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ELECTRIC FIELDS
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A CLOSER LOOK AT CHARGES AND FORCES
Recall Coulombs Law:2
212112
rqqkFF
F12= force exerted by q2on q1 when q1 is placed at a distance
r from q2F21= force exerted by q1 on q2when q2is placed at a distance
r from q1 action at a distance
How does either charge knows that the other charge is placedat a distance r from it?
Ans: One charge sets up an electric fieldaround it.
The second charge placed nearby interacts with the field and
experiences a force due to the field of the first one.
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(Fig b) A charged object sets up an electric field Eat point P
(and everywhere else around it of course).
(Fig a) A charge (positive) q0placed at P interacts with the field
and experiences a force Fin the direction of the field E
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THE ELECTRIC FIELD
defined)field(electric0q
FE
Electric Field : force per unit charge that a charge qo
experiences in a region of space where the electric field is setup by a source charge.
Unit: Newtons per Coulomb; N/C
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Electric Field Lines
used to visualize electric fields
the direction of a straight line or of the tangent to a curved linegives the direction of the field at a given point
the density of the lines is proportional to the magnitude of the
field
applet1
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Electric field lines (a) extend away from positive
charge and (b) toward negative charge.
applet1
http://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.htmlhttp://www.cco.caltech.edu/~phys1/java/phys1/EField/EField.html8/12/2019 02 Electric Fields
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(a) a positive test charge near a portion
of an infinitely large non-conducting
sheet(or plane) with uniform charge
distribution on one side
(b) Efield vector at location of testcharge;
field lines in space near the
sheet.
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Electric dipole
pair of charges of equal magnitude but opposite sign
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A pair of like charges with
field lines around them
2d Field
3d Field
http://www.falstad.com/vector2de/http://www.falstad.com/vector3de/http://www.falstad.com/vector3de/http://www.falstad.com/vector2de/8/12/2019 02 Electric Fields
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Electric Field Due To A Point Charge
2
0
r
qqkF
Force between q and q0 Force on q0due to q
2
0 r
qk
q
FE electric field due to point charge q
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Net force on q0due to charges q1, q2, q3, qn :
F0= F01+ F02+ F03+ .+ F0n
n
n
qqqqq
EEEE
FFFFFE
.......
........
q.......,,q,q,qtodueqofpositionatfieldNet
321
0
0
0
03
0
02
0
01
0
0
n3210
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Electric Field Due To A Dipole
Field at point P(a distance z from dipole midpoint):
2
0
2
0 41
41
rq
rq
EEE
2
2
2
2
21
2
1
z
dzdzr
2
2
2
2
21
2
1
z
dzdzr
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Hence we get:
1/ 22 2
20
20
20
1 1
2 24
2 21 1 (using binomial expansion)
2 1! 2 1!4
1 14
q d dE
z zz
q d d
z zz
q d d
z zz
3
0
2
02
12
4 z
qd
z
d
z
qE
magnitudemomentdipoleelectricqdp
E
dipole)(electric2
13
0 z
pE
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Electric Field Due to A Line Charge
= line charge distribution= Q/L = dq/ds
dq = ds
2220
2
0
2
0
41
4
1
4
1
Rzds
r
ds
r
dqdE
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2122cos
Rz
z
r
z
ds
Rz
zdE
2322
04cos
23220
2
0
2322
0
4
2
4
cos
Rz
Rz
ds
Rz
z
dEE
R
Rq 2
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Find the electric field at point P due to the bent rod with
charge Q uniformly distributed along the bent rod.
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2
0
2
0 41
41
rds
rdqdE
)(rcos4
1cos
4
1cos
2
0
2
0
d
rds
rdEdEx
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r
r
rd
r
dr
dEE x
0
00
0
60
60
0
60
600
60
60
2
0
4
73.1
6060sin4
sin4
cos4
1
cos4
1
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R
drrrzz
dEE0
2/322
0
24
1
.2and,,Setting1
2322
mX
dXX
drrdXmrzXm
m
R
rzzE
021
2/1
22
04
disk)(charged12 220
Rz
z
E
sheet)(infinite2
,RAs0
E
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A Point Charge In An Electric Field
EF q E= electric field
q = point charge
F = electrostatic force; same direction as Efor (+)q;
opposite direction of Efor (-)q
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Dipole In An Electric Field
sinsin2
sin2
:dipoleonnet torqueofMagnitude
Fdd
Fd
F
UsingF=qEandp=qd :
sinpE
dipole)on(torqueEp
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Potential Energy of an Electric Dipole
in the diagram E tends to rotate pclockwise towards = 0
= -pEsinfor clockwise rotation
potential energy is work W done
against Ein rotating pfrom to 900
potential energy is work (-)W doneby Ein the opposite sense, from 900
to .
9090
90
90
sinsin dpEdpE
ddWU
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cos
g,Integratin
pEU
dipole)ofenergy(potentialEp
U