field theory Electrostatics

8/14/2019 field theory Electrostatics

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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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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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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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7/61ENEL2FT Field Theory Electrostatic Fields 7


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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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

==




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



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




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=

=




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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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



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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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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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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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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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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

=

==

=

===

= ==

==

+=

+=

+

+==



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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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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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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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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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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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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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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 +++=

++=



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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



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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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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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