electrostaics coulomb force and electrical dipoles
TRANSCRIPT
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(CharlesD.
Winters)
F IGURE 4.1 Rubbing a balloon against your hairon a dry day causes the balloon and your hair to
become electrically charged.
Electric Forces and Electric
Fields
Chapter 4
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Rubber Rubber
(a)
F F
(b)
F
F
Rubber
+ + +
+
+
+
Glass+
F IGURE 4.2 (a) A negatively charged rubber rod, suspended by
an insulating thread, is attracted to a positively charged glass rod. (b) A
negatively charged rubber rod is repelled by another negatively charged
rubber rod.
F IGURE 4.3 When a glass rod is
rubbed with silk, electrons are
transferred from the glass to the silk.
Because of conservation of charge, each
electron adds negative charge to the silk,
and an equal positive charge is leftbehind on the rod. Also, because the
charges are transferred in discrete
bundles, the charges on the two objects
are e or 2e or 3e, and so on.
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Suspensionhead
Fiber
B
A
F IGURE 4.6 Coulombstorsion balance, which was
used to establish the inverse-
square law for the electrostatic
force between two charges.
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r
(a)F21
F12
q1
q2
(b)
F21
F12
q1
q2
r12
+
+
+
Figure 4.7
Two point charges separated
by a distance r exert a force
on each other given by
Coulombs law. Note that the
force exerted by q2 on
q1 is equal in magnitude andopposite in direction to the
force exerted byq1 on q2.
(a) When the charges are ofthe same sign, the force is
repulsive. (b) When the
charges are of opposite signs,
the force is attractive.
F:
12
F:
21
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2.00 m
x
q1
x
q3q2 F13F23
2.00 x
+
+
F IGURE 4.8 (Example 4.1) Three
point charges are placed along the x
axis. If the net force on q3 is zero, the
force exerted byq1 on q3 must be
equal in magnitude and opposite in
direction to the force exerted byq2 on q3.
F
:
23
F:
13
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(a) (b)
q0
q 0
>> q0
++
F IGURE 4.9 (a) For a small enough test
charge q0, the charge distribution on the
sphere is undisturbed. (b) If the test charge q0
were larger, the charge distribution on thesphere would be disturbed as a result of the
proximity ofq0.
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(b)
E
q
P
r
P
(a)
Fe
q
q0
rP
r
(c)
Feq
q0
P
r
(d)
E
q
r+
+
Figure 4.10
A test charge q0 at pointPis a distance rfrom a
point charge q. (a) Ifq is positive, the force on
the test charge is directed away from q. (b) For
the positive source charge, the electric field at
P points radially outward from q. (c) If q is
negative, the force on the test charge is
directed toward q. (d) For the negative source
charge, the electric field at P points radially
inward toward q.
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P E
y
E1
E2y
r
aq
aq
x+
F I GURE 4.11 (Example4.3) The total electric field
at P due to two equal and
opposite charges (an electric
dipole) equals the vector sum
. The field is dueto the positive charge q, and
is the field due to the
negative charge q.
E:
2
E
:
1E
:
1 E
:
2
E:
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r
qr
P
E
F IGURE 4.12 The
electric field at P due to a
continuous charge distributionis the vector sum of the fields
due to all the elements qof the charge distribution.
E:
E:
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F IGURE 4.13 (Example 4.4) The electric field atPdue to a
uniformly charged rod lying along the xaxis. The field atPdue
to the segment of charge dq is kedq/x2. The total field atP isthe vector sum over all segments of the rod.
x
y
a
Px
dx
dq= dx
E
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(a)
++
+
+
+
+
+
+
++
++
++
++
P dEx
dEdE
x x
r
q
a
(b)
++
+
+
+
+
+
+
+
+
++
+
++
+
dE2
dE1
2
F IGURE 4.14 (Example 4.5)
A uniformly charged ring of radius
a. (a) The field atPon the xaxis
due to an element of charge dq.
(b) The total electric field atP is
along the x axis. The perpendicu
lar component of the electric field
atPdue to segment 1 is canceled
by the perpendicular component
due to segment 2.
BA
F IGURE 4.15 Electric field lines
penetrating two surfaces. The
magnitude of the field is greater on
surface A than on surface B.
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(a)
q
(b)
q+
(DouglasC.
Johnson/CalPolyPomona)
F IGURE 4.16 The electric field lines for a point charge. (a) For a positive point charge, the lines are directed radially
outward. (b) For a negative point charge, the lines are directed radially inward. Note that the figures show only those field
lines that lie in the plane containing the charge. (c) The dark areas are small particles suspended in oil, which align with the
electric field produced by a small charged conductor at the center.
(c)
(a)
+ F IGURE 4.17 (a) The electric field lines for
two charges of equal magnitude and opposite sign
(an electric dipole). Note that the number of lines
leaving the positive charge equals the number
terminating at the negative charge. (b) Small
particles suspended in oil align with the electric
field.(DouglasC.
Johnson/CalPolyPomona)
(b)
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F I GURE 4.18 (a) The electric field lines
for two positive point charges. (The
locations A, B, and Care discussed in Quick
Quiz 4.5.) (b) Small particles suspended in
oil align with the electric field.(DouglasC.
Johnson/CalPolyPomona)
(a)
+ +C
A
B
+2q q+
Figure 4.19
The electric field lines for a
point charge 2q and a
second point charge q.
Note that two lines leave the
charge 2qfor every one that
terminates on q.
(b)
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+
+
+
+
+
+
E
vv= 0
q
x
+ +
F I GURE 4.20 (Example 4.6) A positive
point charge qin a uniform electric field
undergoes constant acceleration in thedirection of the field.
E:
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(0, 0)
E
(x,y)
v
x
y
+ + + + + + + + + + + +
vii
Figure 4.21
An electron is projected horizontally into a uniform
electric field produced by two charged plates. The
electron undergoes a downward acceleration (opposite
), and its motion is parabolic while it is between the
plates.
E:
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Electron
beam
plates
anode
gridgun
CA
plates
Fluorescentscreen
F IGURE 4.22 Schematic diagram of a cathode-ray tube. Electrons leaving the hot cathode C are accelerated to the anode A. In
addition to accelerating electrons, the electron gun is also used to focus the beam of electrons, and the plates deflect the beam.
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A
in Equation 4.18 is then 90.
Area = A
E
F IGURE 4.23 Field lines of a
uniform electric field penetrating aplane of area A perpendicular to the
field. The electric flux E through this
area is equal toEA.
A
A = Acos
E
Normal
F IGURE 4.24 Field lines for a uniform electric
field through an area Awhose normal is at an angle
to the field. Because the number of lines that go
through the shaded area A is the same as thenumber that go through A, we conclude that the
total flux through A is equal to the flux through A
and is given by E EAcos .
Ai
E i
i
F IGURE 4.25 Asmall element
of a surface of area Ai. The
electric field makes an angle iwiththe normal to the surface (the
direction of ), and the flux
through the element is equal to Ei
Aicos i.
A:
i
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Figure 4.26
A closed surface in an electric field. The area
vectors are, by convention, normal to
the surface and point outward. The flux
through an area element can be positive
(element 1), zero (element 2), or negative
(element 3).
A:
i
A1
A3
A2
E
En
En
EE
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y
z
x
E
dA2
dA1
dA3
dA4
F IGURE 4.27 (Example 4.8) A
hypothetical surface in the shape of a cube in
a uniform electric field parallel to the x axis.
The net flux through the surface is zero. Sideis the bottom of the cube and side is
opposite side .
Spherical
surface
r
q
A
E
i+
F IGURE 4.28 A spherical surface
of radius r surrounding a point
charge q. When the charge is at the
center of the sphere, the electric fieldis normal to the surface and constant
in magnitude everywhere on the
surface.
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S3
S2
S1
q
F IGURE 4.29 Closed surfaces of
various shapes surrounding a charge q.
The net electric flux through eachsurface is the same.
F IGURE 4.30 Apoint charge located
outside a closed surface. The number of
lines entering the surface equals the
number leaving the surface.
q
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S
q1
q2
q3 S
S
q4
Figure 4.31
The net electric flux through any closed
surface depends only on the charge
inside that surface. The net flux through
surface S is ql/e0, the net flux throughsurface S9 is (q2 1 q3)/e0, and the net
flux through surface S0 is zero. Charge
q4 does not contribute to the flux
through any surface because it is outsideall surfaces.
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auss ansurface
r
q
dA
E
+
F IGURE 4.32 (Example 4.9)
The point charge q is at the
center of the spherical gaussian
surface, and is parallel to at
every point on the surface.
dA:
E:
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(a)
Gaussiansphere
(b)
Gaussianspherer
a
r
a
F IGURE 4.33(Interactive Example 4.10) Auniformly charged insulating sphere of radius aand total
charge Q. (a) For points outside the sphere, a large,
spherical gaussian surface is drawn concentric with the
sphere. In diagrams such as this one, the dotted line
represents the intersection of the gaussian surface withthe plane of the page. (b) For points inside the sphere, a
spherical gaussian surface smaller than the sphere is
drawn.
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Gaussian
surface
+
++
+++
E
dA
r
(a)
E
(b)
F IGURE 4.35 (Example 4.11) (a) An infinite line of
charge surrounded by a cylindrical gaussian surface
concentric with the line charge. (b) An end view shows
that the electric field on the cylindrical surface is
constant in magnitude and perpendicular to the surface.
a
E
ar
E=keQ
r2
E=keQ
a3r
F IGURE 4.34(Example 4.10) A plot ofEversus rfora uniformly charged insulating sphere. The electric field
inside the sphere (r a) varies linearly with r. The
electric field outside the sphere (r a) is the same as
that of a point charge Qlocated atr 0.
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E
++
++
++
+
++
++
++
++
+
++
++
+
++
+
++
++
+
++
++
++
+
A
Gaussiansurface
E
F IGURE 4.36 (Example 4.12) A
cylindrical gaussian surface penetrating an
infinite sheet of charge. The flux is EA
through each end of the gaussian surface and
zero through its curved surface.
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+
+
++
+
++
+
E E
F I GURE 4.37 Aconducting
slab in an external electric field
. The charges induced on the
surfaces of the slab produce an
electric field that opposes theexternal field, giving a resultant
field of zero inside the
conductor.
E:
Gaussiansurface
F I GURE 4.38 An isolated
conductor of arbitrary shape.
The broken line represents a
gaussian surface just inside the
physical surface of the
conductor.
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A
++ + +
+
+
+
+
+++++++ +
++
+
+
E
F IGURE 4.39 A gaussiansurface in the shape of a small
cylinder is used to calculate the
electric field just outside a
charged conductor. The flux
through the gaussian surface isEA.
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0.500 m
7.00C
2.00C 4.00C
60.0
y
+
+
Figure P4.5
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d
+3q +q
Figure P4.8
0.100 m
x
5.00 nC
0.300 m
6.00 nC
y
aa
a
a
q
3q 4q
2q
Figure P4.15
2a
x
q q
Figure P4.16
1.00 m
2.50C 6.00C
Figure P4.11
Figure P4.13
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y
y
dx
x
P
O
0
Figure P4.21
q2
q1
Figure P4.23
a
q
a a
P +
++