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7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity
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20
21
R 4
QQF
πε=
Chapter 2. CoulombChapter 2. Coulomb’’s Law and Electric Field Intensitys Law and Electric Field Intensity
The Experimental Law of CoulombThe Experimental Law of Coulomb
CoulombCoulomb’ ’ s Laws Law
The magnitude of force between two very small objectsseparated in free space by a distance which is large compared totheir size is given by
Q1 and Q2: positive or negative quantities of charge (coulomb)
R: separation (meters)
ε0: permittivity of free space = 8.854 x 10-12 F/m
= 1/36π x 10-9 F/m
1/4πε0 ≈ 9 x 109
7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity
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In vector form:In vector form:
If QIf Q11 is inis in rr11 and Qand Q22 is inis in rr22, then, then R R 1212 = r= r22 -- rr11 is the directed lineis the directed line
segment from Qsegment from Q11 to Qto Q22..
FF22 = force experienced by Q= force experienced by Q22::
122 aF2
120
21
R 4
πε=
aa1212 is the unit vector in the direction ofis the unit vector in the direction of R R 1212 ==|| 12
12
rrrr
−−
211221 aaFF2
210
122
120
21
R 4
R 4
πε
=
πε
−=−=
r1
Q1 Q
2
x
y
z
r2
F2
F1
R12
7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity
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Electric Field IntensityElectric Field Intensity
Consider a chargeConsider a charge QQii located at Plocated at Pii, and another charge Q, and another charge Qtt situated insituated in
the vicinity ofthe vicinity of QQii. The position of Q. The position of Qtt with respect towith respect to QQii isis PPitit. At. At
any positionany position PPitit, Q, Qtt experiences a force due toexperiences a force due to QQii
itt aF2
it0
ti
P4
πε=
The forceThe force exerted by Qexerted by Qii per unitper unit
charge ischarge is
itt a
F2
it0
i
t P4
Q
Q πε
=
The right side of the equation is a vector field called theThe right side of the equation is a vector field called the electric fieldelectric field
intensity due tointensity due to QQii..
The unit of electric field intensityThe unit of electric field intensity EE is N/C, or V/mis N/C, or V/m
Qi
Qt
Pit
Ft
7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity
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For a charge Q at the origin and using the spherical coordinateFor a charge Q at the origin and using the spherical coordinate
system, the electric fieldsystem, the electric field EE at any given point is given byat any given point is given by
raE 20r4
Qπε
=
For a charge atFor a charge at rrmm,, EE atat rr isis
m
marrE 2
0 ||4
Q
−πε=
aamm is a unit vector in the directionis a unit vector in the direction
ofof rr -- rrmm
x
y
z
Q
rm
r
Note: If Q is positive,Note: If Q is positive, EE “ “radiatesradiates” ” fromfrom
the charge. If Q is negative,the charge. If Q is negative, EE pointspoints
into the charge.into the charge.
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In there an n charges, the electric field atIn there an n charges, the electric field at rr isis
∑= −πε
=n
1i
i2
i0
i
||4
Q)( a
rr
rE
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Example:Example:
A 1uC charge is at the origin and a 2uC charge is at P(0,10,0). A 1uC charge is at the origin and a 2uC charge is at P(0,10,0). WhatWhat
is the electric field at Q(6,4,5)? If ais the electric field at Q(6,4,5)? If a – –0.5 uC charge is placed at Q,0.5 uC charge is placed at Q,
how much force would it experience?how much force would it experience?
ROQ R
PQP
Q
1 uC 2 uC
R R OQOQ = <6,4,5>= <6,4,5> R R PQPQ = <6,4,5>= <6,4,5> -- <0,10,0> = <6,<0,10,0> = <6,--6,5>6,5>
775.877546R 222OQ ==++=
849.9975)6(6R 222PQ ==+−+=
7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity
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zyxzyx
OQOQ 570.0456.0684.0
77.8
546
R αaa
αaaR a
OQ++=
++==
zyx
zyx
PQPQ 508.0609.0609.085.9
566
R αaa
αaaR
a
PQ
+−=
+−
==
V/m53.6623.5384.79
)570.0456.0684.0()77(4
101
zyx
zyx0
6
QO
aaa
aaaE
++=
++πε
×=
−
V/m18.9491.11291.112
)508.0609.0609.0()97(4
102
zyx
zyx0
6
QP
aaa
aaaE
+−=
+−πε
×=
−
V/m71.16068.5975.192 zyx
QPQOQ
aaa
EEE
+−=
=+=
7/26/2019 Ch02_Coulomb's Law and Electric Field Intensity
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EPQ
EOQ P
-0.5 uC
1 uC 2 uC
EQ
FQ
uN36.8084.2938.96
)71.16068.5975.192(105.0F
:EQFthen,QFESince
zyx
zyx6
Q
aaa
aaa
−+−=
+−×−=
==
−v
v v
r
r
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Cathode Ray TubeCathode Ray Tube
The electric fieldsThe electric fields
formed by theformed by theparallel platesparallel plates
deflect the electrondeflect the electron
beam generated atbeam generated at
the back of thethe back of the
tube.tube.
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Electric Field Due to a Continuous Charge DistributionElectric Field Due to a Continuous Charge Distribution
DefineDefine ρρvv = volume charge density == volume charge density =
∆∆vv contains the chargecontains the charge ∆∆QQ
The total charge Q isThe total charge Q is
v
Qlim
0v ∆
∆
→∆
ρ==vol
vvol
dvdQQ
Due to an incremental chargeDue to an incremental charge ∆∆Q, the incremental electric fieldQ, the incremental electric field ∆∆E isE is
||||4
v)(
||||4
Q2
0
v2
0 r'r
r'r
r'r
r'
r'r
r'r
r'r ∆E
−
−
−πε
∆ρ=
−
−
−πε
∆=
rr = position of the point in question= position of the point in question
rr’ ’ = location of= location of ∆∆QQ
To get the field at the point, add theTo get the field at the point, add the
contribution of allcontribution of all ∆∆QsQs
r'
rorigin
∆Q
∆∆∆∆Er - r'
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IfIf ∆∆vv is shrunk so it will approach zero (N will approach infinity),is shrunk so it will approach zero (N will approach infinity), thethe
summation becomes an integral:summation becomes an integral:
∑= −
−
−πε
∆ρ=
N
1i2
0
v
||||4
v)()(
r'r
r'r
r'r
r'rE
∫−
−
−πε
ρ=
vol2
0
v
||||4
dv)()(
r'r
r'r
r'r
r'rE
||||4
v)(
20
v
r'r
r'r
r'r
r'∆E
−
−
−πε
∆ρ=
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Field of a Line ChargeField of a Line Charge
DefineDefine ρρLL = line charge density (C/m)= line charge density (C/m)
Consider a uniform line charge along the zConsider a uniform line charge along the z--axis:axis:
Due to symmetryDue to symmetry, the electric, the electric
fieldfield is a function ofis a function of ρρ and in theand in the
aaρρρρρρρρ direction only.direction only.
dE2
dQ1 = ρ
L dL
dQ2 = ρ
L dL
dE1
z-axis
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r
r'R = r - r'
dE
dEρρρρ
dEz
x
y
z
(0, 0, z) dQ = ρL dz
rr == ρρ aaρρρρρρρρ
rr’ ’ = z= z aazz
R R == rr -- rr’ ’ == ρρ aaρρρρρρρρ -- zz aazz
2222
0
L
z
z
)z(4
dz
+ρ
−ρ
+ρπε
ρ=
zρ aadE
SinceSince EE is directed alongis directed along aaρρρρρρρρ only, theonly, the
zz--component may be ignoredcomponent may be ignored whenwhen
solving for the total electric field:solving for the total electric field:
ρ2 /3220
L
)z(4
dzd aE
+ρπε
ρρ=
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ρaE2 /322
0
L
)z(4
dzd
+ρπε
ρρ= ∫
∞
∞− +ρπε
ρρ== ρρ aEE
2 /3220
L
)z(4
dz
Evaluate the integral using change of variable:Evaluate the integral using change of variable:
Let z =Let z = ρρ tantan αα -->> dzdz == ρρ secsec22 αα ddαα
ρρρ
ρρ
ρ
aaa
aa
aE
ρπε
ρ=α
ρπε
ρ=αα
ρπε
ρ=
αα
α
ρπε
ρ=α
α+ρ
α
πε
ρ=
αρ+ρπεααρρρ=
π
π−
π
π−
π
π−
π
π−
π
π−
∫
∫∫
∫
0
L2 /
2 /0
L2 /
2 /0
L
2 /
2 /3
2
0
L2 /
2 /2 /32
2
0
L
2 /
2 /2 /3222
0
2
L
2sin
4dcos
4
dsec
sec
4d
)tan1(
sec
4
)tan(4d)sec(
α
ρ
z
tan α = z/ρ
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ρaEρπε
ρ=
0
L
2
Note:Note:
1. The electric field due to an infinite line of charge is dire1. The electric field due to an infinite line of charge is directed radiallycted radially
outward or into to the line charge.outward or into to the line charge.
2. The electric field varies inversely with the distance from t2. The electric field varies inversely with the distance from the linehe line
chargecharge3. If the line charge density is positive, the electric field3. If the line charge density is positive, the electric field “ “emanatesemanates” ”
from the charge. If the line charge density is negative, the elfrom the charge. If the line charge density is negative, the electricectric
fieldfield “ “convergeconvergess” ” to the line charge.to the line charge.
Therefore, the electric field due to aTherefore, the electric field due to a uniform uniform line chargeline charge alongalong
the z the z - - axis axis is equal tois equal to
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Field of a Sheet of ChargeField of a Sheet of Charge
DefineDefine ρρSS = surface charge density= surface charge density
Due to symmetry, the electricDue to symmetry, the electric
field at a point is not a functionfield at a point is not a function
of y and z, and does not haveof y and z, and does not havecomponents parallel to the ycomponents parallel to the y--zz
plane.plane.
To simplify the solution, treatTo simplify the solution, treat
the vertical strip as athe vertical strip as a uniformuniformline charge. The sheet ofline charge. The sheet of
charge may now be regarded ascharge may now be regarded as
an infinite number of linean infinite number of line
charges placed beside eachcharges placed beside each
other.other.
(x, 0, 0)
y
dy
ρs
Ex
If the surface charge density isIf the surface charge density is ρρss, the line charge density of a, the line charge density of a
“ “verticalvertical” ” strip isstrip is ρρss dydy
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For a line chargeFor a line charge in the zin the z--axisaxis,,
recall thatrecall that
ρaEρπε
ρ=0
L2
(x, 0, 0)
y
dy
ρs
dEx
β
Therefore:Therefore:
x
x
x x
a
a
adE
)yx(2xdy
yx
x
yx2
dy
cosyx2
dy
220
S
22220
S
220
S
+περ=
++πε
ρ=
β+πε
ρ
=
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x x adE22
0
S
yx
xdy
2 +πε
ρ= dy
yx
1
2
x22
0
S∫∞
∞− +πε
ρ== x x aEE
Note:Note: Cautan
a1
uadu 1
22 +=
+
−∫
x x
aaE
π−−
π
πε
ρ=
πε
ρ=
+∞=
−∞=
− )2
(22x
ytan
x
1
2
x
0
Sy
y
1
0
S
xaE0
S
2ε
ρ−=
At the negative x At the negative x--axis (or at theaxis (or at the “ “back back ” ” of the sheet of charge),of the sheet of charge),
xaE0
S
2ε
ρ=
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In general, the electric field due to an infinite sheet of chargIn general, the electric field due to an infinite sheet of charge is equal toe is equal to
N0
S
2aE
ε
ρ=
NoteNotess::1.1. aaNN is a unit vector perpendicular and pointing away from theis a unit vector perpendicular and pointing away from the
surfacesurface
2.2. The electric fieldThe electric field intensity is constantintensity is constant..
3.3. If theIf the surfacesurface charge density is positive, the electric fieldcharge density is positive, the electric field
“ “emanatesemanates” ” from thefrom the sheet of chargesheet of charge. If the line charge density is. If the line charge density is
negative, the electric fieldnegative, the electric field “ “is intois into” ” to theto the surfacesurface charge.charge.
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Example:Example:
A line charge with charge density equal to 10 nC/m is at x = 4, A line charge with charge density equal to 10 nC/m is at x = 4, z = 3. Az = 3. A
sheet of charge with surface charge density equal tosheet of charge with surface charge density equal to --1 nC/m1 nC/m22 is at the xyis at the xy--
plane. What is the electric field at P(2,3,5)?plane. What is the electric field at P(2,3,5)?
P
EL
ES
x
y
z Side view:Side view: EL
ES
D
x
z
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zxzx
D
22
zx
707.0707.0828.2
22
828.22)2(D
22
aaaa
a
aaD
+−=+−
=
=+−=
+−=E
L
ES
D
x
z
V/m939.44939.44)707.0707.0()828.2(2
1010zxzx
0
9
L aaaaE +−=+−πε
×=
−
V/m472.562
101 zz0
9
S aaE −=ε
×−=−
V/m53.1194.44 zxSLP aaEEE −−=+=
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Streamlines and Sketches of FieldsStreamlines and Sketches of Fields
Given an expression ofGiven an expression of EE, how will one draw (sketch) the field?, how will one draw (sketch) the field?
Take, for example, a point charge.Take, for example, a point charge.
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The arrows show the direction of the field at every point alongThe arrows show the direction of the field at every point along thethe
line, and the separation of the lines is inversely proportionalline, and the separation of the lines is inversely proportional to theto thestrength of the field.strength of the field.
The lines are calledThe lines are called streamlinesstreamlines..
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Given a two dimensional field (Given a two dimensional field (EEzz = 0), the equation of a streamline is= 0), the equation of a streamline is
obtained by solving the differential equationobtained by solving the differential equation
dxdy
E
E
x
y =
E
Ex
Ey
∆y
∆x
x
y
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Example.Example. The electric field intensity is given asThe electric field intensity is given as
EE = 5e= 5e--2x2x (sin 2y(sin 2y aa x x -- coscos 2y2y aa y y) V/m) V/m
Find the equation of the streamline passing through the point P(Find the equation of the streamline passing through the point P(0.5,0.5,
ππ /10, 0). /10, 0).
Solution:Solution: Solving the differential equation :Solving the differential equation :dx
dy
E
E
x
y=
y2coty2sine5
y2cose5
dx
dyx2
x2−=−=
−
−
y2cosKe
y2sece'K e
)y2ln(sec'Cx2
C)y2ln(sec
2
1x
ydy2tandx
ydy2tandx
x2
x2'Cx2
=
==
=+−
+=−
=−
=−
−+−
∫∫
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To solve for K, use the fact that the streamline passes throughTo solve for K, use the fact that the streamline passes through P (0.5,P (0.5,
ππ /10, 0): /10, 0):
KeKe2(0.5)2(0.5) == coscos ππ /5 /5K = 0.298K = 0.298
Therefore, the equation of the streamline through P isTherefore, the equation of the streamline through P is
0.298 e0.298 e2x2x == coscos 2y2y
y2cosKe x2 =