1 engineering electromagnetics essentials chapter 3 basic concepts of static electric fields
TRANSCRIPT
1
Engineering Electromagnetics
EssentialsChapter 3Basic
concepts of static electric
fields
2
Objective
Recapitulation of basic concepts of static electric fields or electrostatics
Topics dealt with
Coulomb’s law Gauss’s law Electric potential Poisson’s and Laplace’s equations
Background
Vector calculus expressions developed in Chapter 2
3
rar
qqF
2
21
4
r
rar
A
B
ABr
1q
2qra
2/1212
212
212 ])()()[( zzyyxxr
zyx azzayyaxxr
)()()( 121212
A: x1, y1, z1
B: x2, y2, z2
Coulomb’s law
:1q
:2q
Point charge at A
Point charge at B
Permittivity of a medium : [C2/N-m2 or F/m]
[Newton or N]
[Coulomb or C]
[Coulomb or C]
[metre or m]
[metre or m]
(That the unit of permittivity is F/m will become clear from the expression for the capacitance of a parallel-plate capacitor to be derived later, the unit of capacitance being Farad or F).
4
Rationalized MKS system of unit is followed.The rationalization factor introduced in Coulomb force expression removes the appearance of this factor from many extensively used expressions derived from Coulomb force expression.
rar
qqF
2
21
4
4
Relative permittivity or dielectric constant
0 r
:0 Permittivity of a free-space medium
r
Coulomb force is maximum in a free-space medium.
1290 10854.810)36/(1 C2/N-m2 or F/m
5
The problem is to find the force on a point charge placed at a corner A of a square on a plane due to three identical point charges each placed at the remaining three corners B , C and D of the square, respectively, located in rectangular system of coordinates, one of the corners, namely, A being at the origin of the coordinates
)0,0(q
)0,(a ),( aa ),0( a
Use Coulomb’s law to find electrostatic force
xa
ya
A
CD
B(0,0) (a,0)
(a,a)(0,a) q q
q q
X
Y
6
xa
ya
A
CD
B(0,0) (a,0)
(a,a)(0,a) q q
q q
X
Y
y
yx
x
aqq
F
aaqqF
aqq
F
20
DA
20
CA
20
BA
)a(4
))((
2
)(
)a2(4
))((
)a(4
))((
yyyx
yxyxyx
xxyx
aaaaa
a
aaaaaaaaa
aaaaa
a
a
a
DA
)0()00(2
)(
a2
)a(
CA
)0()0(a
a
BA
)00()0(
DA
CA
BA
DACABAA FFFF
Force on the point charge at A due to the point charges at B, C and D
AF
7
))()22(
11(
a4 20
2
DACABA yx aaq
FFF
yyx
x aqq
Faaqq
Faqq
F
20
DA20
CA20
BA )a(4
))((;
2
)(
)a2(4
))((;
)a(4
))((
]2
)()[
2
12(
a4)(
20
2
DACABAAyx aaq
FFFF
CAa
CA20
2
A )2
12(
a4a
qF
. (3.10) (magnitude) (unit vector)
Unit vector directed from C to A
Unit vector directed from C to AMagnitude
8
Take up another problem for the use of Coulomb’s lawTwo small identical metallic charged balls, each of charge q and mass m = 1 gm, hung by a long thread of length l = 1 m from a common point (hook), get separated from each other by a distance d = 2 cm due to electrostatic repulsion and is set to an equilibrium on the vertical plane. Find the value of the charge q.
.cossin mgFT
At equilibrium, the tension T in the string balances a component of the Coulomb force and a component of the gravitational force put together as follows:
Also, at equilibrium, a component of the gravitation force will balance a component of the Coulomb force to staify the following relation:
sincos mgF
mgd
q2
0
2
4tan
22)2/(
2/tan
ld
d
From geometry
θθ
θ
θ
F
Fcosθ
Fsinθ+mg cosθ
mg
T
Tsinθ
Tcosθ
q q
l
d
mg sinθ
9
mgd
q2
0
2
4tan
22)2/(
2/tan
ld
d
2/1
22
30
4/
2
ld
mgdq
222
0
2
)2/(
2/
4 ld
d
mgd
q
Equating the right hand sides
dl 2/13
02
l
mgdq
and putting m = 1 gm , l = 1 m and d = 2 cm, you can calculate each charge q as
120 10854.8 Remembering F/m
910088.2 q C nC. 088.2
10
Electric field due to a point charge and electric field due to a
distribution of chargesThe force on a point charge q2 due to the another point charge q1, given by Coulomb’s
law, can be interpreted as the electric field caused by the point charge q1.
rar
qqF
2
21
4
EqF
2
rar
qE
2
1
4
rar
qE
24
(Coulomb’s law)
Electric field due to the point charge q1
In general, the electric field due to point charge q is then given by
Electric field due to a point charge
In general, the electric field due to the point charge q is then given by
EqF
11
An electron moves along x with a velocity 2 cm/s in a region of uniform electric field of 5 kV/cm along z. What is the force on the electron?
Take up the following simple problem
Force on point charge q due to electric field is given by
N 108106.1105 14195zzz aaaEeEeEqF
zaEeE
V/m 10510105kV/cm 5 523 E
C 106.1 19 eq
Answer:Electron carries a negative charge and it may be regarded as a point charge q:
Electric field magnitude is given as 5 kV/cm along z:
12
Electric field due to a long uniform line-charge distribution
Charge is uniformly sprayed over a long line along z axis
Medium is a free-space medium
The length of the line is very large compared to the distance of the point
P is the point where the electric field is sought.
OP = d is the perpendicular distance of P from the long line charge distribution.
O is the foot of perpendicular drawn from P to the long line charge distribution.
A and B are the two points symmetrically located with respect to O on each of which equal amounts of line charge elements are considered (OA = OB).
dzl Charge element each on A and B to be regarded as point charges
l Charge per unit length or line charge density of the line charge distribution
O
A
B
POA = z
OP = d
BPa
na
APa
ldz
l dz
E
13
AP20
AP )AP(4a
dzEd l
BP20
BP )BP(4a
dzEd l
(Element of electric field at P due to charge element at A)
(Element of electric field at P due to charge element at B)
secBPAP d
220
BPAP sec4 d
dzdEdE l
APa
BPa
Unit vector directed from A to P
Unit vector directed from B to P
2222 sec)BP()AP(
tan
d
dz
dddz 2sec
tandz
O
A
B
POA = z
OP = d
BPa
na
APa
ldz
l dz
E
14
cos
)AP(4cos)(
20
APAP
dzdEdE l
)()( BPAP dEdE
Normal components of element of electric field at P due to the charge elements at A and B will add up in the direction of .na
Find out the components of element of electric field, due to a charge element on the line charge distribution, in directions parallel and perpendicular to the line.
Parallel components of element of electric field at P due to the charge elements at A and B equal and in opposite directions cancel out.
cos
)AP(4cos)(
20
BPBP
dzdEdE l
Elements of electric field component will add up in the direction of . na
O
A
B
POA = z
OP = d
BPa
na
APa
ldz
l dz
E
15
Elements of electric field component will add up in the direction of . na
cossec4
cos)AP(4
)(22
02
0AP d
dzdzdE ll
For the length of charge distribution very large compared to the distance d of the point P, you can integrate the above expression between the limits z = 0 and and take twice its value to find the magnitude of the electric field in the direction of . The limits z = 0 corresponds to and z = corresponds to .
na
222
2
sec)AP(
sec
tan
d
dddz
dz
ddd
dddE ll cos
4cos
sec4
sec)(
022
0
2
AP
0 2/
O
A
B
POA = z
OP = d
BPa
na
APa
ldz
l dz
E
16
For the length of charge distribution very large compared to the distance d of the point P, you can integrate the above expression between the limits z = 0 and and take twice its value to find the magnitude of the electric field in the direction of . The limits z = 0 corresponds to and z = corresponds to .
na
0 2/
ddd
dddE ll cos
4cos
sec4
sec)(
022
0
2
AP
2/
00
2/
0 0
cos4
2cos4
2
dd
dd
E ll
dddd
E llll
000
2/0
0 2)01(
2)0sin2/(sin
2sin
2
nl a
dE
02
(Electric field near a long line charge distribution)
Alternatively, the above expression can be very easily derived by Gauss’s law yet to be introduced.
O
A
B
POA = z
OP = d
BPa
na
APa
ldz
l dz
E
17
Electric field due to a uniform planar surface-charge distribution with a circular boundary at a perpendicular distance from the centre of the circle
Uniform surface charge densitys
Radius of the circular boundary of charge distributiona
z ),0,0(P z Perpendicular distance of where to find the electric field
Ed
r Radius of an annular charge ring
Charge is uniformly sprayed over XY plane within a circular boundary
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
18
Approach:
Step 1: Find the sub-element of electric field at the point P due the sub-element of charge on an annular ring sub-element of infinitesimal radial thickness.
Step 2: Find the element of electric field at the point P due to the charge on the entire annular charge ring by integrating the sub-element of electric field obtained in Step 1.
Step 3: Integrate the element of electric field obtained in Step 2 to find the electric field at P due to the entire circular charge distribution.
Ed
Ed
E
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
19
s Surface charge density
Step 1:
drrd Area of the ring sub-element at S
Ed
rar
qE
24
SPsr
drrdq s
SPr aa
Charge on the ring sub-element at S drrds
SPs a
s
drrdEd
2
04
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
20
SPs a
s
drrdEd
2
04
s
azara zr
SP
SP
OPSO
SP
SP
zs
rszrs a
s
drrzda
s
drdr
s
azar
s
drrdEd
30
30
2
20 444
zs
rs ad
s
rzdrad
s
drrEd
2
03
0
2
03
0
2
44
z
sr
s as
drrzda
s
drdrEdEd
3
03
0
2
44
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
Step 2: Ed
21
2
03
0
2
03
0
2
44da
s
rzdrad
s
drrEd z
sr
s
zs
rs ad
s
rzdrad
s
drrEd
2
03
0
2
03
0
2
44
changes with the position of S, that is, from one sub-element of charge to another on the annular charge ring since their directions change even though their magnitudes remain the same as unity. On the contrary does not change so. Therefore,
ra
za
za can be taken outside the integration while cannot be so done.ra
ra
The magnitudes of are the same as unity but their directions are opposite at two diametrically opposite charge sub-elements. That makes the first integral vanish.
2
03
04da
s
rzdrEd z
s
zs
zs
zs
as
rzdra
s
rzdr
as
rzdrEd
30
30
30
22
4
)02(4
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
22
zs
zs a
s
rdrza
s
rzdrEd
30
30 22
tanzr
dzdr 2sec
seczs
zs adEd
sin2 0
za
zs adEdE
/tan
00
1
sin2
The upper limit of integration corresponds to the annular charge ring at the periphery of circular charge distribution at . ar
The lower limit of integration corresponds to the annular charge ring at the centre of circular charge distribution at .
0r
zs
zzas a
z
aaE
0costancos2
]cos[2
1
0
/tan0
0
1
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
Step 3: E
23
zs
zzas a
z
aaE
0costancos2
]cos[2
1
0
/tan0
0
1
zs
zs a
za
za
za
zE
220
220
12
12
zs aE
02
Z
X
Yr
dr
P(0,0 , z)
z
O
Sa
s
ra
za
d
ns aE
02
Alternatively, the above expression can be very easily derived by Gauss’s law yet to be introduced.
The electric field near a large area of charge distributionlarge sheet of charge becomes
az
)( az
nz aa
Interpreting
Electric field near a large sheet of charge becomes
24
Gauss’s law
Life becomes simple while finding electric field in problems, if you can do away with the integrations involved in the application of Coulomb’s law to such problems!
For problems that enjoy geometrical symmetryrectangular, cylindrical or spherical one can apply Gauss’s law to make the solution to such problem very simple. You can avoid carrying out complex integrations to find electric field which would be otherwise necessary if you would apply Coulomb’s law to such problems.
Let us state Gauss’s law and prove it starting from Coulomb’s law.
Gauss’s law can be stated in terms of electric field and electric displacement.
25
Electric displacement is related to electric field .
ED
Dd
E
D
dSaDSdDd nD
dS
na
Element of flux of electric displacement
Element of area
Unit vector normal to the element of area
S
n
SS
DD dSaDSdDd
Element of flux of electric displacement Dd
Flux of electric displacement D
Flux of electric displacement through a closed volume
S S
nD dSaDSdD
D
Involves closed surface integral
na
D
dS
26
S S
nD QdSaDSdD
Flux of electric displacement through a closed volume
S S
nD dSaDSdD
D
Involves closed surface integral
Outward flux of electric displacement D
S S
nE
QdSaESdE
E
Q Charge enclosed
Gauss’s law states that the outward flux of electric displacement through the surface of a volume enclosing a charge —estimated by a closed surface integral of over the area of the volume enclosure—is equal to the charge enclosed, .
QD
D
Q
Outward flux of electric displacement
ED
D of in terms stated becan law sGauss'
ty)permittivi medium ( Gauss’s law
27
• Let us prove Gauss’s law
• Let us apply the law to problems that enjoy geometrical symmetry rectangular, cylindrical or sphericalProof of Gauss’s law
S S
nD dSaDSdD
Take a point charge inside a volume enclosure
q
rar
qE
24
ED
rr ar
qa
r
qED
22 44
S
nrD dSaar
q 24
S S
D dSr
qSdD
cos
4 2
cos nr aa
ra
is directed from O to P.
OP = r
O
P
qSolid angle element dΩ
Area element dS
θ
E
na
ra
28
S S S
nD r
dSqdS
r
qdSaD
22
cos
4cos
4
dr
dS2
cos
Charge inside a volume enclosureq
dq
dSaDSS
nD )4)(4
(4
(Solid angle element subtended by area element at P)
What will be the outward flux of electric displacement if a number of point charges, instead of a single point charge q, are present within the volume enclosure?
S S
nD dSaDSdD
OP = r
O
P
qSolid angle element dΩ
Area element dS
θ
E
na
ra
qdSaDS
n
(Gauss’s law)
29
Outward flux D of electric displacement if a number of point charges q1, q2, q3, ……qn,, instead of a single point charge q, are present within the volume enclosure
S S
nn
S SS
nD dq
dq
dq
dq
dSaD
4
........444 3
32
21
1
dq
S
D )4)(4
(4
S S
n
S S
dddd 4.........321
ly.respective........ charges
pointby subtended angle solid of
elements are,......,
321
321
n
n
qqqq
dddd
Qqqqq
qqqqdSaD
n
n
S
nD
........
44
........44
44
44
321
321
S S
nD dSaDSdD
QdSaDS
n
(Gauss’s law)
ED
Q
dSaES
n
You have thus proved Gauss’s law)!
(Gauss’s law)
30
Gauss’s in terms of volume charge density
nqqqqQ ........321
n
S
nD qqqqQdSaD ........321
dqqqqQdSaD n
S
nD ........321
delement volumethe
over constant regarded is
dqqqqQ n........321
ddSaDS
n
d
dSaES
n
(Gauss’s law in terms of volume charge density)
31
Application of Gauss’s law to a problem that enjoys rectangular symmetry
na
Cross section of planar charge sprayed over a very large area on XY plane. X is perpendicular to the plane of the paper directed away from the reader. Gaussian volume (rectangular parallelepiped is also shown.
Find the electric field at a point near a large planar sheet of charge of uniform surface charge density.
Approach:
Step 1: Appreciate that the direction of electric field at a point P near the sheet of charge is in the direction perpendicular to the the plane of charge in the direction of .
In other words the electric field is along Z axis: .naa
z
Step 2: Find the magnitude of electric field E at P using Gauss’s law.
naEE
Srep 3: . .
Y
Planar sheet of charge
Gaussian rectangular
parallelopiped
ZO P
(b)
A B
CD
Xzn aa
zn aa
32
Cross section of a planar sheet of charge and Gaussian volume (rectangular parallelepiped)
Consider on the planar sheet of charge, supposedly positive, a pair of identical charge elements each treated as a point charge symmetrically placed with respect to the point P.
Appreciate the following with the help of Coulomb’s law,
(i) The magnitudes of electric field at P due to the pair of charge elements are equal.
(ii) The direction of electric field at P due to a charge element of the pair of charge elements is from the charge element to the point.
(iii) The components of electric field due to the charge elements of the pair parallel to the plane of the sheet of charge are equal in magnitude and opposite in direction and they cancel out.
(iv) The components of electric field due to the charge elements of the pair perpendicular to the plane of the sheet of charge add up in the perpendicular direction.
naEE
(Step 1)
Y
Planar sheet of charge
Gaussian rectangular
parallelopiped
ZO P
(b)
A B
CD
Xzn aa
zn aa
33
BC) face(right field Electric naE
The areas of the faces AD (left) and BC (right) are equal and each much smaller than the area of the large sheet of charge considered. Therefore, the magnitude of electric field can be regarded as constant over each of these faces.
The faces AD (left) and BC (right) are equidistant from the sheet of charge. Following step 1
AD) face(left field Electric naE
Cross section of a planar sheet of charge and Gaussian volume (rectangular parallelepiped)
Y
Planar sheet of charge
Gaussian rectangular
parallelopiped
ZO P
(b)
A B
CD
Xzn aa
zn aa
34
Left face AD and right face BC of closed rectangular parallelepiped Gaussian volume are equidistant from the planar charge distribution. Right face BC passes through P.
Cross section of a planar sheet of charge and Gaussian volume (rectangular parallelepiped) passing through P
BC faceright tonormaloutwardly r unit vecto na
AD faceleft tonormaloutwardly r unit vecto na
0)faceleft()faceright(
)()( S
dSaaEdSaaE s
S
nn
S
nn
Q
dSaES
n
Upper and lower faces of Gaussian parallelepiped do not contribute to the surface integral since the electric field is parallel to these faces being perpendicular to the sheet of charge.
enclosed charge ofsheet of area
BC of area AD of area
S
SQ s
Gauss’s law
charge. ofsheet planar the
ofdensity charge surface theis s
Y
Planar sheet of charge
Gaussian rectangular
parallelopiped
ZO P
(b)
A B
CD
Xzn aa
zn aa
35
ns aE
02
Cross section of a planar sheet of charge and Gaussian volume (rectangular parallelepiped) passing through P
0)faceleft()faceright(
)()( S
dSaaEdSaaE s
S
nn
S
nn
(Step 2)
(Step 3)
Y
Planar sheet of charge
Gaussian rectangular
parallelopiped
ZO P
(b)
A B
CD
Xzn aa
zn aa
The areas of the faces AD (left) and BC (right) are equal and each much smaller than the area of the large sheet of charge considered. Therefore, the electric field can be regarded as constant over each of these faces and taken outside the integral.
0
)2)( S
dSEEdSEdS s
02sE
(Electric field near a large planar sheet of charge)
density) charge surface theis ( s
36
Gaussian rectangular parallelepiped enclosure with its left face passing through the conductor and its right face through the point where to find the electric field
Find the electric field at a point near a large planar conductor of uniform surface charge density.
Q
dSaES
n
Follow the same approach and the symbols as those used to find the electric field near a large planar sheet of charge.
(Gauss’s law)
0)faceleft()faceright(
)()( S
dSaaEdSaaE s
S
nn
S
nn
Left face of parallelepiped passes through the conductor where the electric field E is absent and the integrand of the second term is null. Therefore, the second term is null,
0)faceright( S
dSaaE s
S
nn
ty)permittivi spacefree ( 0
Y
Planar conductor
Gaussian rectangular parallelopiped
Z
37
0)faceright( S
dSaaE s
S
nn
0 S
ESdSEEdS s
0sE
ns
n aaEE
0
(Electric field near a large charged planar conductor)
Gaussian rectangular parallelepiped enclosure with its left face passing through the conductor and its right face through the point where to find the electric field
density) charge surface theis ( s
Y
Planar conductor
Gaussian rectangular parallelopiped
Z
38
Application of Gauss’s law to a problem that enjoys cylindrical symmetryFind the electric field near a
long line charge distribution of uniform line charge density.
rl
A B
CD
n ra aP
Z
Line ChargeGaussian Cylinder
Q
dSaES
n
Follow the same approach as that used to find the electric field near a large planar sheet of charge.
(Gauss’s law)
rn aa
rn aEaEE
SS
rr
S
n EdSdSaaEdSaE
enclosed charge lQ l
density) charge lineor length unit per charge ( l
Q
EdSS
l
EdS l
S
39
l
EdS l
S
rl
A B
CD
n ra aP
Z
Line ChargeGaussian Cylinder
ty)permittivi spacefree ( 0 l
rlE l2
rE L
02
rn aEaEE
nL a
rE
02
density) charge lineor length unit per charge ( l
(Electric field near a long line charge distribution)
l
EdS l
S
l
dSE l
S
rldSS
2
40
Application of Gauss’s law to a problem that enjoys spherical symmetryFind the electric field due to a
point charge.Q
dSaES
n
(Gauss’s law)
rn aa
rn aEaEE
SS
rr
S
n EdSdSaaEdSaE
enclosed charge 1 qQ
1q
dSES
Q
EdSS
24 rdSS
21
4 r
qE
rar
qE
2
1
4
raEE
(Electric field due to a point charge)
41
Electric field and potentialGravitational
potential energy
Work required to be done by spending energy to lift a particle to a height from the ground (surface of the earth) against the gravitational force is stored in the form of gravitational potential energy.
The ground surface is taken to be at zero reference gravitational potential.
Electric potential energy
Work required to be done by spending energy to move a charged particlea point
charge Qfrom infinity against the force due to an electric field to a point is stored in the form of electric potential energy W .
Taking the zero reference potential taken as the potential at infinity, the potential V at the point is given by V)or (J/C
Q
WV
The potential at a point is thus numerically equal to the potential energy of a unit point charge placed at the point.
42
O
P
OP r
PPP dr
E
ra
rar
qE
2
04
rr aFar
qQEQF
QF
204
Pat positive supposedly chargepoint on the Force
drdW distance ofelement an through P toP from Q move todone be torequired work ofElement
)(OP distance aat Oat chargepoint todue Pat field Electric
)(OP distance aat Oat chargepoint todue Ppoint at the Potential
rrqE
rrqV
204
r
QqF
P) toO from directedr unit vecto theis ( ra
rar
qE
2
04
Potential at a point due to a point charge
43
drr
QqFdrdW
204
r
r
r
Qqdr
r
Qqdr
r
QqW
rrr 0002
02
0 4
1
4
1
4
1
44
r
q
Q
rQq
Q
WV
0
0
4
4
204
r
QqF
is Ppoint the toupinfinity from chargepoint themove todone be torequired Work QW
) chargepoint thefrom distance aat (Potential qr
O
P
OP r
PPP dr
E
ra
44
Electric field as negative gradient of potential
A
B
V V+dV
dR
AB =dR
AN =dn na
θN
nE Ea
EQF
A and B are nearby points separated by element of distance AB=dr
The potential is V at every point on the equipotential V.
The potential is V+dV at every point on the equipotential V+dV.
The same electric field exists at A and B and the same force is experienced by a point charge at A and B.
Q
WV
(Force on a point charge at A or B)
B)at (potential
A)at (potential
dVVV
VV
B
A
)( dVVQQV
QVQV
B
A
QdVdVVQQVQVQVdW BA )(
positive. makes that negative is or if
A toB from charge themove todone be torequired isWork
dWdVVV
Q
BA
45
A
B
V V+dV
dR
AB =dR
AN =dn na
θN
nE Ea
QdVdW RdFdW
RdFQdV
EQF
RdEdV
RdEQQdV
dnn
VdV
RdEdnn
V
RdadRdn n
cos
RdERdan
Vn
Ean
Vn
nan
VE
VE
nan
VV
Electric field is the negative gradient of potential.
(Definition of gradient of a scalar here V)
46
A
B
V V+dV
dR
AB =dR
AN =dn na
θN
nE Ea
In the limit BA,
0 RdE
Aat ialequipotent the
toal tangentibecomes
V
Rd
0 RdEdV
In the limit BA,
thusandA at ialequipotent the tohence and tonormal becomes VRdE
.capacitors of capacince thefind to
expression apply the ointerest t of be It will VE
Electric field is normal to an equipotential
naEE
47
Capacitance of a parallel-plate capacitor (using the expression for electric field in terms of the gradient of potential)
AV BV
d
A B
Source of Potential
0Z Z d
X
Z
Y
zyx az
Va
y
Va
x
VVE
zaz
VE
l.dimensiona-one becomes problem theand
0//;0/ can take We xxz
capacitor) theinside field electric of (magnitude z
VE
A
B
V
V
z
dz
EdzdV0
Let us consider large dimensions of plates compared to the distance between the plates.
capacitor) theinside field (electric
IntegratingPotential of the plate A at z = 0 is VA
Potential of the plate B at z = d is VB
Distance between plates = d
The geometry of the problem enjoys rectangular symmetry.
dz
dVE .on only depends zV
48
ns
ns
aE
aE
ty permittivi of dielctric a be tomedium thengInterpreti
.
asconducor charged anear field electric eearlier th obtained We
0
side) hand(Left
A
B
A
B
V
V
BAVV VVVdV
sE
A
B
V
V
z
dz
EdzdV0
0z
dz
Edz
dVV sBA
expression in the together sides handright andleft thePutting
side) hand(Right )0)((000
dddzdzEdz ssz
dz
sz
dz
sz
dz
sides. handright andleft theevaluate usLet
49
dVV sBA
obtained weexpression in the
together sides handright andleft thePutting
dA
QdVV s
BA
plateeach of Area
plateeach on Charge
A
QA
Qs
d
A
VV
QC
BA
Capacitance C of the capacitor becomes
The unit of capacitance is Farad or F and accordingly the unit of permittivity is F/m.
(Expression for the capacitance of a parallel-plate capacitor)
50
Take up the following simple problemWhat is the area of the plates required for constructing a parallel-plate
capacitor of capacitance 100 pF using two metal plates using a mica spacer of relative permittivity 5.4 that separates the plates by a distance of 10-2 cm.
d
AC
r
CdCdA
0
Answer:
(given)
F 10100pF 100C
4.5
m 10 cm 10
12
42
r
d
.cm 1.2m101.24.510854.8
1010100 22412
412
0
r
CdCdA
51
Capacitance of a coaxial cable (using the expression for electric field in terms of the gradient of potential)
cable coaxial theofconductor inner theof Radius a
)(2 0
brar
E L
The magnitude of radial electric field in the region between the inner and outer conductors of a coaxial cable can be found using Gauss’s law as in the problem of finding the electric field due to a long line charge distribution already dealt with. Thus we get
cable coaxial theofconductor outer theof Radius b
zr az
Va
V
ra
r
VVE
1
rar
VE
The geometry of the problem enjoys cylindrical symmetry. The length of the cable is taken large compared to the radial dimensions of the cable. The problem becomes one-dimensional. Thus we have here
0//;0/ zr bra
r
VE
52
r
VE
)(2 0
brar
E L
rdr
dV L
02
a
b
L
VV
V
drr
dV1
2 00
0
at 0 and at limits ebetween th gIntegratin 0 brVarVV
abLV rV ln2 0
00
a
babbaV LLL ln
2)ln(ln
2)ln(ln
2 0000
00
ln
2V
abL
conductorsouter andinner ebetween th difference Potential 0 V
rr
V L
02
.on only depends rV
53
conductors ebetween th difference potential the
by conductor theof length unit per charge the
dividingby obtained is cable theoflength unit per eCapacitanc
0VL
00
ln
2V
abL
length)unit per ce(Capacitan ln
2 0
0
a
bVC L
54
A
B
C
D
P(r,θ,Ф)
θ
O
q
- q
r
X
Y
Z
Electric field of a dipole (using the expression for electric field in terms of the gradient of potential)
Two equal and opposite point charges separated by a distance constitutes a dipole.
l = AB = Length of the dipole being the distance between the point charges
q = Charge of the point charge at A
–q = Charge at the point charge at B
r = Distance of the point P (r,,)where to find the electric fieldfrom the middle O of the dipole.
Let us consider a shot dipole for which l<<r.
The problem enjoys azimuhal symmetry: 0/
55
r
qV
04 ) chargepoint thefrom distance aat (Potential qr
A
B
C
D
P(r,θ,Ф)
θ
O
q
- q
r
X
Y
Z
DPBD
1
OCOP
1
4
BP
1
CP
1
4BP
1
AP
1
4
0
00
q
qqV
rOPDP dipole)short afor ( lr
cos2
cosOAOCl
cos2
BDl
rOPDP
rllr
qV
cos)2/(
1
cos)2/(
1
4 0
2220
0
cos)4/(
cos
4
]cos)2/][(cos)2/([
]cos)2/([cos)2/(
4
lr
lq
rllr
lrrlqV
20
20 4
cos
4
cos
r
p
r
qlV
dipole)short afor ( lr
moment) (dipole qlp
56
204
cos
r
pV
a
V
ra
V
ra
r
VVE r
sin
11
scoordinatepolar -sphericalin Expressing V
0/ The problem enjoys azimuhal symmetry:
a
V
ra
r
VE r
1
moment) (dipole qlp
204
cos
r
pV
20
30
4
sin
4
cos2
r
pV
r
p
r
V
a
r
p
ra
r
pE r
2
03
0 4
sin1
4
cos2
57
a
r
p
ra
r
pE r
2
03
0 4
sin1
4
cos2
)sincos2(4
13
0
aap
rE r
A
B
C
D
P(r,θ,Ф)
θ
O
q
- q
r
X
Y
Z
(expression for electric field due to a shot dipole at a distance r and at an angle from the axis of the dipole in terms of the dipole moment p)
(expression for electric field due to a shot dipole at a distance r on the axis of the dipole ( = 0) in terms of the dipole moment p)
rar
pE
3
02
Putting = 0
58
Poisson’s and Laplace’s equations
S
ndSaD
S
ndSaDLt
0
D
E
ED
Let us take an infinitesimal volume element over which we can take the value of volume charge density and that of electric displacement D to be fairly constant. If we apply Gauss’s law to this volume element, we obtain the following approximate relation:
(Gauss’s law)
ddSaDS
n
S
ndSaD The approximation sign of equality is
given since over the volume element , the value of volume charge density and that of electric displacement D have been regarded as constant.
The relation is made exact and the equality sign is given by taking 0 corresponding to the volume element shrinking to a point.
The left hand side is the definition of the divergence of a vector here electric displacement.
Poisson’s equation
,0For
0
0
E
D
Laplace’s equation
59
Laplaciam form of Poisson’s and Laplace’s equations
E
(Poisson’s equation in terms of electric field)
02 V
VE
)( V
V2
With the help of the relation between the electric field in terms of the gradient of electric potential we can find the expression for Poisson’s equation in terms of potential.
Putting the volume charge density equal to zero in Poisson’s equation so found we get Laplace’s equation in electric potential.
We can solve these expressions for potential and subsequently find the electric field from potential with the help of the relation between the electric field in terms of the gradient of potential.
VV 2)(
(Poisson’s equation in terms of electric potential)
0
(Laplace’s equation in terms of electric potential)
60
Electric field in the region between conductors at different potentials from the solution of Laplace’s equation in Laplacian form Take the problem of a parallel-plate capacitor. Find the expression for electric field in the region between plates.
AV BV
d
A B
Source of Potential
0Z Z d
X
Z
Y02 V (Laplace’s equation in terms of electric
potential)
0//;0/ xxz
222 / zVV
0/ 22 zV .on only depends rV
02
2
dz
Vd
Adz
dV
AdzdV
BAzV
Integration
A is integration constant
B is integration constant
61
BAzV
0at zVV A
dzVV B at
AB VAdBAdV
AVB
d
VVA BA
ABA Vz
d
VVV
zadz
dVE
zBA a
d
VVE
(Expression for the electric field in terms of the potential difference VA VB and the distance d between the plates)
62
Take the problem of a pair of large conducting plates at a difference of potential forming a wedge. Find the expression for electric field in the region between plates.
Z
X
Y
V=Vo
V=0
θo
The problem enjoys cylindrical symmetry, allowing us to treat it in cylindrical coordinates.
platesbetween difference Potential 0 V
angle Wedge0
02 V (Laplace’s equation in terms of electric potential)
0//;0/ zr
.on only depends V
011
2
2
22
2
22
d
Vd
r
V
rV
02
2
dVd
63
02
2
dVd
1Cd
dV
21 CCV
Integration
C1 is integration constant
C2 is integration constant
dCdV 1
0at 0 VV
0at 0 V
20 CV
01 VCV
0010 VC
0
01
VC
00
0 VV
V
)1(0
0
VV
64
(Expression for electric field at a point located between a pair of large conducting plates forming a wedge separated by an infinitesimal insulating gap, in terms of the potential difference V0 between the plates and the wedge angle 0)
)1(0
0
VV
a
d
dV
ra
V
rVE
11
0//;0/ zr
0
0V
d
dV
a
V
rE
0
01
65
Basic concepts of static electric field or electrostatics have been recapitulated in this chapter.
Rationalised MKS system has been introduced while writing the expression for the electrostatic force between two point charges called Coulomb’s law. The rationalisation factor 4 has been introduced in Coulomb’s force expression thereby removing the appearance of the factor 4 from many extensively used expressions derived from Coulomb force expression, for instance, Gauss’s law.
Coulomb’s law has been used to find
electrostatic force on a point charge due to the distribution of other point charges; and
electric field due to a point charge as well as the electric field due to a distribution of point charges.
Finding the electric field by Coulomb’s law has been illustrated in examples of charge distribution such as line and surface charge distributions.
Summarising Notes
66
Gauss’s law greatly simplifies the finding of electric field in problems that enjoy rectangular, cylindrical or spherical symmetry, which otherwise would become quite involved if Coulomb’s law were used instead.
Concept of electric potential has been introduced.
Expression for the electric field in terms of the gradient of electric potential has been derived and illustrated in examples, for instance, in the problem of finding electric field due to a short dipole as well as in the problem of finding the capacitance of a parallel-plate capacitor and the capacitance of a coaxial cable.
Poisson’s and Laplace’s equations have been derived from first principles and their use illustrated in the problems of finding the electric field in the region between a pair of conductors at different electric potentials by solving Laplace’s equation expressed in terms of Laplacian of potential.
Poisson’s equation in terms of Laplacian of potential has been solved to deduce Child-Langmuir’s law for a space-charge-limited diode, which relates the anode current of the diode to the anode potential as well as anode-to-cathode distance of the diode.
Readers are encouraged to go through Chapter 3 of the book for more topics and more worked-out examples and review questions.