Post on 31-Dec-2015
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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?
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.
23-2 The Electric field 23-2 The Electric field
Definition of Electric FieldDefinition of Electric Field
0q
FE
0q
FE
23-3 Electric Field Lines23-3 Electric Field Lines
Field lines originate on positive charges and terminate on negative charges.
field lines for positive point field lines for positive point chargecharge
field lines for two equal field lines for two equal positive point chargepositive point charge
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
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
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
Sample problem 32-2 Sample problem 32-2
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:
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
CHECKPOINTCHECKPOINT
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
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
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
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
2
322
0
2
02
322
0
4
2
4cos
Rz
Rz
dsRz
zdEE
R
2/32204 Rz
qE
2/32204 Rz
qE
If z>>R: 204
1
z
qE
204
1
z
qE
Sample problem 23-3Sample problem 23-3
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
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:
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
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
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
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.
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.
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
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:
23-9 A Dipole in an electric Fi23-9 A Dipole in an electric Fieldeld
Ep
Ep
Potential Energy of an Electric Potential Energy of an Electric DipoleDipole
EpU
EpU
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:
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)