Since the flux through A is the same as through A, the flux through A is E = EA cos θ If we define a vector area → A whose magnitude represents the surface area andwhose direction is de
Trang 1Section 20.3 Electric Field of an Electric Dipole
(9) Two point charges q1 = −6 μC and q2= +6 μC are placed at two vertices of
an equilateral triangle, see Fig.20.27 If a = 10 cm, find the electric field at the
(10) A proton and an electron form an electric dipole and are separated by
a distance of 2a= 2 × 10−10m, see Fig.20.28 (a) Use exact formulas to
calculate the electric field along the x-axis at x = −10a, x = −2a, x = −a/2,
Trang 2694 20 Electric Fields
x = +a/2, x = +2a, and x = +10a (b) Show that at both points x = ±10a, the
approximate formula given by Eq.20.13has a very close percentage differencefrom the exact value
Fig 20.28 See Exercise (10)
(11) Rework the calculations of Exercise 10 but on the y-axis at y = −10 a,
y = −2 a, y = −a/2, y = +a/2, y = +2a, and y = +10a In part (b), use
Eq.20.17
Section 20.4 Electric Field of a Continuous Charge Distribution
(12) A non-conductive rod of length L has a total negative charge −Q that is
uni-formly distributed along its length, see Fig.20.29 (a) Find the linear chargedensity of the rod (b) Use the coordinates depicted in the figure to prove that the
electric field at point P, a distance a from the right end of the rod, has the same
form as the one given by Eq.20.27 (c) When P is very far from the rod, i.e.
a L, show that the electric field reduces to the electric field of a point charge
(i.e the rod would look like a point charge) (d) If L = 15 cm, Q = 25 μC, and
a = 20 cm, find the value of the electric field at P.
−
Fig 20.29 See Exercise (12)
(13) A non-conductive rod lies along the x-axis with one of its ends located at x = a
and the other end located at∞, see Fig.20.30 Starting from the definition of
an electric field of a differential element on the rod, find the electric field at the
Trang 320.7 Exercises 695
origin if: (a) the rod carries a uniform positive linear charge density λ (b) the
rod carries a positive varying linear charge densityλ = λ◦ a /x.
Fig 20.30 See Exercise (13)
(14) A uniformly charged ring of radius 15 cm has a total charge of 50μC Find the
electric field on the central perpendicular axis of the ring at: (a) 0 cm, (b) 1 cm,(c) 10 cm, and (d) 100 cm (e) What do you observe about the values you justcalculated?
(15) A charged ring of radius R = 0.5 m has a gap d = 0.1 m, see Fig.20.31
Calculate the electric field at its center C if it carries a uniform charge q = 1 μC.
Fig 20.31 See Exercise (15)
d =0 1m
R =0 5 m
q =1 C
C
(16) Figure20.32shows a non-conductive semicircular arc of radius R that consists
of two quarters The semicircle has a uniform positive total charge Q along its
right half, and a uniform negative total charge−Q along its left half Find the
resultant electric field at the center of the semicircle
Fig 20.32 See Exercise (16)
P R R
R
+Q
-Q
Trang 4696 20 Electric Fields(17) Two non-conductive semicircular arcs, one of a uniform positive charge+Q
and the other of a uniform negative charge−Q, form a circle of radius R,
see Fig.20.33 Find the resultant electric field at the center of the circle, andcompare it with the result of Exercise 16
Fig 20.33 See Exercise (17)
P
R R
+Q -Q
(18) If you consider a uniformly charged ring of total charge Q and a fixed radius
R (as in Fig.20.18), then the graph of Fig.20.34would map the electric field
along the axis of such a ring as a function of z/R Show that the maximum electric field is Emax = 2k Q/3√3R2and occurs at z = R/√2.
Fig 20.34 See Exercise (18)
max
E E
/
z R
(19) An electron is constrained to move along the central axis of a ring of radius R that has a uniform positive charge q, see Fig.20.35 Show that when the position
x of the electron is much less than the radius R (x R), the electrostatic force
exerted on the electron can cause it to oscillate through the center of the ringwith an angular frequency given byω =kqe /mR3, where e and m are the
electronic charge and mass, respectively
(20) Two non-conductive rings having the same radius R are arranged with their central axes along a common horizontal line and separated by a distance of 4 R,
see Fig.20.36 Ring 1 has a uniform positive charge q1 , while ring 2 has a uniform positive charge q2 Given that the net electric field is zero at point P, which is at a distance R from ring 1 and on the common central axis of the two rings, (a) find the ratio between the two charges (b) If only the sign of q1
Trang 5Fig 20.36 See Exercise (20)
R R
P
4R R
C1 1
C2
(21) A disk of radius R = 5 cm has a surface charge density σ = 6 μC/m2 on itssurface Calculate the magnitude of the electric field at points on the centralaxis of the disk located at: (a) 1 mm, (b) 1 cm, (c) 10 cm, and (d) 100 cm (22) A disk of radius R has a charge Q that is uniformly distributed over its surface
area Show that Eq.20.55transforms to:
You may use the binomial expansion(1 + δ) p ≈ 1 + pδ when δ 1.
(23) Compare the obtained results of Exercise 21 to the near-field approximation
E = σ/2◦as well as to the point charge approximation E = k(πR2σ )/a2, and
find which result(s) of Exercise 21 match the two approximations
Trang 6(25) Find the electric field between two oppositely-charged infinite sheets of charge,each having the same charge magnitude and surface charge density σ, but
opposite signs, see Fig.20.38
Fig 20.38 See Exercise (25)
Section 20.5 Electric Field Lines
(26) (a) A negatively charged disk has a uniform charge per unit area Sketch theelectric field lines in the plane of the plane of the disk passing through its center.(b) Redo part (a) taking the disk to be positively charged (c) A negativelycharged rod has a uniform charge per unit length Sketch the electric fieldlines in the plane of the rod (d) Three equal positive charges are placed at thecorners of an equilateral triangle Sketch the electric field lines in the plane
of the charges (e) An infinite linear rod has a uniform charge per unit length.Sketch the electric field lines in the plane of the rod
Trang 720.7 Exercises 699
Section 20.6 Motion of Charged Particles in a Uniform Electric Field
(27) An electron and a proton are released simultaneously from rest in a uniformelectric field of 105N/C Ignore the effect of the fields of the electron and proton
on each other (a) Find the speed and kinetic energy of the electron 50 ns after
it has been released (b) Repeat part (a) for the proton
(28) Figure20.39shows two oppositely charged parallel plates that are separated
by a distance d = 1.5 cm Each plate has a charge per unit area of magnitude
σ = 4 μC/m2 An electron is released from rest at t = 0 from the negative plate.
(a) Calculate the electric field between the two plates (b) Ignoring the effect ofgravity, find the resultant force exerted on the electron? (c) Find the acceleration
of the electron (d) How long does it take the electron to strike the positive plate?(e) What is the speed and kinetic energy of the electron just before striking thepositive plate?
Fig 20.39 See Exercise (28)
travels the distance d = 1.5 cm between the two plates and stops temporarily
before hitting the negatively charged plate (a) Find the magnitude and direction
of its acceleration (b) Find the value of the electron’s initial speed (c) Find thetime before the electron stops temporarily
(30) Two oppositely charged horizontal plates are separated by a distance d= 1 cm
and each has a length L = 3 cm, see Fig.20.40 The electric field between the
plates is uniform and has a magnitude E= 102N/m An electron is projected
between the plates with a horizontal initial speed ofv◦= 106m/s as shown.Assuming this experiment is conducted in a vacuum, where will the electronstrike the upper plate?
(31) Repeat Exercise 30 when a proton replaces the electron
Trang 8700 20 Electric Fields
Fig 20.40 See Exercise (30)
x y
strikes a screen at a distance D = 30 cm (a) What is the acceleration of the
electron in the region between the two plates? (b) Find the time when theelectron leaves the two plates (c) What is the vertical position of the electronjust before leaving the region between the two plates? (d) Find the electron’stotal vertical distance just before hitting the screen
x y
-Fig 20.41 See Exercise (32)
(33) Repeat Exercise 32 when a proton replaces the electron
Trang 9Gauss’s Law 21
Although Coulomb’s law is the governing law in electrostatics, its form does notalways simplify calculations in situations involving symmetry In this chapter, weintroduce Gauss’s law as an alternative method for calculating electric fields of certainhighly symmetrical charge distribution systems In addition to being simpler thanCoulomb’s law, Gauss’s law permits us to use qualitative reasoning
21.1 Electric Flux
Consider a uniform electric field →
E penetrating a small area A oriented
perpendic-ularly to the field as shown in Fig.21.1 Recall fromSect 20.5that the number ofelectric field lines per unit area (measured in a plane perpendicular to the lines) isproportional to the magnitude of →
E Therefore, the total number of lines penetrating
the surface is proportional to EA This product is called the electric flux1 E Thus:
If the area in Fig.21.1is tilted by an angleθ with respect to E→, the flux through
it (the number of electric lines) will decrease To visualize this, Fig.21.2shows an
1 The word “flux” comes from the Latin word meaning “to flow”.
H A Radi and J O Rasmussen, Principles of Physics, 701 Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_21,
© Springer-Verlag Berlin Heidelberg 2013
Trang 10702 21 Gauss’s Law
area A that makes an angle θ with the field The number of lines that cross the area
A is equal to the number that cross the area A, which is perpendicular to E→and
hence A= A cos θ Thus, the flux through A, E (A), is equal to the flux through A,
E (A) But according to Eq.21.1, the flux through Ais defined as E (A) = EA Therefore, the flux through A is:
E
θ
Area A' Area A'
Side view
Area A Area A
E
Fig 21.2 Electric field lines representing a uniform electric field →E penetrating an area A that is at anangleθ with the field (both three dimensional and side views are displayed) Since the flux through A is
the same as through A, the flux through A is E = EA cos θ
If we define a vector area →
A whose magnitude represents the surface area andwhose direction is defined to be perpendicular to the surface area as in Fig.21.3, then
Eq.21.2can be written as:
E=E→•→
A = EA cos θ (21.3)
The flux through a surface of area A has a maximum value EA when the surface is
perpendicular to the field (i.e whenθ = 0◦), and is zero when the surface is parallel
to the field (i.e whenθ = 90◦).
Trang 11Fig 21.3 The definition of a vector area →A whose magnitude represents the surface area and whosedirection is defined to be perpendicular to the surface area
Generally, the electric field may vary over the surface of any shape Let us considerthe general surface depicted by the shape in Fig.21.4and calculate the electric fluxover the whole surface
E , then the differential electric flux d E through this differential areawill be:
Trang 12704 21 Gauss’s Law
According to the definition of the vector area d→
A which always points outwards,the sign of the flux depends on the angle between →
and hence d E=E→• d→
A is negative
The net flux through a surface is proportional to the net number of electric field
lines leaving the surface If more lines are entering than leaving, then the net flux isnegative If more lines are leaving than entering, then the net flux is positive
We can write the net flux through a closed surface as:
E= E→• d→
where the symbol
represents an integral over a closed surface
Example 21.1
Find the electric flux through a sphere of radius r enclosing at its center: (a) a positive charge q, and (b) a negative charge −q.
Solution: (a) When a positive charge q is at the center of such a sphere, the electric
field would be directed outwards and be normal to the surface It would also have
a constant magnitude, E = q/(4π◦r2), see Fig.21.5 Therefore:
d A r
E
+
Trang 1321.1 Electric Flux 705From Eq.21.6and the fact that
dA over a spherical surface gives the area of the
(b) When a negative charge−q is at the center of such a sphere, the electric
field would be directed inwards and be normal to the surface It would also have
a constant magnitude, E = q/(4π◦r2), see Fig.21.6 Therefore:
In this section, we introduce a new foundation of Coulomb’s law, called Gauss’s
law This law can be used to take advantage of symmetry in the problem under
con-sideration Central to Gauss’s law is a hypothetical closed surface called a Gaussian
Trang 14706 21 Gauss’s Law
number of electric field lines passing through the spherical surface S1is the same as
the number of lines passing through the non-spherical surfaces S2 and S3 Therefore,
we conclude that the flux through any closed Gaussian surface surrounding the point
charge q is q /◦.
Fig 21.7 Different closed
Gaussian surfaces enclosing a
point charge q The net electric
flux is the same through all
surfaces
S E
where qin represents the net charge inside the surface and →
E represents the total electric field at any point on the surface, which includes contributions from charges
inside and/or outside (if any)
Note that Gauss’s law is very useful in calculating electric fields in situations wherethe charge distributions have planar, cylindrical, or spherical symmetry In thesecharge distribution systems, one must carefully construct the imaginary Gaussiansurface such that it simplifies the integral in Eq.21.7 This can be done by trying tosatisfy one or more of the following conditions:
(1) The value of the field over the surface is constant, E = constant.
(2) The dot product→
Trang 1521.3 Applications of Gauss’s Law 707
21.3 Applications of Gauss’s Law
Example 21.2
Using Gauss’s law, find the electric field at a distance r from a positive point charge q, and compare it with Coulomb’s law.
Solution: We apply Gauss’s law to the spherical Gaussian surface of radius r
in Fig.21.5 From the symmetry of the problem, we know that at any point, theelectric field →
E is perpendicular to the surface and directed outwards from thespherical center Thus, →
E // d→A and →
E • d→
A = E dA Then, with qin= q, we can
write Gauss’s law as:
Example 21.3
Prove that any excess positive charge q on the isolated conductor of Fig.21.8willlie entirely on its outer surface
Conductor'souter surfaceGaussian
surfaces
Cavity
Conductor'sinner surface
Fig 21.8
Trang 16708 21 Gauss’s Law
Solution: The electric field inside the conductor must be zero If this is not the
case, the field would exert forces on the free electrons and a current would flowwithin the conductor Of course, there are no such currents in an isolated conductor
in electrostatic equilibrium
First, we draw a Gaussian surface surrounding the conductor’s cavity, close toits surface, as seen in Fig.21.8 Since →
E = 0 inside the conductor, then E= 0
and hence according to Gauss’s law, no net charge would exist on the inner walls
of the cavity
Then we draw a Gaussian surface just inside the outer surface of the conductor.Since →
E = 0 inside the conductor, then E= 0 and according to Gauss’s law, no
net charge will exist inside the Gaussian surface If the excess charge is not insidethe Gaussian surface, it must then be outside that surface, on the conductor’ssurface; see Fig.21.9
+
+
Conductor's outer surface
Conductor's inner surface
+
Example 21.4
Using Gauss’s law, find the electric field just outside the surface of a conductorcarrying a positive surface charge densityσ.
Solution: Consider a small section of the conductor’s surface so as to neglect
curvature Then construct a cylindrical Gaussian surface normal to the conductor
as shown in Fig.21.10, where one end of the cylinder is inside the conductor while
the other end is outside Each end has an area A The electric field →
E inside theconductor is zero, and the electric field →
E just outside the conductor’s surface