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ENEL2FT Field Theo Electrostatic Fields 1
ELECTROSTATIC FIELDS
Dr. Thomas Afullo,UKZN, Durban
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ENEL2FT Field Theory Electrostatic Fields 2
ENEL2FT FIELD THEORY REFERENCES 1. M.N. Sadiku: Elements of Electromagnetics , Oxford
University Press, 1995, ISBN 0-19-510368-8. 2. N.N. Rao: Elements of Engineering Electromagnetics,
Prectice-Hall, 1991, ISBN:0-13-251604-7. 3. P. Lorrain, D. Corson: Electromagnetic Fields and Waves ,
W.H. Freeman & Co, 1970, ISBN: 0-7167-0330-0. 4. David T. Thomas: Engineering Electromagnetics ,
Pergamon Press, ISBN: 08-016778-0.
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ENEL2FT Field Theory Electrostatic Fields 3
ELECTROSTATIC FIELDS
COULOMBS LAW
The study of electrostatics begins by investigating twofundamental laws: Coulombs law and Gausss law.
Although Coulombs law is applicable in finding the electricfield due to any charge configuration, it is easier to useGausss law when charge distribution is symmetrical.
Coulombs law is an experimental law formulated in 1785 bythe French colonel, Charles Coulomb. It deals with the force a point charge exerts on another point
charge. By a point charge is meant a charge that is located on a body
whose dimensions are much smaller than other relevantdimensions. For example, the collection of electric charges on a pinhead
may be regarded as a point charge. Charges are generally measured in Coulombs (C).
One Coulomb is approximately equal to 6x1018
electrons; it is avery large unit of charge because the charge of an electron is
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ENEL2FT Field Theory Electrostatic Fields 4
ELECTROSTATIC FIELDS
COULOMBS LAW Coulombs law states that the force F between two point
charges Q 1 and Q 2 is: a) Along the line joining the charges b) Directly proportional to the product Q 1Q 2 of the charges
c) Inversely proportional to the square of the distance Rbetween them.
Mathematically, Coulombs law is expressed as:
Here, k is the proportionality constant. In SI units, charges Q1 and Q2 are in coulombs (C), the
distance R is in metres, and the force F is in newtons (N). A constant o is defined as the permittivity of free space (in
farads/metre).
221
RQkQ
F =
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ELECTROSTATIC FIELDS
COULOMBS LAW
We may re-write Coulombs equation as:
Also note that:
It noted that like charges (charges of the same sign) repeleach other, while unlike charges attract.
The distance R between the two charged bodies Q 1 and Q 2 must be large compared with the linear dimensions of thebodies.
Q 1 and Q 2 must be static (at rest).
The signs of Q 1 and Q 2 must be taken into account.
( )3
12
1221123
2112
44 r r
r r QQ R
R
QQ F
oo ==
1221 F F =
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ELECTROSTATIC FIELDS
COULOMBS LAW: ELECTRIC FIELD INTENSITY We define the electric field intensity or electric field strength as
the force per unit charge when placed in the electric field. That is:
Thus the electric field intensity is in the direction of theforce F and is measured in Volts/metre.
The electric field intensity at point r due to a point chargelocated at r 1 is obtained as:
F Q
E 1=
( )3
1
13 44 r r
r r Q R
RQ
E oo
==
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ELECTROSTATIC FIELDS
COULOMBS LAW: ELECTRIC FIELD INTENSITY
For N point charges Q 1,Q 2,..,Q N located at positions r 1,r 2,.., r N,the electric field intensity at point r is obtained as:
Example: Point charges of 2mC and 4mC are located at (3,2,1) and
(-1,-2,-3), respectively. Calculate the electric force on a 10nC charge located at (0,2,4). Also calculate the electric fieldintensity at that point.
( ) ( ) ( )
( )
=
++
+
=
=
N
k k
k k
o
N
N N
r r
r r Q E
r r
r r Q
r r
r r Q
r r
r r Q E
1 3
332
223
1
11
41
4..
44
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ELECTROSTATIC FIELDS ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS So far, we have only considered forces and electric fields due to
point charges, which are essentially charges occupying verysmall physical space.
At a macroscopic scale, we can disregard the discrete nature of the charge distribution and treat the net charge contained in anelemental volume v as if it were uniformly distributed within it.
Accordingly, we define the volume charge density as:
Where q is the charge contained in v. The variation of v withspatial location is called its spatial distribution . The total chargecontained in volume v is given by:
)/(lim 30
mC dvdq
vq
vv =
=
=v
v CoulombsdvQ
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ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS
In some cases, particularly when dealing with conductors,electric charge may be distributed across the surface of amaterial, in which case the relevant quantity of interest is thesurface charge density, s , defined as:
Where q is the charge present across an elemental surfacearea s . Similarly, if the charge is distributed along a line, we
characterize the distribution in terms of the line charge densityl , defined as:
ds
dq
s
q
s s =
= 0
lim
( )mC dl dq
l q
l l /lim
0=
=
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ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS
The electric field intensity due to each of the chargedistributions l,s ,and v may be regarded as the summation of the field distributed by the numerous point charges making upthe charge distribution.
Thus we replace Q in the equations for E, and integrating, weget:
We shall now apply these formulas to specific chargedistributions.
=
=
=
r Rdv
E
r Rds
E
r Rdl E
o
v
o
s
o
l
4
4
4
2
2
2
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ENEL2FT Field Theory Electrostatic Fields 14
ELECTROSTATIC FIELDS ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE
DISTRIBUTIONS AN INFINITE LINE CHARGE
Consider a line charge with a uniform charge density L extending from - to + along the z-axis, as shown below.
z
r
R
E d
dz
r
r
z RaInfiniteline charge
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ENEL2FT Field Theory Electrostatic Fields 15
ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO AN INFINITE LINE CHARGE The charge element dQ associated with element dz of the
line is:
The electric field intensity at point P a distance r from theline, due to the elemental charge Ldz is given by:
From geometry, we obtain:
dz dQ L =
R R R R
a
R R
dz a
R
dz E d
R
o
L R
o
L
==
==
;
4
4 32
d r dz r d d
r d d
r d dz
r r r Rr z z r R
22
22222
secseccossin
tan
sectantan;
====
=+==+=
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ENEL2FT Field Theory Electrostatic Fields 16
ELECTROSTATIC FIELDS ELECTRIC FIELDS DUE TO AN INFINITE LINE CHARGE Also, for the unit vector we have:
If we now integrate over the entire line, then varies from /2 to + /2 as z varies from - to + ; thus:
In normal cylindrical coordinates, the expression becomes:
( ) ( )[ ]
d z r r r
d r z r E d
z r a
o
L
o
L
R
sincos4sec
secsincos4
sincos
22
2
==
=
( ) [ ] [ ]{ }
r r
E
z r r
d z r r
E
o
L
o
L
o
L
2
cossin4
sincos4
2/2/
2/2/
2/
2/
=
+= =
2 o
L E =
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ENEL2FT Field Theory Electrostatic Fields 17
ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO AN INFINITE LINE CHARGE Alternatively, one can see from the expression for dE that:
One observes that at the observation point P, thecontribution to E z due to the element dz at point +z on theline charge is cancelled by the contribution due to theelement dz at position z along the line charge. Therefore,we could just conclude that:
We shall use a similar argument for surface charge.
( ) ( )[ ] z r o
L
o
L dE z dE r d z r r r
d r z r E d sincos
4secsecsincos
4 222
+===
r z r z E r E z E r E E ;0 =+==
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ENEL2FT Field Theory Electrostatic Fields 18
ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO CIRCULAR RING OF CHARGE
Consider a circular ring of charge of radius , havinguniform charge density l C/m. The ring is placed on the x-yplane.
x
y
z
R
dl
E d
Circular ringof charge
z
Ra
h
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ENEL2FT Field Theory Electrostatic Fields 20
ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO CIRCULAR RING OF CHARGE
Thus dE has both a z- component and a component. However, from symmetry considerations, for every element
dl in the direction giving rise to an elemental field strengthdE , there is a corresponding opposite element dl givingrise to an opposite elemental electric field strength dE .
Therefore the components of dE cancel; this implies thatdE has only a z-component. Thus:
Simplifying, we obtain:
( )( )( )
( )( )
( )
( )
( )22
2
022
22
2cos
4cos
cos4
,0
h z
hd z
E
z h
d dE z E d dE
o
L
o
L
o
L z
+
= +
=
+===
( )( ) ( )
( )( ) ( )2/3222/3222/32222 4
4
2
2
2
cos
h
Qh z
h
h z
h
h z h
z E
oo
L
o
L
o
L
+=
+=
+=
+=
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ENEL2FT Field Theory Electrostatic Fields 21
ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO AN INFINITE SURFACE CHARGE
Let us consider an infinite plane sheet of charge in the xy-plane with uniform surface charge density s C/m 2. We arerequired to find the electric field intensity due to iteverywhere above the sheet.
x
y
z
d d
dA
Rh
Ra z
E d
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ENEL2FT Field Theory Electrostatic Fields 22
ELECTROSTATIC FIELDS ELECTRIC FIELDS DUE TO AN INFINITE SURFACE CHARGE Consider the point P(0,0,h) on the z-axis. The sheet of
surface charge is thus placed a distance h below P. Thecharge contribution due to an elemental area dA is givenby:
We also derive the following relationships from the sketch:
Then the electric field intensity arising from this elementalcharge is:
( ) ( ) d d dQd d dAdAdQ s s === ;
sincos
sec;seccossin
sectan1;tan;
22
222
=
===
=+==+==
z a
d hd hd d
hd d
hh Rhh R R
R
( ) [ ]===
222
22 secsincos)sec)(tan(
4
44
h z d hd h
a R
d d RadQ
E d o
s R
o
s
o
R
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ENEL2FT Field Theory Electrostatic Fields 24
ELECTROSTATIC FIELDS
ELECTRIC FIELDS DUE TO AN INFINITE SURFACE CHARGE
For a point located below the charge sheet, the electric fieldintensity is:
If we consider two infinite parallel, oppositely-charged charge
sheets , one with charge density s , and the other withopposite charge density s C/m 2, the total electric fieldbetween the two plates is given by:
This would therefore be the total electric field between twoplates of a parallel-plate capacitor with (approximately)infinite dimensions.
( )
o s
o
s
o
s
z E
z z E
2
2
=
+=
o
s z E
2
=
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ENEL2FT Field Theory Electrostatic Fields 25
ELECTROSTATIC FIELDS ELECTRIC FLUX DENSITY Let us define a vector field, D, as:
Where is the electrical permittivity of the medium. Thus Dis independent of the medium. Define the electric flux, ,as:
The electric flux is measured in Coulombs, and thereforethe vector D is called the electric flux density, measured inC/m 2.
Thus all formulas derived for E from Coulombs law can beused in calculating D, except we have to multiply thoseresults by o . Thus for a volume charge distribution,
E D =
= S d D.
Rv a Rdv
D 4 2
=
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ENEL2FT Field Theory Electrostatic Fields 26
ELECTROSTATIC FIELDS
GAUSSS LAW Gausss law states that the total electric flux, , flowing out
of a closed surface S equals to the total charge enclosed bythe surface.
That is:
Where Q enc =total charge enclosed.
Gausss law is thus an alternative statement of Coulombslaw.
Gausss law provides an easy means of finding E or D for symmetrical charge distributions such as a point Charge,an infinite line charge, an infinite surface charge, and aspherical charge distribution.
=== S encenc QS d DQ .
==v
vv s
v dvS d DdvQ .
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ENEL2FT Field Theory Electrostatic Fields 28
ELECTROSTATIC FIELDS
APPLICATION OF GAUSSS LAW TO A POINT CHARGE
Applying Gausss law, with a spherical surface as theGaussian surface, we have:
From this, we can determine E to be:
( )( )
2
2
0
2
0
2
4
4sin
sin
..
r Q
r D
r Dd d r DQ
d r rd r S d
S d r DS d DQ
r r
V r
V
=
=
=
=
= =
24
r Q
r E E Do
o
==
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ENEL2FT Field Theory Electrostatic Fields 29
ELECTROSTATIC FIELDS
APPLICATION OF GAUSSS LAW TO A LINE CHARGE
Suppose the infinite line of uniform charge L C/m liesalong the z-axis.
To determine D at a point P a distance from the line, wechoose a cylindrical surface containing P to satisfysymmetry conditions as shown in the figure below.
P
D
z
x
y
L
Gaussiansurface
Line charge L C/m
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ENEL2FT Field Theory Electrostatic Fields 32
ELECTROSTATIC FIELDS
APPLICATION OF GAUSSS LAW TO A UNIFORMLYCHARGED SPHERE
To determine D everywhere, we construct Gaussiansurfaces for cases r a , and r a , separately.
Since the charge has spherical symmetry, it is obvious thata spherical surface is an appropriate Gaussian surface.
For r a , the total charge enclosed by the spherical surfaceof radius r is:
The total flux is given by:
venc
r
r venc
r Q
d drd r dvQ
3
4
sin
3
2
0 0 0
2
=
= == = =
( ) ==== =
2
0 0
22 4sin. r Dd d r DS d D r r
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ELECTROSTATIC FIELDS
APPLICATION OF GAUSSS LAW TO A UNIFORMLY
CHARGED SPHERE Thus we have:
For r a , the charge enclosed by the Gaussian surface isthe entire charge in this case, that is:
Similarly, the flux is given by:
ar r
r D
r Dr Q vr enc
=
==
0,3
34
43
2
=== = = =
2
0 0 0
32
3
4sin
a
r vvvenc
ad drd r dvQ
( r Dr S d D == 24.
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ENEL2FT Field Theory Electrostatic Fields 34
ELECTROSTATIC FIELDS
APPLICATION OF GAUSSS LAW TO A UNIFORMLY
CHARGED SPHERE Hence we obtain,
Thus from the foregoing, D is everywhere given by:
( )
ar r
ar D
a Dr
v
vr
=
=
,3
34
4
2
3
32
=
ar r
ar
ar r
r
Dv
v
,3
,
3
2
3
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ENEL2FT Field Theory Electrostatic Fields 35
ELECTROSTATIC FIELDS
APPLICATION OF GAUSSS LAW TO A AN INFINITE SHEET
OF CHARGE Consider the infinite sheet of uniform charge with charge
density s C/m 2 lying on the z-0 plane (xy-plane).
x
y
z
D
P
Gaussian surface
Area A
Infinite sheet of charge, s C/m 2
D
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ENEL2FT Field Theory Electrostatic Fields 37
ELECTROSTATIC FIELDS
ELECTRIC POTENTIAL
In electric circuits, we work with voltages and currents. The voltage V between two points in the circuitrepresents the amount of work, or potential energy, requiredto move a unit charge between the two points .
In fact, the term voltage is a shortened version of theterm voltage potential and is the same as electric
potential . Even though when we solve a circuit problem we usuallydo not consider the electric fields present in the circuit,in fact it is the existence of an electric field between twopoints that gives rise to the voltage difference betweenthem, such as across a resistor or capacitor.
The relationship between the electric field, E, and theelectric potential, V, is the subject of this section.
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ELECTROSTATIC FIELDS
ELECTRIC POTENTIAL
The force exerted is in the negative y-direction. If we attempt to move the charge along the positive y-
direction, against the force F e , we will need to provide anexternal force F ext to counteract F e , which requires an
expenditure of energy. To move q without any acceleration (at a constant speed), it
is necessary that the net force acting on the charge bezero. This means that:
The work done, or energy expended, in moving any object avector differential distance dl under the influence of forceF ext is:
qE y E q F e ==
E q F F eext ==
l d E ql d F dW ext .. ==
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ELECTROSTATIC FIELDS
ELECTRIC POTENTIAL If the charge is moved a distance dy along y, then:
The differential electric potential energy dW per unit charge
is called the differential electric potential , or differentialvoltage, dV.
That is,
The unit of V is the volt (V), and therefore the electric fieldis expressed in volts per metre (V/m).
( ) qEdydy y E yqdW == .
)/(. V or C J l d E q
dW dV ==
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ELECTROSTATIC FIELDS ELECTRIC POTENTIAL Thus the potential difference between any two points P 2
and P 1 is obtained by integrating dV along the path betweenP 1 and P 2. That is:
Where V 1 and V 2 are the electric potentials at points P 1 andP 2, respectively.
The result of the line integral above should be independent
of the specific path of integration between points P 1 and P 2. It is also readily seen that the integral of the electrostatic field E
around any closed contour is zero :
= ==
=2
1
2
11221 .
P
P
P
P l d E dV V V V
dV V
0..
0..2
2
2
2
2222
= =
= = = ==
E xS d E xl d E But
l d E l d E dV V V V
S C
C
P
P
P
P
ELECTROSTATIC FIELDS
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ELECTROSTATIC FIELDS
ELECTRIC POTENTIAL We now define what is meant by the electric potential V at a
point in space. Whenever we talk of a voltage V in a circuit, we do so in
reference to a voltage of some conveniently chosen pointto which we have assigned a reference voltage of zero , whichwe call ground .
The same principle applies to electric potential V. Usually,the reference potential point is chosen to be at infinity. Thatis, if we assume V 1=0 when P 1 is at infinity, the electricpotential at any point P is given by:
=
P l d E V .
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ENEL2FT Field Theory Electrostatic Fields 44
ELECTROSTATIC FIELDS
ELECTRIC POTENTIAL DUE TO POINT CHARGES The principle of superposition that has been applied
previously to the electric field E also applies to theelectric potential V.
For N discrete point charges q 1, q 2, ..,q N havingposition vectors R 1, R 2, .., R N, the electric potential is:)(
41
1V
R Rq
V N
i i
i
o
==
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ENEL2FT Field Theory Electrostatic Fields 45
ELECTROSTATIC FIELDS
ELECTRIC POTENTIAL DUE TO CONTINUOUSCHARGE DISTRIBUTIONS
For a continuous charge distribution specified over a givenvolume V, across a surface S, or along a line l, we replacethe q i with:
Then, converting the summation into integration, weobtain:
dl dsdv l sv ;;
)(4
1)(
)(41
)(
)(4
1)(
ondistributilinedl R
RV
ondistributi surfacedS R RV
ondistributivolumedv R
RV
L
l
o
S s
o
V
v
o
=
=
=
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ENEL2FT Field Theory Electrostatic Fields 46
ELECTROSTATIC FIELDS
ELECTRIC FIELD AS A FUNCTION OF ELECTRIC
POTENTIAL We have seen that:
If we resolve E and dl into rectangular coordinates, wehave:
Thus
l d E dV .=
( )( )
;;;
..
;
z V E
yV E
xV E
dz z V
dy yV
dx xV
dV
dz E dy E dx E dz z dy ydx x E z E y E xl d E
dz z dy ydx xl d E z E y E x E
z y x
z y x z y x
z y x
===
+
+=
++=++++=++=++=
V E =
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ENEL2FT Field Theory Electrostatic Fields 48
ELECTROSTATIC FIELDS
SOLUTION:
( )
( )
( ) ( ) J W
x
r r
Q
V V Ql d E QQV W
mC xr E D
mV r mV r r E
r r r
r
V r
V r
r r V
V E
oooo
B
A A B AB
o
oooo
5-
56
)120,30,1(
2
)60,90,4(
2
2119
0,2/,2
333
2.8125x10
410
3210
10120cos30sin1
1060cos90sin
1610
1010
cossin10
cossin10
.
/1021.28
2036
10
/5.2/8
20008
20
sin10coscos
10cossin
20
sin11
=
==
=
===
= ==
==
+=
+=
+
+==
ELECTROSTATIC FIELDS
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ENEL2FT Field Theory Electrostatic Fields 49
ELECTROSTATIC FIELDS THE ELECTRIC DIPOLE An electric dipole is formed when two point charges of
equal but opposite sign are separated by a small distance,as shown below.
P
x
y
z
+Q
-Q
d
r 1
r 2
r
An Eectric Dipole
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ENEL2FT Field Theory Electrostatic Fields 50
ELECTROSTATIC FIELDS THE ELECTRIC DIPOLE The potential at point P(r, , ) is given by:
Where r 1 and r 2 are the distances between P and +Q and Q,respectively.
If r>>d, then:
==21
12
21 411
4 r r r r Q
r r Q
V oo
( )( ) ( )( )( )( )[ ]
221
12
222
221
12
22
22
222
21
22
22
1
4
cos4
cos)/(1
cos)/(1cos)/(1cos)/(1cos)/(1
cos
cos)2/(cos)/(1cos)2/(2
cos)2/(2cos)2/(22
cos)2/(cos)/(1cos)2/(2
cos)2/(2cos)2/(22
r
Qd r r
r r QV
r r d r
r d r d r r d r r d r r r
d r r
d r r d r d r r r
d r r d r d
r r
d r r d r d r r r
d r r d r d
r r
oo
==
=
+=+=
++=+=
++
+=
==
+=
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ENEL2FT Field Theory Electrostatic Fields 51
ELECTROSTATIC FIELDS THE ELECTRIC DIPOLE
Define the dipole moment p as:
The electric field due to the dipole with centre at theorigin, is:
22 4
.
4
cos
r
r p
r
Qd V
d Q p
oo
==
=
[ ]
sincos24
4
sin2
cos
1
3
33
+=
+=
+
==
r r
p E
r
Qd r r
Qd
V r
r r V
V E
o
oo
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ENEL2FT Field Theory Electrostatic Fields 52
ELECTROSTATIC FIELDS
THE ELECTRIC DIPOLE Notice that a point charge is a monopole, and its electric
filed varies inversely as r 2, while its potential variesinversely as r.
For the dipole, we notice that the electric field variesinversely as r 3, while its potential varies inversely as r 2.
The electric fields due to the presence of a quadrupole(consisting of two dipoles) vary inversely as r 4, while thecorresponding potential varies inversely as r 3.
EXAMPLE:
Two dipoles have dipole moments p 1 and p 2 are located atpoints (0,0,2) and (0,0,3), respectively. Find the potential atthe origin if: Cm z x pCm z x p 109;105 92
91
==
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ENEL2FT Field Theory Electrostatic Fields 54
ELECTROSTATIC FIELDS EXAMPLE:
An electric dipole of dipole moment p is located at theorigin, where:
Find the electric filed intensity E and potential V at thefollowing points:
A) (0,0,10). B) ( 1,/3, /2)
ANS:
Cm x p 1210100 =
( ) V V mV xr E BV xV mV r x E A
45.0;/1078.09.0)
109;/108.1)3
33
=+= ==
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ENEL2FT Field Theory Electrostatic Fields 55
ELECTROSTATIC FIELDS ENERGY DENSITY IN ELECTROSTATIC FIELDS To determine the energy present in an assembly of
charges, we must first determine the amount of worknecessary to assemble them.
Suppose we wish to position three point charges Q 1, Q 2,and Q 3 in an initial empty space shown below.
P1
P2
P3
Q1
Q2
Q3
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ENEL2FT Field Theory Electrostatic Fields 56
ELECTROSTATIC FIELDS ENERGY DENSITY IN ELECTROSTATIC FIELDS No work is required to transfer Q 1 from infinity to P 1
because the space is initially charge free and there is noelectric field.
The work done in transferring Q 2 from infinity to P 2 is equalto the product of Q 2 and the potential V 21 at P 2 due to Q 1.
Similarly, the work done in positioning Q 3 at P 3 is equal toQ 3 (V32+V31), where V 32 and V 31 are the potentials at P 3 dueto Q 2 and Q 1, respectively.
Hence the total work done in positioning the three chargesis:
If the charges were positioned in reverse order, then:
( )32313212321
0 V V QV Q
W W W W E +++=
++=
( )13121232123
0 V V QV Q
W W W W E +++=
++=
ELECTROSTATIC FIELDS
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ENEL2FT Field Theory Electrostatic Fields 57
ELECTROSTATIC FIELDS ENERGY DENSITY IN ELECTROSTATIC FIELDS Here, V 23 is the potential at P 2 due to Q 3, V12 and V 13 are
respectively the potentials at P 1 due to Q 2 and Q 3. Thus thetwo equations give:
Where V 1, V2, and V 3 are the potentials at P 1, P 2, and P 3,
respectively. In general, if there are n point charges, the above equationbecomes:
( ) ( ) ( )
( )332211
332211
323132321213121
2
1
2
V QV QV QW
V QV QV Q
V V QV V QV V QW
E
E
++=
++=+++++=
==n
k k k E V QW
121
ELECTROSTATIC FIELDS
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ENEL2FT Field Theory Electrostatic Fields 58
ELECTROSTATIC FIELDS ENERGY DENSITY IN ELECTROSTATIC FIELDS If, instead of point charges, the region has a continuous
charge distribution, the above summation becomes asintegration:
We can further refine the expression using volume chargedensity by using the vector identities:
( )
( )
( )echvolumeVdvW
ech surfaceVdS W
echlineVdl W
V E
S E
L E
arg21
arg21
arg21
=
=
=
( )( ) V A AV AV
AV V A AV
Dv
=+=
=
...
...
.
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ENEL2FT Field Theory Electrostatic Fields 59
ELECTROSTATIC FIELDS ENERGY DENSITY IN ELECTROSTATIC FIELDS Therefore we obtain:
By applying the divergence theorem to the first term on theright-hand side of the equation, we have:
For point charges, V varies as 1/r, and D varies as 1/r 2; for dipoles, V varies as 1/r 2 and D varies as 1/r 3; and so on.
( )
( ) ( ) ( )
==
==
dvV Ddv DV Vdv D
Vdv DVdvW V E
.21
.21
.21
.21
21
( ) ( ) =V S
E dvV DS d DV W .21
.21
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