-to determine the force on the 2.00-mC charge
15. Two charged particles are located on the x axis The
first is a charge Q at x a The second is an unknown
charge located at x 3a The net electric field these
charges produce at the origin has a magnitude of
2k e Q /a2 Explain how many values are possible for theunknown charge and find the possible values
16. Two 2.00-mC charged particles are located on the x axis One is at x 1.00 m, and the other is at x 1.00 m (a) Determine the electric field on the y axis at y 0.500 m (b) Calculate the electric force on a 3.00-mC
charge placed on the y axis at y 0.500 m
17. Four charged particles are at the corners of a square of
side a as shown in Figure P23.17 (a) Determine the
mag-nitude and direction of the electric field at the location of
charge q (b) What is the total electric force exerted on q?
2= intermediate; 3= challenging; = SSM/SG; = ThomsonNOW; = symbolic reasoning; = qualitative reasoning
d
Figure P23.8
9. Two identical conducting small spheres are placed with
their centers 0.300 m apart One is given a charge of
12.0 nC and the other a charge of 18.0 nC (a) Find the
electric force exerted by one sphere on the other
(b) What If? The spheres are connected by a conducting
wire Find the electric force each exerts on the other after
they have come to equilibrium
charge q, are fixed in space and separated by a distance
d A third particle with charge Q is free to move and lies
initially at rest on the perpendicular bisector of the two
fixed charges a distance x from the midpoint between the
two fixed charges (Fig P23.10) (a) Show that if x is small
compared with d, the motion of Q is simple harmonic
along the perpendicular bisector Determine the period
of that motion (b) How fast will the charge Q be
mov-ing when it is at the midpoint between the two fixed
charges if initially it is released at a distance a V d from
Figure P23.10
atom, an electron moves in a circular orbit about a
pro-ton, where the radius of the orbit is 0.529 1010 m
(a) Find the electric force exerted on each particle (b) If
this force causes the centripetal acceleration of the
elec-tron, what is the speed of the electron?
Section 23.4 The Electric Field
12. In Figure P23.12, determine the point (other than
infin-ity) at which the electric field is zero
18. Consider the electric dipole shown in Figure P23.18
Show that the electric field at a distant point on the x axis is E x 4k e qa/x3
2a
x
– q
q y
Figure P23.18
19. Consider n equal positive charged particles each of magnitude Q/n placed symmetrically around a circle of radius R (a) Calculate the magnitude of the electric field
at a point a distance x from the center of the circle and
on the line passing through the center and perpendicular
to the plane of the circle (b) Explain why this result isidentical to the result of the calculation done in Example23.7
Section 23.5 Electric Field of a Continuous Charge Distribution
20. A continuous line of charge lies along the x axis, ing from x x0 to positive infinity The line carriescharge with a uniform linear charge density l0 What are
Trang 5extend-the magnitude and direction of extend-the electric field at extend-the
origin?
21. A rod 14.0 cm long is uniformly charged and has a total
charge of 22.0 mC Determine the magnitude and
direc-tion of the electric field along the axis of the rod at a
point 36.0 cm from its center
22. Show that the maximum magnitude Emaxof the electric
field along the axis of a uniformly charged ring occurs
at x (see Fig 23.16) and has the value
23. A uniformly charged ring of radius 10.0 cm has a total
charge of 75.0 mC Find the electric field on the axis of
the ring at (a) 1.00 cm, (b) 5.00 cm, (c) 30.0 cm, and
(d) 100 cm from the center of the ring
24. A uniformly charged disk of radius 35.0 cm carries charge
with a density of 7.90 103C/m2 Calculate the electric
field on the axis of the disk at (a) 5.00 cm, (b) 10.0 cm,
(c) 50.0 cm, and (d) 200 cm from the center of the disk
25. Example 23.8 derives the exact expression for the
elec-tric field at a point on the axis of a uniformly charged
disk Consider a disk of radius R 3.00 cm having a
uni-formly distributed charge of 5.20 mC (a) Using the
result of Example 23.8, compute the electric field at a
point on the axis and 3.00 mm from the center What If?
Explain how this answer compares with the field
com-puted from the near-field approximation E s/2P0
(b) Using the result of Example 23.8, compute the
elec-tric field at a point on the axis and 30.0 cm from the
cen-ter of the disk What If? Explain how this answer
com-pares with the electric field obtained by treating the disk
as a 5.20-mC charged particle at a distance of 30.0 cm
26. The electric field along the axis of a uniformly charged
disk of radius R and total charge Q was calculated in
Example 23.8 Show that the electric field at distances x
that are large compared with R approaches that of a
parti-cle with charge Q spR2 Suggestion: First show that
x/(x2 R2)1/2 (1 R2/x2)1/2and use the binomial
expansion (1 d)n 1 nd when d V 1.
27. A uniformly charged insulating rod of length 14.0 cm is
bent into the shape of a semicircle as shown in Figure
P23.27 The rod has a total charge of 7.50 mC Find the
magnitude and direction of the electric field at O, the
center of the semicircle
Q> 1613pP0a> 12a22
668 Chapter 23 Electric Fields
through its volume Use the result of Example 23.8 tofind the field it creates at the same point
2= intermediate; 3= challenging; = SSM/SG; = ThomsonNOW; = symbolic reasoning; = qualitative reasoning
O
Figure P23.27
28. (a) Consider a uniformly charged thin-walled right
circu-lar cylindrical shell having total charge Q , radius R, and
height h Determine the electric field at a point a distance
d from the right side of the cylinder as shown in Figure
P23.28 Suggestion: Use the result of Example 23.7 and
treat the cylinder as a collection of ring charges (b) What
If? Consider now a solid cylinder with the same
dimen-sions and carrying the same charge, uniformly distributed
R d
dx h
Figure P23.28
29. A thin rod of length and uniform charge per unit
length l lies along the x axis as shown in Figure P23.29 (a) Show that the electric field at P, a distance y from the rod along its perpendicular bisector, has no x component and is given by E 2k el sin u0/y (b) What If? Using your
result to part (a), show that the field of a rod of infinite
length is E 2k e l/y Suggestion: First calculate the field at
P due to an element of length dx, which has a charge l dx.
Then change variables from x to u, using the relationships
x y tan u and dx y sec2u du, and integrate over u.
u
y y
31. Eight solid plastic cubes, each 3.00 cm on each edge, areglued together to form each one of the objects (i, ii, iii,and iv) shown in Figure P23.31 (a) Assuming each objectcarries charge with uniform density 400 nC/m3through-out its volume, find the charge of each object (b) Assum-ing each object carries charge with uniform density15.0 nC/m2everywhere on its exposed surface, find thecharge on each object (c) Assuming charge is placed
Figure P23.31
Trang 6only on the edges where perpendicular surfaces meet,
with uniform density 80.0 pC/m, find the charge of each
object
Section 23.6 Electric Field Lines
32. A positively charged disk has a uniform charge per unit
area as described in Example 23.8 Sketch the electric
field lines in a plane perpendicular to the plane of the
disk passing through its center
33. A negatively charged rod of finite length carries charge
with a uniform charge per unit length Sketch the electric
field lines in a plane containing the rod
34. Figure P23.34 shows the electric field lines for two
charged particles separated by a small distance (a)
Deter-mine the ratio q1/q2 (b) What are the signs of q1and q2?
Problems 669
38. Two horizontal metal plates, each 100 mm square, arealigned 10.0 mm apart, with one above the other Theyare given equal-magnitude charges of opposite sign sothat a uniform downward electric field of 2 000 N/Cexists in the region between them A particle of mass 2.00 1016kg and with a positive charge of 1.00 106Cleaves the center of the bottom negative plate with an ini-tial speed of 1.00 105 m/s at an angle of 37.0° abovethe horizontal Describe the trajectory of the particle.Which plate does it strike? Where does it strike, relative toits starting point?
39. The electrons in a particle beam each have a kinetic
energy K What are the magnitude and direction of the electric field that will stop these electrons in a distance d ?
40. Protons are projected with initial speed v i 9.55 km/sinto a region where a uniform electric field
N/C is present as shown in Figure P23.40.The protons are to hit a target that lies at a horizontal dis-tance of 1.27 mm from the point where the protons crossthe plane and enter the electric field Find (a) the twoprojection angles u that will result in a hit and (b) thetime of flight (the time interval during which the proton
is above the plane in Fig P23.40) for each trajectory
35. Three equal positive charges q are at the corners of an
equilateral triangle of side a as shown in Figure P23.35.
(a) Assume the three charges together create an electric
field Sketch the field lines in the plane of the charges
Find the location of one point (other than ) where the
electric field is zero (b) What are the magnitude and
direction of the electric field at P due to the two charges
at the base?
a a
q
P +
Figure P23.35 Problems 35 and 58
Section 23.7 Motion of Charged Particles in a Uniform
Electric Field
36. A proton is projected in the positive x direction into a
region of a uniform electric field N/C
at t 0 The proton travels 7.00 cm as it comes to rest
Determine (a) the acceleration of the proton, (b) its
ini-tial speed, and (c) the time interval over which the
pro-ton comes to rest
37. A proton accelerates from rest in a uniform electric field
of 640 N/C At one later moment, its speed is 1.20 Mm/s
(nonrelativistic because v is much less than the speed of
light) (a) Find the acceleration of the proton (b) Over
what time interval does the proton reach this speed?
(c) How far does it move in this time interval? (d) What is
its kinetic energy at the end of this interval?
E = (–720 j) N/C
Protonbeam
u
Figure P23.40
41. A proton moves at 4.50 105m/s in the horizontal tion It enters a uniform vertical electric field with a mag-nitude of 9.60 103 N/C Ignoring any gravitationaleffects, find (a) the time interval required for the proton
direc-to travel 5.00 cm horizontally, (b) its vertical displacementduring the time interval in which it travels 5.00 cm hori-zontally, and (c) the horizontal and vertical components
of its velocity after it has traveled 5.00 cm horizontally
Additional Problems
42. Two known charges, 12.0 mC and 45.0 mC, and a
third unknown charge are located on the x axis The
charge 12.0 mC is at the origin, and the charge 45.0 mC
is at x 15.0 cm The unknown charge is to be placed sothat each charge is in equilibrium under the action of theelectric forces exerted by the other two charges Is this sit-uation possible? Is it possible in more than one way?Explain Find the required location, magnitude, and sign
of the unknown charge
43. A uniform electric field of magnitude 640 N/C existsbetween two parallel plates that are 4.00 cm apart A pro-ton is released from the positive plate at the same instant
an electron is released from the negative plate (a) mine the distance from the positive plate at which the two
Trang 7Deter-pass each other (Ignore the electrical attraction between
the proton and electron.) (b) What If? Repeat part (a)
for a sodium ion (Na) and a chloride ion (Cl)
44. Three charged particles are aligned along the x axis as
shown in Figure P23.44 Find the electric field at (a) the
position (2.00, 0) and (b) the position (0, 2.00)
670 Chapter 23 Electric Fields
effects of the gravitational and buoyant forces on it, eachballoon can be modeled as a particle of mass 2.00 g, withits center 50.0 cm from the point of support To show offthe colors of the balloons, Inez rubs the whole surface ofeach balloon with her woolen scarf, making them hangseparately with gaps between them The centers of thehanging balloons form a horizontal equilateral trianglewith sides 30.0 cm long What is the common charge eachballoon carries?
2= intermediate; 3= challenging; = SSM/SG; = ThomsonNOW; = symbolic reasoning; = qualitative reasoning
0.800 m
y
3.00 nC5.00 nC
0.500 m
– 4.00 nC
x
Figure P23.44
45. A charged cork ball of mass 1.00 g is suspended on a light
string in the presence of a uniform electric field as shown in
the ball is in equilibrium at u 37.0° Find (a) the charge
on the ball and (b) the tension in the string
E
S
13.00 iˆ 5.00 jˆ2 105
x y
E
q
u
Figure P23.45 Problems 45 and 46
46. A charged cork ball of mass m is suspended on a light
string in the presence of a uniform electric field as shown
and B are positive numbers, the ball is in equilibrium at
the angle u Find (a) the charge on the ball and (b) the
tension in the string
47. Four identical charged particles (q 10.0 mC) are
located on the corners of a rectangle as shown in Figure
P23.47 The dimensions of the rectangle are L 60.0 cm
and W 15.0 cm Calculate the magnitude and direction
of the total electric force exerted on the charge at the
lower left corner by the other three charges
y
x L
W
Figure P23.47
48. Inez is putting up decorations for her sister’s quinceañera
(fifteenth birthday party) She ties three light silk ribbons
together to the top of a gateway and hangs a rubber
bal-loon from each ribbon (Fig P23.48) To include the
Figure P23.48
a frictionless horizontal surface are connected by a light
metallic spring having a spring constant k and an unstretched length L ias shown in Figure P23.49a A total
charge Q is slowly placed on the system, causing the spring to stretch to an equilibrium length L as shown in Figure P23.49b Determine the value of Q , assuming all
the charge resides on the blocks and modeling the blocks
50. Consider a regular polygon with 29 sides The distance
from the center to each vertex is a Identical charges q are placed at 28 vertices of the polygon A single charge Q is
placed at the center of the polygon What is the tude and direction of the force experienced by the
magni-charge Q ? Suggestion: You may use the result of Problem
60 in Chapter 3
51. Identical thin rods of length 2a carry equal charges Q
uniformly distributed along their lengths The rods lie
along the x axis with their centers separated by a distance
Trang 852. Two small spheres hang in equilibrium at the bottom
ends of threads, 40.0 cm long, that have their top ends
tied to the same fixed point One sphere has mass 2.40 g
and charge 300 nC The other sphere has the same
mass and a charge of 200 nC Find the distance between
the centers of the spheres You will need to solve an
equa-tion numerically
53. A line of positive charge is formed into a semicircle of
radius R 60.0 cm as shown in Figure P23.53 The
charge per unit length along the semicircle is described
by the expression l l0cos u The total charge on the
semicircle is 12.0 mC Calculate the total force on a
charge of 3.00 mC placed at the center of curvature
Problems 671
2.00 mC is suspended vertically on a 0.500-m-long lightstring in the presence of a uniform, downward-directed
electric field of magnitude E 1.00 105 N/C If theball is displaced slightly from the vertical, it oscillates like
a simple pendulum (a) Determine the period of thisoscillation (b) Should the effect of gravitation beincluded in the calculation for part (a)? Explain
58. Figure P23.35 shows three equal positive charges at
the corners of an equilateral triangle of side a 3.00 cm
Add a vertical line through the top charge at P, bisecting the triangle Along this line label points A, B, C, D, E, and
F, with A just below the charge at P ; B at the center of the
triangle; B, C, D, and E in order and close together with E
at the center of the bottom side of the triangle; and F close below E (a) Identify the direction of the total elec- tric field at A, E, and F Identify the electric field at B Identify the direction of the electric field at C (b) Argue
that the answers to part (a) imply that the electric field
must be zero at a point close to D (c) Find the distance from point E on the bottom side of the triangle to the point around D where the electric field is zero You will
need to solve a transcendental equation
59. Eight charged particles, each of magnitude q, are located
on the corners of a cube of edge s as shown in Figure P23.59 (a) Determine the x, y, and z components of the
total force exerted by the other charges on the charge
located at point A (b) What are the magnitude and
direc-tion of this total force?
2= intermediate; 3= challenging; = SSM/SG; = ThomsonNOW; = symbolic reasoning; = qualitative reasoning
b y
54. Two particles, each with charge 52.0 nC, are located
on the y axis at y 25.0 cm and y 25.0 cm (a) Find
the vector electric field at a point on the x axis as a
func-tion of x (b) Find the field at x 36.0 cm (c) At what
location is the field 1.00 kN/C? You may need to solve
an equation numerically (d) At what location is the field
16.0 kN/C? (e) Compare this problem with Question 7
Describe the similarities and explain the differences
55. Two small spheres of mass m are suspended from
strings of length that are connected at a common point
One sphere has charge Q and the other has charge 2Q.
The strings make angles u1 and u2 with the vertical
(a) Explain how u1and u2are related (b) Assume u1and
u2are small Show that the distance r between the spheres
is approximately
56. Two identical beads each have a mass m and charge q.
When placed in a hemispherical bowl of radius R with
frictionless, nonconducting walls, the beads move, and at
equilibrium they are a distance R apart (Fig P23.56).
Determine the charge on each bead
R m
q
q q
q
q
s s s
Figure P23.59 Problems 59 and 60
60. Consider the charge distribution shown in Figure P23.59.(a) Show that the magnitude of the electric field at the
center of any face of the cube has a value of 2.18k e q/s2.(b) What is the direction of the electric field at the center
of the top face of the cube?
placed at the center of a uniformly charged ring, where
the ring has a total positive charge Q as shown in ple 23.7 The particle, confined to move along the x axis,
Trang 9Exam-is moved a small dExam-istance x along the axExam-is (where x V a)
and released Show that the particle oscillates in simple
harmonic motion with a frequency given by
62. A line of charge with uniform density 35.0 nC/m lies
along the line y 15.0 cm between the points with
coordinates x 0 and x 40.0 cm Find the electric field
it creates at the origin
field is displaced slightly from its equilibrium position as
shown in Figure P23.63, where u is small The separation
of the charges is 2a, and the moment of inertia of the
dipole is I Assuming the dipole is released from this
posi-tion, show that its angular orientation exhibits simple
har-monic motion with a frequency
f 12pB
2qaE
I
f 12pak e qQ
ma3b1>2
672 Chapter 23 Electric Fields
64. Consider an infinite number of identical particles, each
with charge q, placed along the x axis at distances a, 2a, 3a, 4a, , from the origin What is the electric field at the origin due to this distribution? Suggestion: Use the fact
–q
2a
u
Figure P23.63
Answers to Quick Quizzes
23.1 (a), (c), (e) The experiment shows that A and B have
charges of the same sign, as do objects B and C
There-fore, all three objects have charges of the same sign We
cannot determine from this information, however, if the
charges are positive or negative
23.2 (e) In the first experiment, objects A and B may have
charges with opposite signs or one of the objects may be
neutral The second experiment shows that B and C
have charges with the same signs, so B must be charged
We still do not know, however, if A is charged or neutral
23.3 (b) From Newton’s third law, the electric force exerted
by object B on object A is equal in magnitude to the
force exerted by object A on object B and in the site direction
oppo-23.4 (a) There is no effect on the electric field if we assumethe source charge producing the field is not disturbed
by our actions Remember that the electric field is ated by the source charge(s) (unseen in this case), notthe test charge(s)
cre-23.5 A, B, C The field is greatest at point A because that is
where the field lines are closest together The absence of
lines near point C indicates that the electric field there is
zero
Trang 10In a tabletop plasma ball, the colorful lines emanating from the sphere
give evidence of strong electric fields Using Gauss’s law, we show in this
chapter that the electric field surrounding a uniformly charged sphere is
identical to that of a point charge (Getty Images)
24.1 Electric Flux 24.2 Gauss’s Law 24.3 Application of Gauss’s Law to Various Charge Distributions 24.4 Conductors in Electrostatic Equilibrium
In Chapter 23, we showed how to calculate the electric field due to a given charge
distribution In this chapter, we describe Gauss’s law and an alternative procedure
for calculating electric fields Gauss’s law is based on the inverse-square behavior
of the electric force between point charges Although Gauss’s law is a consequence
of Coulomb’s law, it is more convenient for calculating the electric fields of highly
symmetric charge distributions and makes it possible to deal with complicated
problems using qualitative reasoning.
The concept of electric field lines was described qualitatively in Chapter 23 We
now treat electric field lines in a more quantitative way.
Consider an electric field that is uniform in both magnitude and direction as
shown in Figure 24.1 The field lines penetrate a rectangular surface of area A,
whose plane is oriented perpendicular to the field Recall from Section 23.6 that
the number of lines per unit area (in other words, the line density) is proportional
to the magnitude of the electric field Therefore, the total number of lines
pene-trating the surface is proportional to the product EA This product of the
magni-tude of the electric field E and surface area A perpendicular to the field is called
the electric flux E(uppercase Greek letter phi):
Figure 24.1 Field lines representing
a uniform electric field penetrating a
plane of area A perpendicular to the
field The electric flux Ethrough
this area is equal to EA.
Trang 11From the SI units of E and A, we see that Ehas units of newton meters squared per coulomb (Nm2/C) Electric flux is proportional to the number of electric field lines penetrating some surface.
If the surface under consideration is not perpendicular to the field, the flux through it must be less than that given by Equation 24.1 Consider Figure 24.2,
where the normal to the surface of area A is at an angle u to the uniform electric field Notice that the number of lines that cross this area A is equal to the number
of lines that cross the area A›, which is a projection of area A onto a plane
ori-ented perpendicular to the field Figure 24.2 shows that the two areas are related
by A› A cos u Because the flux through A equals the flux through A›, the flux
We assumed a uniform electric field in the preceding discussion In more eral situations, the electric field may vary over a large surface Therefore, the defi- nition of flux given by Equation 24.2 has meaning only for a small element of area over which the field is approximately constant Consider a general surface divided into a large number of small elements, each of area A It is convenient to define
gen-a vector whose magnitude represents the area of the ith element of the large surface and whose direction is defined to be perpendicular to the surface element as
shown in Figure 24.3 The electric field at the location of this element makes an angle uiwith the vector The electric flux Ethrough this element is
where we have used the definition of the scalar product of two vectors ( ; see Chapter 7) Summing the contributions of all elements gives
an approximation to the total flux through the surface:
If the area of each element approaches zero, the number of elements approaches infinity and the sum is replaced by an integral Therefore, the general definition
of electric flux is
(24.3)
Equation 24.3 is a surface integral, which means it must be evaluated over the
sur-face in question In general, the value of Edepends both on the field pattern and
on the surface.
We are often interested in evaluating the flux through a closed surface, defined as
a surface that divides space into an inside and an outside region so that one not move from one region to the other without crossing the surface The surface
can-of a sphere, for example, is a closed surface.
Consider the closed surface in Active Figure 24.4 The vectors point in ferent directions for the various surface elements, but at each point they are nor- mal to the surface and, by convention, always point outward At the element labeled , the field lines are crossing the surface from the inside to the outside and u 90°; hence, the flux through this element is positive For element , the field lines graze the surface (perpendicular to the vector ); therefore, u 90° and the flux is zero For elements such as , where the field lines are crossing the surface from outside to inside, 180° u 90° and the flux
dif-is negative because cos u dif-is negative The net flux through the surface dif-is
propor-tional to the net number of lines leaving the surface, where the net number
Figure 24.2 Field lines representing
a uniform electric field penetrating
an area A that is at an angle u to the
field Because the number of lines
that go through the area A›is the
same as the number that go through
A, the flux through A›is equal to
the flux through A and is given by
Figure 24.3 A small element of
surface area A i The electric field
makes an angle uiwith the vector ,
defined as being normal to the
sur-face element, and the flux through
the element is equal to E i A icos ui
Trang 12means the number of lines leaving the surface minus the number of lines entering the
sur-face If more lines are leaving than entering, the net flux is positive If more lines
are entering than leaving, the net flux is negative Using the symbol to represent
an integral over a closed surface, we can write the net flux Ethrough a closed
surface as
(24.4)
where Enrepresents the component of the electric field normal to the surface.
Quick Quiz 24.1 Suppose a point charge is located at the center of a spherical
surface The electric field at the surface of the sphere and the total flux through
the sphere are determined Now the radius of the sphere is halved What happens
to the flux through the sphere and the magnitude of the electric field at the
sur-face of the sphere? (a) The flux and field both increase (b) The flux and field
both decrease (c) The flux increases, and the field decreases (d) The flux
decreases, and the field increases (e) The flux remains the same, and the field
increases (f) The flux decreases, and the field remains the same.
Sign in at www.thomsonedu.comand
go to ThomsonNOW to select anysegment on the surface and see therelationship between the electric fieldvector and the area vector ES ¢ASi
E X A M P L E 2 4 1
Consider a uniform electric field oriented
in the x direction in empty space Find the
net electric flux through the surface of a
cube of edge length , oriented as shown in
Figure 24.5.
SOLUTION
Conceptualize Examine Figure 24.5
care-fully Notice that the electric field lines pass
through two faces perpendicularly and are
parallel to four other faces of the cube.
parallel to the x axis Side
is the bottom of the cube,and side is opposite side
Trang 13
24.2 Gauss’s Law
In this section, we describe a general relationship between the net electric flux
through a closed surface (often called a gaussian surface) and the charge enclosed
by the surface This relationship, known as Gauss’s law, is of fundamental
impor-tance in the study of electric fields.
Consider a positive point charge q located at the center of a sphere of radius r as
shown in Figure 24.6 From Equation 23.9, we know that the magnitude of the
elec-tric field everywhere on the surface of the sphere is E keq/r2 The field lines are directed radially outward and hence are perpendicular to the surface at every point
on the surface That is, at each surface point, is parallel to the vector senting a local element of area Aisurrounding the surface point Therefore, and, from Equation 24.4, we find that the net flux through the gaussian surface is
repre-where we have moved E outside of the integral because, by symmetry, E is constant over the surface The value of E is given by E keq/r2 Furthermore, because the surface is spherical, dA A 4pr2 Hence, the net flux through the gaussian surface is
Recalling from Section 23.3 that ke 1/4pP0 , we can write this equation in the form
(24.5)
Equation 24.5 shows that the net flux through the spherical surface is
propor-tional to the charge inside the surface The flux is independent of the radius r because the area of the spherical surface is proportional to r2, whereas the electric
field is proportional to 1/r2 Therefore, in the product of area and electric field,
the dependence on r cancels.
676 Chapter 24 Gauss’s Law
Categorize We evaluate the flux from its definition, so we categorize this example as a substitution problem The flux through four of the faces ( , , and the unnumbered ones) is zero because is parallel to the four faces and therefore perpendicular to d A on these faces.
For face , is constant and directed inward but is
directed outward (u 180°) Find the flux through this
1E 1cos 180°2 dA E 1dA EA E/2
For face , is constant and outward and in the same
direction as (u 0°) Find the flux through this
Find the net flux by adding the flux over all six faces: £E E/2 E/2 0 0 0 0 0
KARL FRIEDRICH GAUSS
German mathematician and astronomer
(1777–1855)
Gauss received a doctoral degree in
mathemat-ics from the University of Helmstedt in 1799
In addition to his work in electromagnetism,
he made contributions to mathematics and
science in number theory, statistics,
non-Euclidean geometry, and cometary orbital
mechanics He was a founder of the German
Magnetic Union, which studies the Earth’s
magnetic field on a continual basis.
Trang 14Now consider several closed surfaces surrounding a charge q as shown in Figure
24.7 Surface S1is spherical, but surfaces S2 and S3 are not From Equation 24.5,
the flux that passes through S1 has the value q/P0 As discussed in the preceding
section, flux is proportional to the number of electric field lines passing through a
surface The construction shown in Figure 24.7 shows that the number of lines
through S1 is equal to the number of lines through the nonspherical surfaces S2
and S3 Therefore, the net flux through any closed surface surrounding a point
charge q is given by q/P0and is independent of the shape of that surface.
Now consider a point charge located outside a closed surface of arbitrary shape
as shown in Figure 24.8 As can be seen from this construction, any electric field
line entering the surface leaves the surface at another point The number of
elec-tric field lines entering the surface equals the number leaving the surface
There-fore, the net electric flux through a closed surface that surrounds no charge is
zero. Applying this result to Example 24.1, we see that the net flux through the
cube is zero because there is no charge inside the cube.
Let’s extend these arguments to two generalized cases: (1) that of many point
charges and (2) that of a continuous distribution of charge We once again use the
superposition principle, which states that the electric field due to many charges is
the vector sum of the electric fields produced by the individual charges.
There-fore, the flux through any closed surface can be expressed as
where is the total electric field at any point on the surface produced by the
vec-tor addition of the electric fields at that point due to the individual charges
Con-sider the system of charges shown in Active Figure 24.9 The surface S surrounds
only one charge, q1; hence, the net flux through S is q1/P0 The flux through S due
to charges q2, q3, and q4 outside it is zero because each electric field line from
these charges that enters S at one point leaves it at another The surface S
rounds charges q2 and q3; hence, the net flux through it is (q2 q3 )/P0 Finally,
the net flux through surface S is zero because there is no charge inside this
sur-face That is, all the electric field lines that enter S at one point leave at another.
Charge q4does not contribute to the net flux through any of the surfaces because
it is outside all the surfaces.
Gauss’s law is a generalization of what we have just described and states that the
net flux through any closed surface is
(24.6)
where represents the electric field at any point on the surface and qinrepresents
the net charge inside the surface.
r
q
A E
i
Figure 24.6 A spherical gaussian
surface of radius r surrounding a
point charge q When the charge is at
the center of the sphere, the electric
field is everywhere normal to the
sur-face and constant in magnitude
S3
S2
S1q
Figure 24.7 Closed surfaces of
vari-ous shapes surrounding a charge q.
The net electric flux is the samethrough all surfaces
q
Figure 24.8 A point charge located
outside a closed surface The number
of lines entering the surface equalsthe number leaving the surface
through surface S is zero Charge
q4does not contribute to the fluxthrough any surface because it is out-side all surfaces
Sign in at www.thomsonedu.comand
go to ThomsonNOW to change thesize and shape of a closed surface andsee the effect on the electric flux ofsurrounding combinations of chargewith that surface
Gauss’s law
Trang 15When using Equation 24.6, you should note that although the charge qinis the net charge inside the gaussian surface, represents the total electric field, which
includes contributions from charges both inside and outside the surface.
In principle, Gauss’s law can be solved for to determine the electric field due
to a system of charges or a continuous distribution of charge In practice, however, this type of solution is applicable only in a limited number of highly symmetric sit- uations In the next section, we use Gauss’s law to evaluate the electric field for charge distributions that have spherical, cylindrical, or planar symmetry If one chooses the gaussian surface surrounding the charge distribution carefully, the integral in Equation 24.6 can be simplified.
Quick Quiz 24.2 If the net flux through a gaussian surface is zero, the following four statements could be true Which of the statements must be true? (a) There are
no charges inside the surface (b) The net charge inside the surface is zero (c) The electric field is zero everywhere on the surface (d) The number of electric field lines entering the surface equals the number leaving the surface.
Zero Flux Is Not Zero Field
In two situations, there is zero flux
through a closed surface: either
there are no charged particles
enclosed by the surface or there
are charged particles enclosed, but
the net charge inside the surface is
zero For either situation, it is
incor-rect to conclude that the electric
field on the surface is zero Gauss’s
law states that the electric flux is
proportional to the enclosed
charge, not the electric field.
CO N C E P T UA L E X A M P L E 2 4 2
A spherical gaussian surface surrounds a point charge q Describe what happens to the total flux through the surface
if (A) the charge is tripled, (B) the radius of the sphere is doubled, (C) the surface is changed to a cube, and (D) the
charge is moved to another location inside the surface.
SOLUTION
(A) The flux through the surface is tripled because flux is proportional to the amount of charge inside the surface.
(B) The flux does not change because all electric field lines from the charge pass through the sphere, regardless of its radius.
(C) The flux does not change when the shape of the gaussian surface changes because all electric field lines from the charge pass through the surface, regardless of its shape.
(D) The flux does not change when the charge is moved to another location inside that surface because Gauss’s law refers to the total charge enclosed, regardless of where the charge is located inside the surface.
Flux Due to a Point Charge
Charge Distributions
As mentioned earlier, Gauss’s law is useful for determining electric fields when the charge distribution is highly symmetric The following examples demonstrate ways of choosing the gaussian surface over which the surface integral given by Equation 24.6 can be simplified and the electric field determined In choosing the surface, always
take advantage of the symmetry of the charge distribution so that E can be removed
from the integral The goal in this type of calculation is to determine a surface for which each portion of the surface satisfies one or more of the following conditions:
1. The value of the electric field can be argued by symmetry to be constant over the portion of the surface.
2. The dot product in Equation 24.6 can be expressed as a simple algebraic
product E dA because and are parallel.
3. The dot product in Equation 24.6 is zero because and are perpendicular.
4. The electric field is zero over the portion of the surface.
Different portions of the gaussian surface can satisfy different conditions as long as every portion satisfies at least one condition All four conditions are used
in examples throughout the remainder of this chapter.
Gaussian Surfaces Are Not Real
A gaussian surface is an imaginary
surface you construct to satisfy the
conditions listed here It does not
have to coincide with a physical
sur-face in the situation