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8/8/2019 Field Theory Compiled Basvaraj
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e-Notes by Prof.T.Basavaraj Sri Revanna Siddeshwara Institute of Technology, Bangalore
FIELD THEORY
Sub Code : EC44 IA Marks : 25Hrs/Week : 04 Exam Hrs. : 03
Total Hrs. : 52 Exam Marks : 100
1. Electric Fields 18 hours
a. Coulombs law and Electric field intensity
b. Electric flux density, Gauss law and divergence
c. Energy and potential
d. Conductors, dielectrics and capacitance
e. Poissons and Laplaces equations
2. Magnetic fields 14 hoursa. The steady magnetic field
b. Magnetic forces, materials and inductance
3. Time varying fields and Maxwells equations 5 hours
4. Electromagnetic waves 15 hours
Text Books :
William H Hayt Jr and John A Buck, Engineering Electromagnetics, Tata McGraw-Hill,
6th Edition, 2001
Reference books :
John Krauss and Daniel A Fleisch, Electromagnetics with Application, McGraw-Hill,
5th Edition, 1999
Guru and Hiziroglu, Electromagnetics Field theory fundamentals, Thomson Asia Pvt. Ltd
I Edition, 2001
Joseph Edminster, Electromagnetics, Schaum Outline Series, McGraw-Hill
Edward C Jordan and Keith G Balmain, Electromagnetic Waves and Radiating Systems,
Prentice-Hall of India, II Edition, 1968, Reprint 2002.
David K Cheng, Field and Wave Electromagnetics, Pearson Education Ais II Edition, 1989,
Indian Repr-01
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Introduction to Field Theory
The behavior of a physical device subjected to electric field can be studied either by Field
approach or by Circuit approach. The Circuit approach uses discrete circuit parameters like
RLCM, voltage and current sources. At higher frequencies (MHz or GHz) parameters would no
longer be discrete. They may become non linear also depending on material property andstrength of v and i associated. This makes circuit approach to be difficult and may not give very
accurate results.
Thus at high frequencies, Field approach is necessary to get a better understanding of
performance of the device.
FIELD THEORY
The Vector approach provides better insight into the various aspects of Electromagnetic
phenomenon. Vector analysis is therefore an essential tool for the study of Field Theory.
The Vector Analysis comprises of Vector Algebra and Vector Calculus.
Any physical quantity may be Scalar quantity or Vector quantity. A Scalar quantity is
specified by magnitude only while for a Vector quantity requires both magnitude and direction
to be specified.
Examples :
Scalar quantity : Mass, Time, Charge, Density, Potential, Energy etc.,
Represented by alphabets A, B, q, t etc
Vector quantity : Electric field, force, velocity, acceleration, weight etc., represented by alphabets
with arrow on top.
etc.,B,E,B,A
Vector algebra : If C,B,A are vectors and m, n are scalars then
(1) Addition
AssociativC)BA()CB(A
CommutativABBA++=++
+=+
(2) Subtraction
)B(-AB-A +=
(3) Multiplication by a scalar
Am)BA(m
nAmAn)(m
)A(mn)A(nm
mAAm
A vector is represented graphically by a directed line segment.
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A Unit vector is a vector of unit magnitude and directed along that vector.
Aa is a Unit vector along the direction of A .
Thus, the graphical representation of A and Aa are
AaAorA/AaAlso
ctorUnit veAVector
A
AA
==
Product of two or more vectors :
(1) Dot Product ( . )
0,B}COSA{ORCOSB(AB.A =
B B
CosA
A CosB A
A . B = B . A (A Scalar quantity)
(2) CROSS PRODUCT (X)
C = A x B = nSINBA
xA)CB(xA
AxB-BxA
BAsuchthatdirected
perpendicurunitvectoisnand
betweenangleis''whereEx.,
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CO-ORDINATE SYSTEMS :
For an explicit representation of a vector quantity, a co-ordinate system is essential.
Different systems used :
Sl.No. System Co-ordinate variables Unit vectors
1. Rectangular x, y, z ax , ay , az2. Cylindrical , , z a , a , az3. Spherical r, , ar , a , a
These are ORTHOGONAL i.e., unit vectors in such system of co-ordinates are mutually
perpendicular in the right circular way.
r,z,zyxi.e.,
RECTANGULAR CO-ORDINATE SYSTEM :
Z
x=0 plane
azp
y=0 Y
plane ay
ax z=0 plane
X
yxz
xzy
zyx
xzzyyx
aaxa
aaxa
aaxa
0a.aa.aa.a
=
=
=
===
az is in direction of advance of a right circular screw as it is turned from a x to ay
Co-ordinate variable x is intersection of planes OYX and OXZ i.e, z = 0 & y = 0
Location of point P :
If the point P is at a distance of r from O, then
If the components of r along X, Y, Z are x, y, z then
arazayaxrrzyx =++=
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Equation of Vector AB :
componentsareB&B,Band careA&A,Awhere
ABorBABA
aBBOBand
AaAAOAIf
zys zys
xx
xx
+
=
=
Dot and Cross Products :
wegrouping,andby termtermproducts'Cross'Taking
aBa(Bx)aAaAa(ABxA
aBa(B.)aAaAa(AB.A
yxxzzyyxx
yyxxzzyyxx
+++=
+++=
zyx
zyx
zyx
BBB
AAA
aaa
BxA =
ABlengthVector
where
AB
ABa
alongVectorUnit
represents)CxB(.A(ii)
then0BxA
Costhen0B.A(i)
nonareCandB,AIf
CBA)CxB(.A
AB
x
x
x
Differential length, surface and volume elements in rectangular co-ordinate systems
zyx
zyx
adzadyadxrd
z
rdy
rdx
x
rrd
azayaxr
+=
+
=
++=
y
Differential length 1-----]d zd yd x[rd 1 /2222 ++=
Differential surface element, sd
1. zadxdy:ztor
2. zadxdy:ztor ------ 2
3. zadxdy:ztor
Differential Volume element
dv = dx dy dz ------ 3
z
dx p
5
B
B AB
0 A A
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p dz
dy
rrdr+
0 y
x
Other Co-ordinate systems :-
Depending on the geometry of problem it is easier if we use the appropriate co-ordinate system
than to use the Cartesian co-ordinate system always. For problems having cylindrical symmetry
cylindrical co-ordinate system is to be used while for applications having spherical symmetry
spherical co-ordinate system is preferred.
Cylindrical Co-ordiante systems :-z
P(, , z) x = Cos y = Sin
az r z = z
ap r yzz
y / xtan
yx
1-
22
==
+=
x
ar
aSin-r SinaCosr rdrrd SiaCosr azayaxr
z
x
x
xyx
z
Thus unit vectors in (, , z) systems can be expressed in (x,y,z) system as
2222
z
zzz
y
xyx
(dz))d(drdand
adzadadrd,Further
orthogonalareaanda,a;aa
SinaaCosaSin-a
CosaaSinaCosa
++=
++=
=
=+=
=+=
yx
Differential areas :
zz
adz)(dads
3-------a.)d((dz)ads
a.)d()(dads
=
=
=
Differential volume :
6
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4-----dzdddor
(dz))d()(dd
==
7
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Spherical Co-ordinate Systems :-
Z X = r Sin Cos
Y = r Sin Sin z p Z = r Cos R
r
0 y Y
x r Sin
X
dddrSinrvd
ddrrSd
ddrSinrSd
ddSinrSd
adSinradradrRd
dR
dR
drr
RRd
aCosaSin-R
/R
a
aSinaSinCosaCosCosR
/R
a
aCosaSinSinaCosSinr
R/
r
Ra
aCosraSinSinraCosSinrR
2
2
2
r
r
yx
zyx
zyxr
zyx
=
=
=
=
++=
+
+
=
+=
=
+=
=
++=
=
++=
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General Orthogonal Curvilinear Co-ordinates :-
z u1 a3 u3
a1 u2a2
y
x
Co-ordinate Variables : (u1 , u2, u3) ;
Here
u1 is Intersection of surfaces u2 = C & u3 = C
u2 is Intersection of surfaces u1 = C & u3 = Cu3 is Intersection of surfaces u1 = C & u2 = C
2
1
1
321
1
1
x
321
h,u
Rh
h,h,hwhere
duh
u
RRdthen
yaxRIfOrthogonalisSystem ubnitvectarea,a,a
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Co-ordinate Variables, unit Vectors and Scale factors in different systems
Systems Co-ordinate Variables Unit Vector Scale factors
General u1 u2 u3 a1 a2 a3 h1 h2 h3
Rectangular x y z ax ay az 1 1 1
Cylindrical z a a az 1 1
Spherical r ar a a 1 r r sin
Transformation equations (x,y,z interms of cylindrical and spherical co-ordinate system
variables)
Cylindrical : x = Cos , y = Sin , z = z ; 0, 0 2 - < z <
Spherical
x = r Sin Cos , y = r Sin Cos , z = r Sin r 0 , 0 , 0 2
10
)u,u,(uAAand)u,u,(uA
)u,u,(uAAwherefieldVectoraisaAaAaAA
fieldScalara)u,u,u(VVere
AhAhAhuuu
ahahah
hhh
1Ax
)Ah(hu
)Ah(hu
)Ah(hu
hhh
1A
au
v
h
1a
u
v
h
1a
u
v
h
1V
3213332122
32111332211
321
332211
321
332211
321
321
3
231
2
132
1321
3
33
2
22
1
11
==
=++=
=
=
+
+
=
+
+
=
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Vector Transformation from Rectangular to Spherical :
11
= ++=+=++=
z
y
x
zyx
zyx
rzryrxr
zyxr
rr
RrrRS
zzyyxxR
A
A
A
a.aa.aa.a
a.aa.aa.a
a.aa.aa.a
A
A
A
asA,A,AtorelatedareA,A,Awe
aAaAaA
a)a.A(a)aA(a)aA(A:cal
aAaAaAA:rRangula
R
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Field Theory
A field is a region where any object experiences a force. The study of performance in the
presence of Electric field )E( , Magnetic field ( ) is the essence of EM Theory.
P1 : Obtain the equation for the line between the points P(1,2,3) and Q (2,-2,1)
zyx a2-a4-aPQ =
P2 : Obtain unit vector from the origin to G (2, -2, 1)
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Problems on Vector Analysis
Examples :-
1. Obtain the vector equation for the line PQ between the points P (1,2,3)m and Q (2,
-2, 1)mZ
PQ P (1,2,3)
Q(2,-2,-1)
0
Y
X
)a2-a4-a(a3)-(-1a2)-(-2a1)-(2
a)z-(za)y-(ya)x-(xPQvectorThe
zyx
zyx
zpqypqxpq
=
++=
++=
2. Obtain unit vector from origin to G (2,-2,-1)
G
G
0
a0.667-a(0.667a
(-2)2G
G
Ga,orunit vectThe
a2-a(2
a)0-(xGvectorThe
yxg
2
g
x
xg
=
=
=
3. Given
zyx
zyx
a5a2-a4-B
aa3-a2A
==
BxA(2)andB.A(1)find
Solution :
)a5a2-a(-4.)aa3-a(2B.A(1)zyxzyx
++=
= - 8 + 6 + 5 = 3
Since ax . ax = ay . ay = az . az = 0 and ax ay = ay az = az ax = 0
(2)524
132
aaa
BxA
zyx
=
= (-13 ax -14 ay - 16 az)
4. Find the distance between A( 2, /6, 0) and B = ( 1, /2,2)Soln : The points are given in Cylindrical Co-ordinate (, , z). To find the distance betweentwo points, the co-ordinates are to be in Cartesian (rectangular). The corresponding
rectangular co-ordinates are ( Cos , Sin , z)
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1.73AB)(
a1.73-
a1.73-
)A-(BAB
a6
CosB&
a6
Cos2A
2
x
x
xx
x
5. Find the distance between A( 1, /4, 0) and B = ( 1, 3 /4, )Soln : The specified co-ordinates (r, , ) are spherical. Writing in rectangular, they are
(r Sin Cos , r Sin Sin , r Cos ).Therefore, A & B in rectangular co-ordinates,
0.5(2 AB.AB(AB a1.414- )A-(BAB a0.707( Cos43Sin(B a0.707( Cos4Sin(1A
x
xx
xx
6. Find a unit vector along AB in Problem 5 above.
AB
ABa AB = = [ - 1.414 ax + (-0.707) ay + (-0.707) az]
1.732
1
= )a408.0a0.408-a0.816-( zyx
7. Transform ordinates.-ColCylindricainFinto)a6a8-a(10F zyx +=
Soln :
a)a.F(a)a.F(a)a.F(F zzppCyl ++=
yx
yx
yx
Siny
Cosx
Sin8-Cos(10
a8-a(10[a8-a10([ a8-a[(10
)a6a(12.8
a6a38.66)](-Cos8-38.66)(-Sin10[-a]38.66)-(Sin8-38.66)(-Cos10[F
z
zpCyl
+=
++=
8. Transform azax-ayB zyx += into Cylindrical Co-ordinates.
zxxCyl
x
aza- CosSin[ -aSin([ -aSin([ (B.a)a(B.B
Cos-aSinB
Siny,Cosx
9. Transform xa5 into Spherical Co-ordinates.
SinCos5aCosSin5
)]aCosaSin(-.a5[
CosaCos(Cos.a5[
SinaCos(Sin.a5[
a(A.a)a(A.a)a(A.A
r
yxx
xx
xx
rrSph
+=+++= ++=
10. Transform to Cylindrical Co-ordinates z),,(Qata4x)-(y-ay)x(2G yx +=
Soln :
14
x
y
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Siny,Cosx
Siny)x(2-[
-Cosy)2x([
(y-ay)x(2[
-(y-ay)x(2[G
(G.a)a(G.G
x
xCyl
Cyl
=
a)CosSin3-SinCos4(
a)Sin-CosSin5Cos2(
a]Cos4CosSin-Sin-CosSin2-[
a]CosSin4Sin-CosSinCos2[
a]Cos)Cos4-Sin(-Sin)SinCos2(-[
a]Sin)Cos4-Sin(-Cos)SinCos2([G
22
22
22
22
Cyl
+
+=
++
++=
+++=
11. Find a unit vector from ( 10, 3 /4, /6) to (5, /4, )Soln :
A(r, , ) expressed in rectangular co-ordinates
ABBAa 9.65AB A-BAB .a6.12A 4Sin5B Sin10A osSinrOA
AB
x
12. Transform a6a8-a10F zyx += into F in Spherical Co-orindates.
)a0.783a5.38a(11.529F
a0.781)x8-0.625-x(-10
a(0.625))x0.42x8-0.781x0.42x10(
a(-0.625))x0.9x8-0.781x0.9x(10F
0.781(-38.66)CosCos0.4264.69CosCos
0.625-(-38.66)SinSin0.964.69SinSin
38.66-10
8-tan
64.89200
6Cos
r
zCos;2006810r
a)Cos8-Sin10(-
a)Sin6-SinCos8-CosCos(10a)Cos6SinSin8-CosSin(10
a)a.F(a)a.F(a)a.F(F
aCosaSin-a
aSin-aSinCosaCosCosa
aCosaSinSinaCosSina
r
r
01-
1-1-222
r
rrSph
yx
zyx
zyxr
++=
+
+
=
====
====
==
====++=
+
+ +=
++=
+=
+=
++=
Line Integrals
In general orthogonal Curvilinear Co-ordinate system
++=++=
++=
C
333
C
222
C C
111
332211
333222111
duFhduFhduFhdl.F
aFaFaFF
aduhaduhaduhdl
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Conservative Field A field is said to be conservative if it is such that 0dl.C
=
contour.closedaaroundtakenisitifzeroisandaandbbetweenpotentialtherepresentdl.
andintensityfieldelectrictherepresent-E
thenflux,ticelectrostaisIf).!pathon thedependnot(does(a)-b)(ddl.
b
a
b
a
=
==
= 0dl.i.e.,Therefore ES flux field is Conservative.
EXAMPLES :
13. Evaluate line integral = dl.aI
where yx ax)-(yay)(xa ++=
(4,2)Bto(1,1)Afromxyalong
2 =
Soln : adyadxdl yx +=
++===
++=
2
1
2
1
22
2
dy)y-(ydy2yy)(ydl.a
dxdydy2orxy
dy)x-(ydxy)(xdl.a
++=2
1
223dy)y-yy2y(2
++=2
1
23dyy)yy(2
3
111
3
11-
3
212
3
4-2
3
88
2
1
3
1
2
1-
2
2
3
2
2
2
2
y
3
y
24
y2
234
2
1
234
==
++=
++
++=
++
//
=
16
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14. Evaluate the Integral =S
ds.EI where aradiusofhunisphereisSandaxE x=
Soln:
If S is hemisphere of radius a, then S is defined by
3
a2x
3
2xadCosdSinads.E
20,2/0
ddCosSinads.E
ddSina.a)CosSin(ads.Eds.E
CosSinax;aCosSinxE
a)a(E.a)a(E.a)a.(EE
addSinads
ad)Sin(a)d(ads;0z,azyx
3/2
0
2
0
3233
23
2
r
2
r
rr
rr
r
2
r
2222
===
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where r, r1 , r2 .. rm are the vector distances of q, q1 , qm from origin, 0.
mr-r is distance between charge qm and q.
ma is unit vector in the direction of line joining qm to q.
Electric field is the region or vicinity of a charged body where a test charge experiences a
force. It is expressed as a scalar function of co-ordinates variables. This can be illustrated by
drawing force lines and these may be termed as Electric Flux represented by and unit iscoulomb (C).
Electric Flux Density )D(
is the measure of cluster of electric lines of force. It is the
number of lines of force per unit area of cross section.
i.e.,
==S
2surfacetonormalorunit vectisnwhereCdsnDorc/m
A
D
Electric Field Intensity )E( at any point is the electric force on a unit +ve charge at that
point.
i.e., c/Nar4
q
q
FE 12
10
1
==
CEDorc/ND
c/Na
r4
q
1 0
0
12
1
1
0
=
=
=
in
vacuum
In any medium other than vacuum, the field Intensity at a point distant r m from + Q C is
Car4
QDorCEDand
m)/Vor(c/Nar4
QE
r20
r2
0
==
=
r
r
Thus D is independent of medium, while E depends on the property of medium.
r E+QC q = 1 C (Test Charge)
Source charge
E
E
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0 r , m
Electric Field Intensity E for different charge configurations
1. E due to Array of Discrete charges
Let Q, Q1 , Q2 , Qn be +ve charges at P, P1 , P2 , .. Pn . It is required to find E
at P.
Q1 1r nE
P1
Q2 2r P 2E
1E
P2 1rQn 0
Pr nr
m/Var-r
Q
4
1E m2
m
m
0
r =
2. E due to continuous volume charge distribution
Ra R P
v C / m3
The charge is uniformly distributed within in a closed surface with a volume charge density
of v C / m3 i.e,
vd
QdanddvQ
V
V
V ==
C/Na)r-(r4
)(rE
aR4
V
aR4
Q
E
R210
1
r
R20
R20
1
=
=
=
V
V
V
Ra is unit vector directed from source to filed point.
3. Electric field intensity E due to a line charge of infinite length with a line charge
densityof l C / m Ra
P
R
dl l C / m
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L
C/NaR
dl
4
1E R
L
2
0
p =l
4. E due to a surface charge with density of S C / m2
Ra
P (Field point)
ds R
(Source charge)
C/NaR
ds
4
1E R
S
2
0
p =S
Electrical Potential (V) The work done in moving a unit +ve charge from Infinity to that is
called the Electric Potential at that point. Its unit is volt (V).
Electric Potential Difference (V12) is the work done in moving a unit +ve charge from one
point to (1) another (2) in an electric field.
Relation between E and V
If the electric potential at a point is expressed as a Scalar function of co-ordinate variables
(say x,y,z) then V = V(x,y,z)
V-E(2)and(1)From
----dl.VdV
dyy
Vdx
x
VdVAlso,
----dl.E-dlq
f-dV
== +===
Determination of electric potential V at a point P due to a point charge of + Q C
la
dRR+
0 + Q R P Ra
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At point P, C/NaR4
QE R2
0=
Therefore, the force f on a unit charge at P.
NaR4
QEx1f R2
0
p
==
The work done in moving a unit charge over a distance dl in the electric field is
scalar(aVoltR4
QV
-)a.a(R4
dlQ-V
dl.E-dl.f-dV
2
0
P
lR
R
2
0
p
====
Electric Potential Difference between two points P & Q distant Rp and Rq from 0 is
voltR
1-
R
1
4
Q)V-(VV
qp0
qppq
==
Electric Potential at a point due to different charge configurations.
1. Discretecharges
. Q1
. PQ2 Rm
Qm
VR
Q
4
1V
n
1 m
m
0
1P
=
2. Linecharge
x P Vdl
R
l
4
1V
l
l
0
2P
=
l C / m
3. Surfacecharge
x P VR
ds
4
1V
S
S
0
3P =
s C / m2
4. Volumecharge
x P
21
v
C/ m3
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R VR
dv
4
1V
V
V
0
4P =
5. Combination of above V5P = V1P + V2P + V3P + V4P
Equipotential Surface : All the points in space at which the potential has same value lie on a
surface called as Equipotential Surface.
Thus for a point change Q at origin the spherical surface with the centre of sphere at the
origin, is the equipotential surface.
Sphere of
Radius , R
R
P
equipotential surfaces
Q
V
0 R
Potential at every point on the spherical surface is
potentialsurfaceialequipotenttwopotentialofdifferenceisV
voltR4
QV
PQ
0
R
=
Gausss law : The surface integral of normal component of D emerging from a closed
surface is equal to the charge contained in the space bounded by the surface.
i.e., =S
(1)CQdsn.D
where S is called the Gaussian Surface.
By Divergence Theorem,
=S V
dvD.dsn.D ----------- (2)
Also, =V
V dvQ ---------- (3)
From 1, 2 & 3,D. = ----------- (4) is point form (or differential form) of Gausss law while
equation (1) is Integral form of Gauss law.
Poissons equation and Laplace equation
22
0
+Q
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In equation 4, ED 0=
equationLaplace0V0,If
equationPoisson-V
/V)(-.or/E.
2
0
2
00
==
==
Till now, we have discussed (1) Colulombs law (2) Gauss law and (3) Laplace equation.
The determination of E and V can be carried out by using any one of the above relations.
However, the method of Coulombs law is fundamental in approach while the other two use
the physical concepts involved in the problem.
(1) Coulombs law : Here E is found as force f per unit charge. Thus for the simple case
of point charge of Q C,
=
=
l
20
VoltdlEV
MV/R
Q
4
1E
(2) Gausss law : An appropriate Gaussian surface S is chosen. The charge enclosed is
determined. Then
=
=
l
S
enc
voltdlEVAlso
determinedareEhenceandDThen
QdsnD
(3) Laplace equation : The Laplace equation 0V2 = is solved subjecting to differentboundary conditions to get V. Then, V-E =
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Solutions to Problems on Electrostatics :-
1. Data : Q1 = 12 C , Q2 = 2 C , Q3 = 3 C at the corners of equilateral triangle d m.To find : 3QonF
Solution :
y
23
2323
y
13
1313
232
2132
1
0
33
231333
zy3
y2
1
3
2
1
321
21321
a0.866a0.5-r-r
r-ra
d
0.866ad0.5
r-r
r-ra
ad
Qa
d
Q
4
QF
FFFisFforceThea
0.866a
d0.5r
adr
0r
md)0.866d,0.5(0,P
m0)d,(0,P
m(0,0,0)P
norigintheat
withplane,YZinliePandP,PIf
andP,PatlieQandQ,QLet
+==
+==
+
=
+=
+=
=
=
=
=
=
Substituting,
)a0 . 9 2 4a( 0 . 3 8aw h e r eNa0 . 3 5 4F
1 3 . 1 11 2 . 1 25
a1 2 . 1 2a5
d
1 0x2 7
0 . 8a0 .5-(d
1 0x2)a0 . 8 6 6a0 .5(
d
1 0x1 21 0x9)1 0x( 3F
zyFF3
22
zy
2
3-
y2
- 6
zy2
- 696-
3
+==
+
+=
+++=
2. Data : At the point P, the potential is V)zy(xV222
p++=
To find :
(1)
VforexpressiongeneralusingbyV(3)(1,1,2)QandP(1,0.2)givenV(2)EPQPQp
Solution :
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/mV]a3zay2ax2[-
az
Va
y
Va
x
V-V-E)1(
z
2
yx
z
p
y
p
x
p
pp
++=
+
+
==
]V1-V-VV(3)
V1-0y0
dz3zdy2ydx2xdl.E-V)2(
PQPQ
02
1
1
0
1
2
2
2
P
Q
pPQ
==
=++=
++==
3. Data : Q = 64.4 nC at A (-4, 2, -3) m A
To find : m(0,0,0)0atE 0E Solution : 0
20a29
9x64.4E
)AO(29
1
AO
AOa
2)-(0a4)(0AO
[
(AO)36
10x4
10x64.4
/Na(AO)4
QE
AO0
AO
x
29-
9-
AO2
0
0
=
==
+=
=
=
4. Q1 = 100 C at P1 (0.03 , 0.08 , - 0.02) mQ2 = 0.12 C at P2 (- 0.03 , 0.01 , 0.04) mF12 = Force on Q2 due to Q1 = ?
Solution :
Na9F
0.11
0.121x10x100F
0.636-a0.545-(a
0.06-(
(-0.03R-RR
aR4
QQF
1212
2
6-
12
x12
1212
122
120
2112
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5. Q1 = 2 x 10-9 C , Q2 = - 0.5 x 10
-9 C C
(1) R12 = 4 x 10-2 m , ?F12 =
(2) Q1 & Q2 are brought in contact and separated by R12 = 4 x 10-2 m ?F`12 =
Solution :
(1)
(repulsiveN12.66F
x4(x36
10x4
)10x1.5(F
contactintobroughtWhen(2)
10x4(x36
10x4
10x0.5-x10x2F
`
12
9-
29-`
12
-9-
--9
12
=
=
=
6. Y
P3x x
P1 P2x x
0 X
Q1 = Q2 = Q3 = Q4 = 20 CQP = 200 C at P(0,0,3) m
P1 = (0, 0 , 0) m P2 = (4, 0, 0) m
P3 = (4, 4, 0) m P4 = (0, 4, 0) m
FP = ?
Solution :
R
Q
36104
QF
a3a4-R
a4-a4-R
a3a4-R
Ra3R
FFF
2
1p
1
9-
p
p
zy4p
yx3p
zx2p
1pz1p
2p1pp
=
=
=
=
=
=
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6-
zy2
zyx2zx2z296- 12
)a
0 . 6a
0 . 8-(5
1
)a0 .a0 . 6-a( - 0 . 66 . 4
1
)a0 . 6a0 . 8-(5
1
a3
1
1 0x9 x1 0x0 02
++
++++
=
)a0 . 6a0 . 8-(
2 5
1 0 0
)a0 .a0 .-a( - 0 .4 0 . 9
1 0 0)a0 . 6a0 . 8( -
2 5
1 0 0a
9
1 0 0
1 0xx 1 0x 1 0x 9 x 1 00 0 x 1 02
zy
zyxzxz2-6-996-
++
++++
=
Na17.23N)a17a1.7-a1.7(-
(11.11a3.2)-(-1.5266.4
1a)526.12.3(36.0
pzyx
y2x
=+=
++=
7. Data : Q1 , Q2 & Q3 at the corners of equilateral triangle of side 1 m.
Q1 = - 1 C, Q2 = -2 C , Q3 = - 3 C
To find : E at the bisecting point between Q2 & Q3 .Solution :
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Z
P1 Q1
Q2 P E1P Q3
YP2 E2P E3P P3
[
[
a410x9
1.3310x9
36
104
1E
a0.5-R
a0.5R
a0.866-R
R
4
1
EEE
3
3
9-P
y3P
y2P
z1P
10
2P1PP
=
=
=
=
+=
=
=
+=
Z
E1P EP ( EP ) = 37.9 k V / m
Y
E2P (E3P E2P) E3P
8. Data Pl = 25 n C /m on (-3, y, 4) line in free space and P : (2,15,3) m
To find : EPSolution :
Z l = 25 n C / m
A
R (2, 15, 3) m
P
Y
28
P1 : (0, 0.5, 0.866) m
P2 : (0, 0, 0) m
P3 : (0, 1, 0) m
P : (0, 0.5, 0) m
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X
The line charge is parallel to Y axis. Therefore EPY = 0
m/Va88.23E
36
102
25a2E
0.167-a(0.834R
Ra
(3a(-3))-(2APR
RP
R
0
lP
xR
x
=
=
==
==
R
9. Data : P1 (2, 2, 0) m ; P2 (0, 1, 2) m ; P3 (1, 0, 2) m
Q2 = 10 C ; Q3 = - 10 CTo find : E1 , V1
Solution :
R
Q
4
1V
14.14]aa[10
(0.679
1010x9E
a2a2aR
a2-aa(2R
4
1EEE
21
2
0
1
yx
3
6-9
1
zyx31
zyx21
0
21211
= += = ++=+=
=+=
V3000VmV/14.14E11
==
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10. Data : Q1 = 10 C at P1 (0, 1, 2) m ; Q2 = - 5 C at P2 (-1, 1, 3) mP3 (0, 2, 0) m
To find : (1) 0Efor0)0,(0,atQ(2)E 3x3 =Solution :
[[ a1.23-
16-a(8
1010x9E
0.3RRa
(R
Ra
a1)(0R
a1)-(2R
R
4
1E(1)
x
y
9
3
23
23
23
13
13
13
x23
y13
10
3
=
=
=
=
=
+=
=
=
zerobecannotE
a1.23-E
a
R
Q
a
R
Q
a
R
Q
10x9E(2)
3x
x3x
03203
23223
2
13213
19
3
=
++=
11. Data : Q2 = 121 x 10-9 C at P2 (-0.02, 0.01, 0.04) m
Q1 = 110 x 10-9 C at P1 (0.03, 0.08, 0.02) m
P3 (0, 2, 0) m
To find : F12Solution :
R]a[
10x7.8x36
104
10x110x10x121F
-a0.05-R;NaR4
QQF
1212
3-9-
9-9-
12
x12122
120
21
12
=
=
=
Na0.015F1212
=
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12. Given V = (50 x2yz + 20y2) volt in free space
Find VP , m3)-2,(1,PataandE npP
Solution :
[
/Va6.562a150a600 a(-3)(2)100-E -azyx100-E Vx-V-E
(-3)(2)(1)50V
P
x
P
x
2
P
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Additional Problems
A1. Find the electric field intensity E at P (0, -h, 0) due to an infinite line charge of density
l C / m along Z axis.
+Z
A dz
APR
z
dEPy P Y
dEPz h 0
d PE
Pa X
-
Solution :
Source : Line charge l C / m. Field point : P (0, -h, 0)
[
aR
h
R4
dz-dE
aR
h-
R4
dzdE
-ah-R
1
R
Ra
aR4
dQdE
2
0
lPy
y2
0
lP
yR
R2
0
P
=
=
==
=
Expressing all distances in terms of fixed distance h,
h = R Cos or R = h Sec ; z = h tan , dz = h sec2 d
0]Cos[h4
E
dSinh4
-Sech
tanhxSech4
dSech-dE
ah2
-2x
h4
-]Sin[
h4
-E
dCosh4
-Cosx
Sech4
dSech-dE
2/
2/-
0
lPz
0
l
22
0
2l
Pz
y
0
l
0
l2/
2/-
0
lPy
0
l
22
0
2
lPy
=
=
=
/
/
=
=
=+
=
=
=
mV /ah2
-E y
0
l
=
An alternate approach uses cylindrical co-ordinate system since this yields a more general
insight into the problem.
Z +
A dz
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z R
P ( , / 2, 0)0 Y
P
/ 2
APX
-
4E dE(ii)
4E
4
dE(i)
dz,tanz
Taking
4dE(i)
4
dzdE
dQ
Rwhere
4
dQdE
intensityfieldThe
isdzdQ
P P
P
lP
P
0
lP
0
P
l
z z
m/Va2
E
0
lP
=
Thus, directioninradialisE
A2. Find the electric field intensity E at (0, -h, 0) due to a line charge of finite length along
Z axis between A (0, 0, z1) and B(0, 0, z2)
Z
B (0, 0, z2)
dz
P 2 A(0, 0, z1) 1
Y
X
Solution :
[
2-,
2
t-fromextendingislinetheIf
)Sin-(Sinh4
E
)Sin(-h4
dCosh4
-EdE
aR
z-a
R
h
R4
dzdE
12
21
0
lP
0
l
0
l
z
z
PP
z2
0
lP
2
1
2
1
==
+=
+=
==
=
y
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mV /ah2
-E y
lP
0=
A3. Two wires AB and CD each 1 m length carry a total charge of 0.2 C and are disposedas shown. Given BC = 1 m, find BC.ofmidpointP,atE
P
A B . C
1 m
1 m
D
Solution :
(1)
1 = 1800 2 = 1800
A B P
1 m
[ ] [ ]{ } n a t( I n d e t e r0
0aC o s-C o sa)S i n-( S i n-
h4
E z12y12
0
lPA B
=+
=
az(2) Pay
C
1 1 = - tan-15.0
1= - 63.430
2 = 0
D
[ ]
[ ]
[ ] )a1989.75a(-3218a0.447)-(1a0.894-10x3.6E
a63.43)Cos-0(Cosa(-63.43))(Sin-
0.536
104
10x0.2
a)Cos-(Cosa)Sin-(Sin-h4
E
zyzy
3
P
zy9-
6-
z12y12
0
lP
CD
CD
+=+=
+=
+
=
Since EAB
is indeterminate, an alternate method is to be used as under :
Z
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dEPz
d
dy
y B Y P dEPyA
L R
t4
E
dtt4
-dE
;t-y-dLLet
(L4
adE
;ay)-d(LR
aR4
dydE
d
1
L0
l
P
2
0
lP
0
yl
Py
R
2
0
lP
mV /dL
1-
d
1
4
E lP
+=
0
mV/a2400a0.67]-2[1800E
a1.5
1-
0.5
1
36
104
10x0.2E
yyP
y9-
-6
P
AB
AB
==
=
a0.381(-awhere
EEE
yP
PPP CDAB
A4. Develop an expression for E due to a charge uniformly distributed over an infinite
plane with a surface charge density of S C / m2.
Solution :
Let the plane be perpendicular to Z axis and we shall use Cylindrical Co-ordinates. The
source charge is an infinite plane charge with S C / m2 .
dEP Z
RAP =z P
0 Y
d
X A
)aza-(R1a
)aza-(R
OPOA-OPAOAP
zR
z
+=
+=
+=+=
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The field intensityPdE due to dQ = S ds = S (d d) is along AP and given by
dd)aza-(R4
a
R4
dddE z3
0
R2
0
P
+
=
= SS
Since radial components cancel because of symmetry, only z components exist
dR
z2x
4
R
dzd
4
dEE
ddR4
zdE
0
3
00
3
2
00S
PP
3
0
P
=
==
=
SS
S
z is fixed height of above plane and let APO = be integration variable. All distances
are expressed in terms of z and
= z tan , d = z Sec2 d ; R = z Sec ; = 0, = 0 ; = , = / 2
(na2
[-2
dSin
2
dSecz
Secz
tanzz
2
E
z
0
S
0
S
2/
00
S2
0
33
0
SP
=
=
=
=
A5. Find the force on a point charge of 50 C at P (0, 0, 5) m due to a charge of 500 Cthat is uniformly distributed over the circular disc of radius 5 m.
Z
P
h =5 m
0 Y
X
Solution :
Given : = 5 m, h = 5 m and Q = 500 CTo find : fp & qp = 50 C
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Na56.55f
x50xa10x1131f
EwhereqxEf
zP
z
3
P
PPPP
=
=
=