23 Electric Fields
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23 Electric Fields23 Electric Fields
Suppose we fix a positive charge q1 in place and then put a second positive point charge q2 near it.Since the charges do not touch,how can q1 exert a force on q2?
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23-1 Charges and Forces:A Clos23-1 Charges and Forces:A Close Looke Look
The question about action at a distance
can be answered by saying that q1 sets
up an electric field in the space
surrounding it.
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23-2 The Electric field 23-2 The Electric field
Definition of Electric FieldDefinition of Electric Field
0q
FE
0q
FE
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23-3 Electric Field Lines23-3 Electric Field Lines
Field lines originate on positive charges and terminate on negative charges.
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field lines for positive point field lines for positive point chargecharge
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field lines for two equal field lines for two equal positive point chargepositive point charge
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field lines for a positive and a negative field lines for a positive and a negative point charge that are equal in point charge that are equal in
magnitudemagnitude
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23-4 The Electric Field Due T23-4 The Electric Field Due To a Point Chargeo a Point Charge
20
||
4
1
r
qE
20
||
4
1
r
qE
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If we place a positive test charge q0 nearn point charges q1,q2 ,…,qn,then,fromEq.22-7,the net force from the n pointCharges acting on the test charge is
0F
.002010 nFFFF
.002010 nFFFF
0
0
0
02
0
01
0
0
q
F
q
F
q
F
q
FE n
0
0
0
02
0
01
0
0
q
F
q
F
q
F
q
FE n
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Sample problem 32-2 Sample problem 32-2
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20
1
2
4
1
d
QE
20
1
2
4
1
d
QE
241
2
20 d
QE
241
2
20 d
QE
20
3
4
4
1
d
QE
20
3
4
4
1
d
QE
Step one:Step one:
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20
20
20
21
4
4
12
4
12
4
1
d
Q
d
Q
d
QEE
20
20
20
21
4
4
12
4
12
4
1
d
Q
d
Q
d
QEE
Step two:Step two:
Step three:Step three:
22
0
33
04
93.6866.0
4
4
12
30cos22
d
Q
d
Q
EEE ox
22
0
33
04
93.6866.0
4
4
12
30cos22
d
Q
d
Q
EEE ox
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CHECKPOINTCHECKPOINT
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23-5 The Electric Field Due to 23-5 The Electric Field Due to an Electric Dipole an Electric Dipole
2
0
2
0
20
20
21
421
4
4
1
4
1
dz
q
dz
q
r
q
r
q
EEE
2
0
2
0
20
20
21
421
4
4
1
4
1
dz
q
dz
q
r
q
r
q
EEE
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22
20 2
12
14 z
d
z
d
z
qE
22
20 2
12
14 z
d
z
d
z
qE
...1...1
4 20 z
d
z
d
z
qE
...1...1
4 20 z
d
z
d
z
qE
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302
1
z
pE
302
1
z
pE
30
20 2
12
4 z
qd
z
d
z
qE
30
20 2
12
4 z
qd
z
d
z
qE
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23-6 The Electric Field Due to 23-6 The Electric Field Due to a Line of Chargea Line of Charge
220
20
20
4
1
4
1
4
1
Rz
ds
r
ds
r
dqdE
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2
322
0
2
02
322
0
4
2
4cos
Rz
Rz
dsRz
zdEE
R
2/32204 Rz
qE
2/32204 Rz
qE
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If z>>R: 204
1
z
qE
204
1
z
qE
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Sample problem 23-3Sample problem 23-3
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20
20 4
1
4
1
r
ds
r
dqdE
2
02
0 4
1
4
1
r
ds
r
dqdE
dsr
dEdEx
cos4
1cos
20
dsr
dEdEx
cos4
1cos
20
Step one:Step one:
Step two:Step two:
Step three:Step three:
Step four:Step four:
dsdq dsdq
rdds rdds
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rrd
rdEE x
0
60
60
20 4
73.1cos
4
10
0
rrd
rdEE x
0
60
60
20 4
73.1cos
4
10
0
r
Q
r
Q
length
ech 477.0
3/2
arg
r
Q
r
Q
length
ech 477.0
3/2
arg
Step five:Step five:
Step six:Step six:
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Step seven:Step seven:
2
02
0 4
83.0
4
477.073.1
r
Q
r
QE
20
20 4
83.0
4
477.073.1
r
Q
r
QE
ir
QE ˆ
4
83.02
0 i
r
QE ˆ
4
83.02
0
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23-7 The Electric Field 23-7 The Electric Field Due to a Charged DiskDue to a Charged Disk
rdrdAdq 2 rdrdAdq 2
23
220
23
220
2
4
4
2
rz
rdr
rz
rdrzdE
z
23
220
23
220
2
4
4
2
rz
rdr
rz
rdrzdE
z
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drrrzdEERz
24
23
0
22
0
drrrzdEERz
24
23
0
22
0
Taking the limits,we fine:
220
12 Rz
zE
220
12 Rz
zE
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If we let R→∞ while keeping z finite:
02
E02
E
This is the electric field produced by an infinite sheet of uniform charge located onone side of a nonconductor such as plastic.
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Field Due to a Continuous Field Due to a Continuous Charge DistributionCharge Distribution
The electric field due to a continuous charge distribution is found by treating charge elements as point charges and then summing,via integration the electric field vectors produced by all the charge elements.
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23-8 A Point Charge in an Ele23-8 A Point Charge in an Electric Fieldctric Field
EqF
EqF
The electrostatic force acting on a chargedparticle located in an external electric field has the direction of if the charge q of theparticle is positive and has the opposite direction if q is negative.
F
E
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Sample problem 23-4Sample problem 23-4
m
QE
m
Fay
m
QE
m
Fay
2
2
1tay y 2
2
1tay y
tvL x tvL x
mmmv
QELy
x
64.02
2
mmmv
QELy
x
64.02
2
Step one:Step one:
Step two:Step two:
Step three:Step three:
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23-9 A Dipole in an electric Fi23-9 A Dipole in an electric Fieldeld
Ep
Ep
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Potential Energy of an Electric Potential Energy of an Electric DipoleDipole
EpU
EpU
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Sample problem 23-5Sample problem 23-5
))(10( deqdp ))(10( deqdp
pmm
C
mC
e
pd
9.3109.3
1060.110
102.6
1012
19
30
pmm
C
mC
e
pd
9.3109.3
1060.110
102.6
1012
19
30
(a)
Step one:Step one:
Step two:Step two:
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mN
CNmC
pE
26
0430
103.9
90sin/105.1102.6
sin
mN
CNmC
pE
26
0430
103.9
90sin/105.1102.6
sin
J
CNmCpE
pEpE
UUWa
25
430
0
0
109.1
/105.1102.622
0cos180cos
0180
J
CNmCpE
pEpE
UUWa
25
430
0
0
109.1
/105.1102.622
0cos180cos
0180
(b)
(c)