Electrostatics - College of Engineering and Applied...

Chapter 4Electrostatics

Contents4.1 Maxwell’s Equations . . . . . . . . . . . . . . . . . . . 4-34.2 Charge and Current Distributions . . . . . . . . . . . 4-4

4.2.1 Charge Densities . . . . . . . . . . . . . . . . . 4-4

4.2.2 Current Densities . . . . . . . . . . . . . . . . . 4-6

4.3 Coulomb’s Law . . . . . . . . . . . . . . . . . . . . . 4-84.3.1 Field Due to N Point Charges . . . . . . . . . . 4-9

4.3.2 Field Due to a Charge Distribution . . . . . . . . 4-11

4.4 Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . 4-224.5 Electric Scalar Potential . . . . . . . . . . . . . . . . . 4-324.6 Conductors . . . . . . . . . . . . . . . . . . . . . . . . 4-41

4.6.1 Drift Velocity . . . . . . . . . . . . . . . . . . . 4-42

4.6.2 Resistance . . . . . . . . . . . . . . . . . . . . 4-43

4.6.3 Joule’s Law . . . . . . . . . . . . . . . . . . . . 4-46

4.7 Dielectrics . . . . . . . . . . . . . . . . . . . . . . . . 4-474.7.1 Polarization Field . . . . . . . . . . . . . . . . . 4-48

4.7.2 Dielectric Breakdown . . . . . . . . . . . . . . 4-49

4.8 Electric Boundary Conditions . . . . . . . . . . . . . 4-49

4-1

CHAPTER 4. ELECTROSTATICS

4.8.1 Dielectric-Conductor Boundary . . . . . . . . . 4-53

4.8.2 Conductor-Conductor Boundary . . . . . . . . . 4-55

4.9 Capacitance . . . . . . . . . . . . . . . . . . . . . . . 4-564.10 Electrostatic Potential Energy . . . . . . . . . . . . . 4-604.11 Image Method . . . . . . . . . . . . . . . . . . . . . . 4-62

4-2

4.1. MAXWELL’S EQUATIONS

4.1 Maxwell’s Equations

� The chapter opens as the first pure fields chapter

� The remaining chapters of the text focus on Maxwell’s equa-tions (4 total)

r �D D �v (4.1)

r � E D �@B

@t(4.2)

r � B D 0 (4.3)

r �H D J D@D

@t; (4.4)

where

E D electric field internsityD D �E D electric flux densityH D magnetic field intensityB D �H D magntic flux densityJ D convection or conduction current density

� In Chapter 4 and 5 we will only consider static conditions,which means terms of the form @=@t D 0

� What remains of the four Maxwell’s equations is two pairs ofsimplified equations:

– Electrostatics (Chapter 4)

r �D D �v (4.5)r � E D 0 (4.6)

4-3


– Magnetostatics (Chapter 5)

r � B D 0 (4.7)r �H D J (4.8)

� The above pairs of equations are said to be decoupled, whichholds only for the static case

4.2 Charge and Current Distributions

With regard to electrostatics, working with charge current distribu-tions is common place.

4.2.1 Charge Densities

� Charge densities are similar to probability densities studied inprob and stats and mass densities found in mechanics

� There are three basic forms:

– Volume distribution

�v D lim�V!0

�q

�V Ddq

dV .C/m3/

Note:

Q D

ZV�v dV (C)

– Surface distribution

�s D lim�s!0

�q

�sDdq

ds.C/m2/

4-4

4.2. CHARGE AND CURRENT DISTRIBUTIONS

Note:

Q D

ZS

�s ds (C)(a) Line charge distribution

(b) Surface charge distribution

Surface charge ρs

z

x

y

3 cm

rϕ

z

x

y

Line charge ρl10 cm

Figure 4-1 Charge distributions for Examples 4-1and 4-2.

Figure 4.1: Circular surface charge �s.

– Line distribution

�` D lim�l!0

�q

�lDdq

dl.C/m/

Note:

Q D

Zl

�` dl (C)

(a) Line charge distribution

(b) Surface charge distribution

Surface charge ρs

z

x

y

3 cm

rϕ

z

x

y

Line charge ρl10 cm

Figure 4-1 Charge distributions for Examples 4-1and 4-2.

Figure 4.2: Linear line charge �`.

4-5


Example 4.1: Nonuniform Surface Charge

� Consider the surface charge density

�s D

(4y2 .�C/m2/; �3 � x; y � 3 m

0; otherwise

� Find the total charge

Q D

Z 3

�3

Z 3

�3

4y2 dy dx D

Z 3

�3

�4y3

3

�3�3

dx

D 72 � Œx�3�3 D 72 �

�3 � .�3/

�D 432 .�C/

4.2.2 Current Densities

� Current is related to charge density, except we have to put thecharge into motion

� Consider charge in a tube having volume density �v and mov-ing from left to right with velocity u

� In �t s the charge moves �l D u�t , creating a charge flowacross the tube’s surface area, �s0 of

�q0 D �v � .�l ��s0/„ ƒ‚ …

V ��t

alsoD �vu ��s

0��t

4-6

4.2. CHARGE AND CURRENT DISTRIBUTIONS

Volume charge ρv

u

∆l

∆q' = ρvu ∆s' ∆t

∆s'

∆sρv

u ∆q = ρvu • ∆s ∆t = ρvu ∆s ∆t cos θ

∆s = n ∆s

(a)

(b)

θ

ˆ

Figure 4-2 Charges with velocity u moving through across section ∆s′ in (a) and ∆s in (b).

Figure 4.3: (a) Charges flowing in a tube with cross section �s0

moving with velocity u m/s and (b) dealing with a surface normaldiffernt from the flow velocity.

� For the general case of charge flow across a surface (not par-allel to the velocity u) the normal to the surface �s, On, can beused to write�s D On�s, and then describe the general chargeincrement �q (prime is dropped) as

�q D �vu ��s�t

� Since current is charge flow per unit time, we have

�I D�q

�tD �vu„ƒ‚…

(C/s)/m2

��s D J ��s

where J is the current density

J D �vu .A/m2/

4-7


� Integrating over an arbitrary surface S yields the total current

I D

ZS

J � ds .A/

– For the movement of charged matter, J represents a con-vection current

– For the movement of charged particles (e.g., electrons ina conductor), J represents a conduction current

– Note: Conduction current obeys Ohm’s law, while con-vection current does not!

4.3 Coulomb’s Law

� First introduced in Chapter 1

� Now its time to get serious about studying it and working withit!

� Review: For an isolated charge q the induced electric field is

E D ORq

4��R2.V/m/;

where OR points from q to the field point P

� Review: A test charge q0 placed in electric field E at point Pexperiences force

F D q0E .N/

Note: It would appear that the units of E is also (N/C), i.e.,(N/C) = (V/m)

4-8

4.3. COULOMB’S LAW

� Review: When a material with permittivity � D �0�r is presentthe electric flux density and electric field intensity are relatedby

D D �E

Note: The �0 D 8:85 � 10�12 F/m

– As long as � is independent of the amplitude of E, thematerial is linear

– A material is said to be isotropic if � is independent ofthe direction of E; some PCB materials are anisotropic,meaning � takes on one value in the .x; y/ plane and an-other value in the z direction (sheet thickness)

4.3.1 Field Due to N Point Charges

� For N point charges q1; q2; : : : ; qN with corresponding posi-tion vectors Ri , i D 1; 2; : : : ; N connecting the charge loca-tion with the field point P , is the vector sum of the field due tothe individual charges

E D1

4��

NXiD1

.R �Ri/

jR �Ri j3

.V/m/

Example 4.2: Two Point Charges in Python and TI nspire

� Consider two point charges as described in text Example 4-3

� In Cartesian coordinates we have

R1 D .1; 3;�1/; R2 D .�3; 1;�2/; R D .3; 1;�2/

4-9


with

q1 D 2 � 10�5 (C) and q2 D �4 � 10

�5 (C)

� We calculate E by plugging the 3D vector coefficients into

E D1

4��0

�q1

R �R1

jR �R1j3C q2

R �R2

jR �R2j3

�(V/m)

� In Python the calculation is straight forward using numpy ndarrays

Figure 4.4: Python calculation of the two charge electric field at(3,1,-2).

� The nested list [[1,2,3],[4,5,6]] creates a 2D array havingdimensions 2 by 3

� A 1D array is formed by indexing just one row, e.g., Ri[i,:],;not the use of the colon operator to span all columns

� The norm function finds the length of an array in Cartesiancoordinates

� Using the TI nspire calculator a symbolic solution can be ob-tained and then converted to a numerical form, no problem

4-10


Page 1 of 1ECE3110_Chapter4a

r1:= 1 3 −1 1 3 −1

r2:= −3 1 −2 −3 1 −2

r:= 3 1 −2 3 1 −2

14· π· ε0

· 2·r-r1

norm r-r1 3+−4·

r-r2

norm r-r2 31

108· ε0· π−1

27· ε0· π−1

54· ε0· π

1

4· π· 8.85· 10−12· 2· 10−5·

r-r1

norm r-r1 3+−4· 10−5·

r-r2

norm r-r2 33330.3 −13321.2 −6660.6

r_array:= 1 3 −1−3 1 −2

1 3 −1−3 1 −2

r_array 1 1 3 −1

r_array 1,2 3

norm r_array 1 11

norm r_array 2 14

Two Charge CalculationOn the nspire you can enter an array of three element lists usingr_array:={{1,3,-1},{-3,1,-2}}. Once entered it looks like the following:r_array:= 1 3 −1

−3 1 −2 ▸ 1 3 −1

−3 1 −2r_point:= 3 1 −2 ▸ 3 1 −2

Now the calculation of the E field less the 10-5 factor and ε₀ :1

4·π· ε0· 2·

r_point-r_array 1

norm r_point-r_array 1 3+−4·

r_point-r_array 2

norm r_point-r_array 2 3

▸ 1

108· ε0· π−1

27· ε0· π−1

54· ε0· πNow include the missing terms to get a pure numerical answer:

10-5

4·π·8.85·10-12· 2·

r_point-r_array 1

norm r_point-r_array 1 3+−4·

r_point-r_array 2

norm r_point-r_array 2 3

▸ 3330.3 −13321.2 −6660.6

Figure 4.5: TI nspire calculation of the two charge electric field at(3,1,-2).

� The units (not shown in the figures) is of course (V/m)

4.3.2 Field Due to a Charge Distribution

� A practical extension to Coulomb’s law is consider charge dis-tributions: (1) volume, (2) surface, or (3) line distributions

Volume Distributon

� Consider a volume V 0 that contains charge density �v

4-11


� The electric field at a point P due to a differential charge dq D�v dV 0, is

dE D OR0dq

4��R02D OR0

�vdV4��R02

;

where R0 is the vector pointing from the differential chargeto the field point P and OR0 is the corresponding unit vectorR0=jR0j

� Since superposition holds for discrete charges it holds here, soin integral form

E D

ZV 0dE D

1

4��

ZV 0OR0�v dV 0R02

.V/m/

� This formula is nice and compact, but R0 and R0 are likelyfunctions of the integration variables used to describe V 0

� Furthermore, there are no examples or homework problems inthe book for a volume charge distribution

Surface Distribution

� For the case of a surface charge density dq D �sds0, we can

write

E D1

4��

ZS 0

OR0�s ds

0

R02.V/m/

� With the charge distribution limited to just two dimensions,problem set-up and integration become easier

4-12


Line Distribution

� Finally for the case of a line charge density dq D �`dl0, we

can write

E D1

4��

Zl 0

OR0�` dl

0

R02.V/m/

� Here the charge distribution one dimensional, but may lie alonga curve

� The math is manageable in many cases

Example 4.3: A Ring of Charge in Air

� Here we consider a circular line charge lying in the x�y planeof radius b uniform positive density �`

– Note: This is a classical, yet also important, example

� The field point for determining E is along the z-axis at P D.0; 0; h/

� The problem symmetry makes cylindrical coordinates the ob-vious choice

4-13


(a)

(b)

φ

+

+

+

+

+

+

+

+

+

+

+

+

+b

h

z

P = (0, 0, h)

R'1 ρl

dE1r

dE1dE1z

dφ

1

y

x

dl = b dφ

R'1R'2

+

+

+

++

+

+

+

+

+

++ y

x

dE1dE2

dE1rdE2r

1

2

dE = dE1 + dE2

φ + π

φ

z

Figure 4-6 Ring of charge with line density ρℓ. (a) Thefield dE1 due to infinitesimal segment 1 and (b) the fieldsdE1 and dE2 due to segments at diametrically oppositelocations (Example 4-4).

Figure 4.6: Setting up the ring of charge field calculation.

� The differential line charge segment length is dl D bd� andthe differential charge density is dq D �`dl D �`b d�

� From Figure 4.6 we see that

R0 D �Or b C Oz h

R0 D jR0j Dpb2 C h2

OR0 DR0

jR0jD�Or b C Oz hpb2 C h2

and

dE D1

4��0�`b�Or b C Oz h�b2 C h2

�3=2 d�� With the field point along the z-axis further simplification is

possible, namely the radial (Or) field contributions from charge

4-14


segments on opposite sides of the ring cancel, leaving only theaxial or Oz component (a)

(b)

φ

+

+

+

+

+

+

+

+

+

+

+

+

+b

h

z

P = (0, 0, h)

R'1 ρl

dE1r

dE1dE1z

dφ

1

y

x

dl = b dφ

R'1R'2

+

+

+

++

+

+

+

+

+

++ y

x

dE1dE2

dE1rdE2r

1

2

dE = dE1 + dE2

φ + π

φ

z

Figure 4-6 Ring of charge with line density ρℓ. (a) Thefield dE1 due to infinitesimal segment 1 and (b) the fieldsdE1 and dE2 due to segments at diametrically oppositelocations (Example 4-4).

Figure 4.7: Opposing line charge segments result in radial compo-nent cancellation and constructive combining of the axial compo-nents due to the segment 1 and 2 semicircles.

� The charge ring is broken into two semicircles, each definedover 0 � � � �

� We know the radial field components cancel and reason thatthe axial components constructively add, thus

E D Oz 2 ��`bh

4��0�b2 C h2

�3=2 Z �

0

d�

D Oz�`bh

2�0�b2 C h2

�3=2 alsoD Oz

h

4��0�b2 C h2

�3=2 Q;4-15


where Q D 2�b�` is the total charge on the ring

� A curiosity is how does the diameter of the ring, relative to thefield point distance h alter the field strength, and when doesthe ring look like a point charge?

� A simple plot (here using Python), helps explain

� In the code and corresponding plot shown below, the scale fac-tors of Q and �0 are not included (normalized out)

� For comparison purposes the field due to a point charge of Qlocated at the origin is also included in the plot

Figure 4.8: Python code for calculating the charge ring axial field.

4-16


1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Axial Field Point h (m)

0.0

0.2

0.4

0.6

0.8

1.0N

orm

aliz

ed A

xia

l Fi

eld

Inte

nsi

tyPlot of Ez · 4πε0/Q versus h with b a Parameter

Point charge

b= 1/8 (m)

b= 1/4 (m)

b= 1/2 (m)

b= 1 (m)

Figure 4.9: Axial field component of charge ring compared withequivalent point charge.

� The results are not too surprising:

– The ring of charge looks like a point charge at not toogreate a distance

– For a larger ring radius the field stength on axis is reducedcompared with the a smaller ring

– Note: As h approaches zero the axial field component dueto the ring is zero! why?

4-17


Example 4.4: Disk of Charge in the x � y-Plane

� Another classical example, that extends from the ring of charge,is a uniform circular disk of charge

�s.r; �; z/ D

(�s C/m2; 0 � r � a; z D 0

0; otherwise

� Field point is again P D .0; 0; h/

� The elemental field contribution for the ring of charge can beextended to the case of the disk by adding in an integration overthe charge differential dq D 2��sr dr as shown in Figure 4.10

z

P = (0, 0, h)

h

y

x

a

a

r

dr

dq = 2πρsr drρs

E

Figure 4-7 Circular disk of charge with surface chargedensity ρs. The electric field at P= (0,0,h) points alongthe z direction (Example 4-5).

Figure 4.10: Set-up for calculating E due to a disk of charge in thex � y-plane.

4-18


� The charge disk is composed of concentric rings of charge

� The total charge on the disk is

Q D

Z 2�

0

Z a

0

�sr dr d� D 2��sr2

2

ˇ̌a0D ��s a

2

� Due to symmetry the radial component of E is again zero

� Putting the pieces together we have

E D Oz�sh

2�0

Z a

0

r dr�r2 C h2

�3=2� Recall/look up Z

rdr�r2 C h2

�3=2 D �1pr2 C h2

;

so

E D

8̂̂̂̂<̂̂ˆ̂̂̂:Oz �s2�0

�1 �

jhjpa2Ch2

�; h > 0

�Oz �s2�0

�1 �

jhjpa2Ch2

�; h < 0

0; h D 0

� Writing in terms of the total charge Q amounts to replacing�s=.2�0/ with Q=.��0a2/, so

E D

8̂̂̂̂<̂̂ˆ̂̂̂:Oz Q

2��0a2

�1 �

jhjpa2Ch2

�; h > 0

�Oz Q

2��0a2

�1 �

jhjpa2Ch2

�; h < 0

0; h D 0

4-19


� Infinite Sheet of Charge: By letting a!1 we have the fielddue to an infinite sheet of charge being

E D

(Oz �s2�0; z > 0

�Oz �s2�0; z < 0

– Notice no dependence on distance from the plane!

� As a final sanity check, we compare the axial electric field ver-sus h while keeping the total charge, Q, the same in all cases

� Keeping the first two terms binomial in the expansion for .1Cx/�1=2 yields 1 � x=2, so assuming h� b or h� a we have

Ez;ring ' Q.h2 � b2=2/3

4��0h8and Ez;disk ' Q

1

4��0h2

Figure 4.11: Python code for calculating the charge disk axial field.

4-20


1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Axial Field Point h (m)

0.0

0.2

0.4

0.6

0.8

1.0N

orm

aliz

ed A

xia

l Fi

eld

Inte

nsi

tyPlot of Ez · 4πε0/Q versus h with a and b Parameters

Point charge

Ring b= 1/2 (m)

Disk a= 1/2 (m)

Figure 4.12: Axial field component of a charge disk compared witha ring and an equivalent point charge (total charge help constant forall three).

� The asymptotic behavior is as expected and is supported by thebinomial expansion results too

4-21


4.4 Gauss’s Law

� From physics you may remember this useful result

� Gauss’s law is arrived at by starting from Maxwell’s equation

r �D D �v

(differential form since partial derivatives are involved in thedivergence calculation)

� The integral form of the above, which is obtained by integrat-ing both sides over an arbitrary volme V , isZ

Vr �D dV D

ZV�v dV D Q

� Now invoke the divergence theorem from Chapter 3 which saysZVr �D dV D

IS

D � ds;

where S is encloses V (S is known as a Gaussian surface)

� Finally, arrive at Gauss’s Law:IS

D � ds D Q

In words, the flux passing through S equals the enclosed chargeQ

4-22

4.4. GAUSS’S LAW

Q D • ds

Gaussian surface Senclosing volume v

Total chargein v

v

Figure 4-8 The integral form of Gauss’s law states thatthe outward flux of D through a surface is proportional tothe enclosed chargeQ.

Figure 4.13: Gauss’s law illustrated.

Example 4.5: Classical Point Charge at the Origin

� If you remember anything about Gauss’s law, it is likely howyou can calculate E by setting up a Gaussian surface at theorigin that encloses a point charge q

Gaussian surface

q

R

D

ds

R̂

Figure 4-9 Electric field D due to point charge q.Figure 4.14: Point charge at the origin and the appropriate Gaussiansurface.

4-23


� Suppose we do not know the exact form of D (or E), but weknow from symmetry that

D D ORDR;

and a sphere centered at the origin has ds D OR ds, soIS

D � ds D

ZS

ORDR �OR ds D

IS

DR ds D .4�R2/DR D q

� Clearly,DR D

q

4�R2

orE D

D

�D OR

q

4��R2

A familiar result!

Gaussian Surface Tips

� Picking the Gaussian surface is key to finding D using Gauss’slaw

– Make use of symmetry so the form of D can be deduced(at least for each component)

– Choose S so the form of D is normal or parallel to thesurface, making integration trivial

– Be clever in selecting S (OK, you need to practice at this)

4-24

4.4. GAUSS’S LAW

Example 4.6: Classical Infinite Line Charge

� Given an infinite length line charge along the z axis (or parallelto it so you can shift the cylindrical coordinate frame) havinguniform charge density �`

� The logical Gaussian surface is a cylinder axially centered overa portion of the line charge

� Suppose the cylinder has radius r and length/height h

� We deduce that an infinite line charge can only have a radialcomponent D D OrDr ; Why?

z

r

D

Gaussian surface

uniform linecharge ρl

h ds

Figure 4-10 Gaussian surface around an infinitely longline of charge (Example 4-6).

Figure 4.15: Gaussian surface for infinite line charge.

� Apply Gauss’s law using the cylinder as S and notice that noflux passes through the ends of the cylinder, so the surface in-

4-25


tegral is just over the cylinder proper

IS

D � ds D

Z h

zD0

Z 2�

�D0

OrDr � Orr d� dz D 2� hDr ralsoD �`h

� So, we can solve for Dr and hence D and/or E

E DD

�0D Or

Dr

�0D Or

�`

2��0r

– The electric field from an infinite length line charge is in-versely proportional to the radial (perpendicular) distanceto the line

Example 4.7: Multiple Line Charges

� Find E when more than one line charge is parallel to the z axis

� As a special case consider a line charges at .x; y/ D .1; 0/

and .x; y/ D .�1; 0/ each having density �` as shown below(looking down frlom theCz axis)

4-26

4.4. GAUSS’S LAW

𝑦𝑦

𝑥𝑥

1

-1Line chargesparallel to the𝑧𝑧 axis on the 𝑥𝑥axis

Field point: 𝑃𝑃 = (0, 𝑦𝑦𝑝𝑝, 0)

Radial field componentsfrom each line charge

𝜌𝜌𝑙𝑙

𝜌𝜌𝑙𝑙

E1

E−1

Sum the vectors at the field point

Figure 4.16: Configuration of two line charges and the vector addi-tion of the fields.

� The field point we consider is P D .0; yp; 0/

� A vector sum is needed to combine the offset radial compo-nents from each line charge and of course we need unit vectorsat .0; yp; 0/

� Denote the commponents E1 and E�1

E1 D

unit vector‚ …„ ƒ�OxC Oyypq1C y2p

jE1j‚ …„ ƒ�`

2��0

q1C y2p

E�1 DOxC Oyypq1C y2p

�`

2��0

q1C y2p

4-27


� Add the components

E D E1 C E�1 DOy2yq1C y2p

�`

2��0

q1C y2p

DOyyp�`

��0�1C y2p

�� Note the Ox component is zero due to symmetry

� Would moving the field point off the y axis make the Ox com-ponent nonzero?

Example 4.8: A Uniform Surface Charge Density Sphere

� Given a thin shell of radius a contains a uniform surface chargeof �s, find E everywhere

� Due to symmetry the electric field will be of the form E DORER

𝑎𝑎

Surface charge density 𝜌𝜌𝑠𝑠

𝐄𝐄 = �𝐑𝐑𝐸𝐸𝑅𝑅

Gaussian surface for 𝑅𝑅 < 𝑎𝑎

Gaussian surface for R > 𝑎𝑎

Side view of spheres

Figure 4.17: Gaussian surfaces for findng E inside and outside asphere with surface charge density.

4-28

4.4. GAUSS’S LAW

� From Figure 4.17 the appropriate Gaussian surface is a spherecentered at the origin having radius R > a or R < a

� R < a: The Gaussian surface does not enclose any charge, soE D 0

� R > a: Here the Gaussian surface encloses the surface chargeof the thin sphere of radius a, allowing us to writeI

s

D � ds D DR

�4�R2

�D

Zs

�s ds D 4�a2 �s

) E DD

�0D OR

�sa2

�0R2.V/m/

� In summary,

E D

(0; R < a

OR �sa2

�0R2 ; R > a

.V/m/

0 1 2 3 4 5

Radial Field Point R (m)

0.0

0.2

0.4

0.6

0.8

1.0

Norm

aliz

ed R

adia

l Fi

eld

Inte

nsi

ty

Plot of ER · ε0/ρs versus R with parameter a

a= 1

a= 0. 5

a= 2. 0

Figure 4.18: Normalized radial field component.

4-29


Example 4.9: A Uniform Volume Charge Density Sphere

� Given a spherical volume radius a contains a uniform volumecharge of �v, find E everywhere

� Due to symmetry the electric field will be of the form E DORER

𝑎𝑎

Volume charge density 𝜌𝜌𝑣𝑣


Gaussian surface for 𝑅𝑅 < 𝑎𝑎

Gaussian surface for R > 𝑎𝑎

Side view of spheres

Figure 4.19: Gaussian surfaces for findng E inside and outside asphere with volume charge density.

� From Figure 4.19 the appropriate Gaussian surface is a spherecentered at the origin having radius R > a or R < a

� R < a: The Gaussian surface encloses a portion of the totalvolume charge, soI

s

D � ds D DR

�4�R2

�D

Zv

�v dv D4

3�R3 �v

) E DD

�0D OR

�vR

3�0.V/m/

4-30

4.4. GAUSS’S LAW

� R > a: Here the Gaussian surface encloses the surface chargeof the thin sphere of radius a, allowing us to writeI

s

D � ds D DR

�4�R2

�D

Zv

�v dv D4

3�a3 �v

) E DD

�0D OR

�va3

�0R2.V/m/

� In summary,

E D

(OR�vR

3�0; R < a

OR �va3

3�0R2 ; R > a

.V/m/

0 1 2 3 4 5

Radial Field Point R (m)

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Norm

aliz

ed R

adia

l Fi

eld

Inte

nsi

ty

Plot of ER · ε0/ρv versus R with parameter a

a= 1

a= 0. 5

a= 2. 0

Figure 4.20: Normalized radial field component.

4-31


4.5 Electric Scalar Potential

� Associated with the electric field there is also an electric po-tential V or simply voltage V

� This is the same voltage you measure in a circuit

� Similar to the concept of voltage drop across an electrical com-ponent, in field theory we are interested in the potential differ-ence between two points in space, e.g. V21 D V2 � V1 is thepotential difference observed as you move from field point P1to P2

� The definition of the potential difference lies in the work done(units of joules of (J)) in moving a charge from P1 o P2

� In physics work W is force times distance (N� m = J), i.e., tomove a charge q a differential distance d l in the electric fieldE requires work

dW D �qE � d l;

where the minus sign comes from the fact that energy is ex-pelled when we move the charge q in the opposite (or against)direction of the field

� The work per unit charge defines the potential difference, thatis 1 V = 1 J/C, i.e.,

dV DdW

qD �E � d l

4-32

4.5. ELECTRIC SCALAR POTENTIAL

� Finally, the potential difference in moving from field point P1to P2 is defined to be

V21 D V2 � V1 D �

Z P2

P1

E � d l

� In an electrical circuit the sum of the voltage drops around aclosed loop (Kirchoff’s voltage law) is zero, so too in a staticelectric field I

C

E � d l D 0

� This behavior means that the field is conservative or irrota-tional

� It is Maxwell’s second equation, for @=@t D 0,

r � E D 0

that makes this conservation property concrete

� Furthermore, via Stokes theoremZS

.r � E/ � ds D

IC

E � d l D 0

where C surrounds the surface S

� What do we use as a voltage reference?

� In circuits it is ground

� In fields it is typical to assume V1 D 0 when P1 is at infinity,i.e., V at P is

V D �

Z P

1

E � d l .V/

4-33


Electric Potential Due to Point Charges

� A point charge at the origin produces potential at radial dis-tance R given by

V D

Z R

1

�OR

q

4��R2

�� OR d R D

q

4��R.V/

� Generalizing to a point charge at location R1, we have

V Dq

4��jR �R1j.V/

� For an arbitrary configuration of point charges

V D1

4��

NXiD1

qi

jR �Ri j.V/

Electric Potential Due to Continuous Distributions

� The electric potential can be solved with the three forms ofcharge distributions we have been using

� In particular

V D1

4��

ZV 0

�v

R0dV 0

V D1

4��

ZS 0

�s

R0ds0

V D1

4��

Zl 0

�`

R0dl 0

4-34


Electric Field from the Electric Potential

� The subject of this subsection is finding the electric field afterfirst finding the electric potential

� The needed relationship comes about by first recalling fromChapter 3 that a scalar function V obeys

dV D rV � d l;

and for the just established scalar potential,

dV D �E � d l;

so putting the above equations together gives the key result

E D �rV;

which says the gradient of the potential function is the electricfield!

� Finding E via V is a two-step process, but the integrals for thescalar potential are likely easier to evaluate

Example 4.10: Charge Ring Potential

� Consider a ring of charge in the x � y plane having radius band line charge density �`

� Calculate the potential along the z axis at the point P.0; 0; h)and then E

4-35


𝑥𝑥

𝑦𝑦

𝑧𝑧

𝑏𝑏

𝜌𝜌𝑙𝑙

𝑅𝑅𝑅

(0, 0, ℎ)

𝜙𝜙 𝑟𝑟

Figure 4.21: Charge ring configuration for potential along the z axis.

� The scalar potential follows easily from Figure 4.21

V D1

4��0

Zl

�`

R0dl 0 D

�`

4��0

Z 2�

0

1pb2 C z2

b d�„ƒ‚…dl 0

D�`b

2�0pb2 C z2

ˇ̌̌̌zDh

D�`b

2�0pb2 C h2

.V/

� The electric field follows easily as well

E D �rV D Oz@

@z

�`b

2�0pb2 C z2

D Oz��`b

2�0

2z.�1=2/�b2 C z2

�3=2 ˇ̌̌̌zDh

D Oz�`bh

2�0�b2 C h2

�3=2 .V/m/

� This is consistent with the earlier Coulomb’s law calculation

4-36


Example 4.11: Electric Dipole

� An electric dipole along the z axis is formed by placing pair ofcharges˙q at z D ˙d=2 respectively

– In the study of antennas, structures for propagating elec-tromagnetic energy, another type of dipole is studied

� Calculate V and E at the field point P.R; �; �/

(a) Electric dipole

(b) Electric-field pattern

+q

–q

R1

R2

R

d y

x

z

θ

d cos θ

P = (R, θ, φ)

E

Figure 4-13 Electric dipole with dipole moment p= qd(Example 4-7).

Figure 4.22: Electric dipole configuraton and far field approxima-tion.

� The potential is

V Dq

4��0

�1

R1�1

R2

�D

q

4��0

�R2 �R1

R1R2

�.V/

� The intent of this example to obtain a far field approximation,which means for R� d , so

4-37


– R2 �R1 � d cos �

– R1R2 � R2

� Using these approximations V becomes

V �qd cos �4��0

Dp � OR4��0R2

where p D qd is the dipole moment with d and also p pointingin the direction from �q to Cq, and OR is the unit vector fromthe dipole center to the field point

� The electric field follows from �rV using the spherical coor-dinates form to calculate the gradient

E D �rV �qd

4��0R3

�OR2 cos � C O� sin �

�.V/m/

(a) Electric dipole

(b) Electric-field pattern

+q

–q

R1

R2

R

d y

x

z

θ

d cos θ

P = (R, θ, φ)

E

Figure 4-13 Electric dipole with dipole moment p= qd(Example 4-7).

Figure 4.23: Exact electric field pattern for the electric dipole.

4-38


Poisson and Laplace’s Equation in Electrostatics

� Gauss’s law was originally presented in differential form as

r �D D �v or r � E D�v

�

� If we set E D �rV we arrive at

r �

a vector‚…„ƒ�rV

�„ ƒ‚ …a scalar

D r2V D �

�v

�.Poisson’s equation/

� Recall from Chapter 3 that in Cartesian coordinates

r2V D

@2V

@x2C@2V

@y2C@2V

@z2

� If there is no charge present in the medium, Poisson’s equationreduces to Laplace’s equation

r2V D 0 .Laplace’s equation/

� Laplace’s equation in particular pops up when we want to solvefor the electrostatic potential when boundaries (boundary con-ditions), such as the plates of a capacitor, have a known poten-tial

� A course in partial differential equations considers problemsof this sort

4-39


Example 4.12: Potential Inside A Spherical Shell

� Consider a spherical shell of radius a with uniform surfacecharge density �s

� Find the potential and electric field at P.0; 0; 0/

𝑎𝑎

Surface charge density 𝜌𝜌𝑠𝑠


Field point 𝑃𝑃(0,0,0)

Side view of sphere

𝑅𝑅𝑅

Figure 4.24: Set up for finding the potential inside a spherical shellof radius a.

� From Figure 4.24 we see that

V D1

4��

Zs0

�s

ads0 D

�s

4��

Z 2�

0

Z �

0

1

aa2 sin � d� d�

D�s

4��4�a D

�sa

�.V/

� To find E we form

E D �rV D 0;

as there is no variation with R, � , or �

4-40

4.6. CONDUCTORS

4.6 Conductors

� In the section the focus is conductors and the conduction cur-rent J , introduced earlier

� Again the material constitutive parameters of permittivity, �,permeability, �, and conductivity, � are of interest

� From a materials consideration, we have conductors (metals)or dielectrics (insulators)

� What an electric field is applied to a conductor conduction cur-rent flows in the same direction as the electric field:

J D �E .A/m2/;

where � is the material conductivity in (S/m)

� This relationship is known in this context as Ohm’s law

� We expect:

– A perfect dielectric to have � D 0 (for good insulators10�17 � � � 10�10 S/m)

– A perfect conductor to have � D1 (for good conductors106 � � � 107 S/m)

Perfect dielectric: J D 0 since J D �E and � D 0Perfect conductor: E D 0 since E D J=� and � D1

4-41


4.6.1 Drift Velocity

� In a conductor electrons have a drift velocity of ue such that

ue D ��eE;

where �e is the electron mobility in (m2/V� s)

� In semiconductors there is also hole velocity and hole mobility(positive charge carriers)

uh D �hE

� The total conduction current is

J D Je C Jh D �veue C �vhuh .A/m2/;

where �ve and �vh are volume charge densities

� In particular, �ve D �Nee and �vh D Nhe, where Ne and Nhare the number of free electrons and holes respectively, per unitvolume, and e D 1:6 � 10�19 C

� Because ue=h is related to E via �e=h,

J D .��ve�e C �vh�h/„ ƒ‚ …�

E

� For a good conductor

� D ��ve�e D Ne�ee .S/m/;

4-42

4.6. CONDUCTORS

4.6.2 Resistance

� Consider the resistance,R, of a conductor of length l and crosssection A

� We assume the conductor has uniform cross section and liesalong Ox, making E D OxEx

� The voltage applied across the terminals is V , so relative toreference points x1 and x2

V D �

Z x1

x2

E � d l D Exl .V/

x1 x2l

1 2I I

A

J E

+ –

V

y

x

Figure 4-14 Linear resistor of cross section A andlength l connected to a dc voltage source V .

Figure 4.25: A resistor from a materials view point.

� The current flowing is

I D

ZA

J � ds D

ZA

�E � ds D �ExA .A/

4-43


� From Ohm’s law it follows that

R DV

ID

l

�A

� For any conductor shape R can be found as

R DV

ID�RlE � d lR

sJ � ds

D�RlE � d lR

s�E � ds

4-44

4.6. CONDUCTORS

Example 4.13: Coax Cable with Finite �

� The coax cable was studied in Chapter 2 and equations for tlineparameters were given

� Here we establish the conductance per unit length using fieldtheory and the equation for R D 1=G

–

+Vab

l

E

r

a

bσ

Figure 4-15 Coaxial cable of Example 4-9.Figure 4.26: Finding the shunt conductance of a coax cable.

� The dielectric filling is assumed to have conductivity �

� Given Vab is connected from the center conductor to the outershield, we need to find the corresponding current flow I

� For a length l section of line, the surface area at some a < r <b through which current flows is A D 2�rl

� The current density J D �E is outward radially i.e., Or, as thepotential is higher on the center conductor

J D OrI

AD Or

I

2�rlor E D Or

I

2��rl

4-45


� The voltage between the conductors, Vab, must be

Vab D �

Z a

b

E � d l D �I

2��l

Z a

b

Or � Ordrr

D �I

2��lln�r�ˇ̌̌̌ab

DI

2��lln�b

a

�

� Finally,

R DVab

ID

1

2��lln�b

a

�.�/

G 0 D1

RlD

2��

ln.b=a/.S/m/

4.6.3 Joule’s Law

� The power dissipated in a conducting medium is

P D

ZV

E � J dV DZV� jEj2 dV .W/

� For the case of a simple resistor as cylindrical conductor, Joule’slaw reduces to

P D I 2R .W/

� Power Dissipated in Coax Dielectric: For a length l line sec-tion

P D I 2R D I 2 ln.b=a/=.2��l/

4-46

4.7. DIELECTRICS

4.7 Dielectrics

� When a dielectric material is subject to an electric field, theatoms or molecules of the material become polarized

(a) External Eext = 0

(b) External Eext ≠ 0 (c) Electric dipole

––

––––

–

––

AtomNucleus

Electron

– ––

–

–––

– –

NucleusE E

Center of electron cloud

d

q

–q

Figure 4-16 In the absence of an external electricfield E, the center of the electron cloud is co-located withthe center of the nucleus, but when a field is applied, thetwo centers are separated by a distance d.

Figure 4.27: Impact of E on material atoms and the creation of adipole moment.

� When no field is present the electron cloud is symmetricalabout the nucleus (a) in Figure 4.27

� In a dielectric when the field is applied (b) in Figure 4.27, ashift occurs and E is said to polarize the atoms and create adipole

4-47


– The dipole creates its own electric field known as the po-larization field, P

� Molecules such as water have a permanent dipole moment, butthe dipoles are randomly aligned until an applied field is ap-plied

+–

+–

+–

+–

+–

+–

+–

+– +

–

+–

+–

+–

+– +

–+–

+–

+–

+–

+– +

–+–

+–

+–

+– +

– +–

+–

+–

+–

+– +

– +–

+–

+–

+–

+– +

–+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+–

+– +

–+–

+–

+–

+–

+–

+–+

–+–

+–+

–+–

+–

+–

EEEEE

Polarized moleculePositive surface charge

Negative surface charge

Figure 4-17 A dielectric medium polarized by anexternal electric field E.

Figure 4.28: Applying an electric field polarizes the molecules cre-ating an effective surface charge.

4.7.1 Polarization Field

� In a dielectric material the total flux density under the influenceof an external E is

D D �0EC P

4-48

4.8. ELECTRIC BOUNDARY CONDITIONS

where P is the electric polarization field

� In a linear, isotropic, and homogeneous medium, P is propor-tional to the applied E via

P D �0�eE

where �e is the electric susceptibility of the material

� In the end it is �e that defines the material permittivity, as

D D �0EC �0�eE D �0.1C �e/E D �E

so it must be that �r D �=�0 D 1C �e

4.7.2 Dielectric Breakdown

� Real materials are subject to dielectric breakdown

� The electric field magnitudeEds, known as the dielectric strength,is the largest field strength a material can handle without break-down

� For air Eds is about 3 (MV/m)

� For mica Eds is about 200 (MV/m), hence mica is a strongdielectric

4.8 Electric Boundary Conditions

� A vector field does not experience abrupt changes in it magni-tude or direction unless is passes from one medium to another,e.g., a dielectric interface or a metal conductor

4-49


� Boundary conditions among E, D, and J dielectric and con-ductor interfaces are now established with the aid of Figure 4.29

Δh2

Δh2

E1E1n

E1t

E2

E2n

E2t

}}

c

b a

dΔl

ε1

ε2

Medium 1

Medium 2

Δh2

Δh2

Δs

{{

ρs

n2

n1

D1n

D2nˆ

ˆl̂l1

l̂l2

Figure 4-18 Interface between two dielectric media.Figure 4.29: Establishing electric field/flux boundary conditions be-tween two mediums.

� To establish these relationships, we rely on:IC

E � d l D 0, r � E D 0 .conservation prop./IS

D � ds D Q, r �D D �v .divergence prop./

� Working through the details sketched out in Figure 4.29, weestablish from the conservation of E that at an interface thetangential electric field component is continuous, i.e.,

E1t D E2t .V/m/

� Similarly working from the divergence property the flux den-sity normal to the interface is continuous, subject any addedcharge density that may be present, i.e.,

On2 � .D1 �D2/ D �s .C/m2/

4-50


orD1n �D2n D �s .C/m2/

� Any abrupt change in the flux density, D, normal to the in-terface, is due to a surface charge density being present at theinterface

� The above results in Table 4.1, including a specialization for adielectric conductor interface, yet to be explained

Table 4.1: Electric field/flux boundary conditions.Table 4-3 Boundary conditions for the electric fields.

Field Component Any Two Media Medium 1Dielectric ε1

Medium 2Conductor

Tangential E E1t = E2t E1t = E2t = 0

Tangential D D1t/ε1 = D2t/ε2 D1t = D2t = 0

Normal E ε1E1n− ε2E2n = ρs E1n = ρs/ε1 E2n = 0

Normal D D1n−D2n = ρs D1n = ρs D2n = 0

Notes: (1) ρs is the surface charge density at the boundary; (2) normalcomponents of E1,D1, E2, andD2 are along n̂2, the outward normal unit vectorof medium 2.

Example 4.14: A Dielectric–Dielctric Interface

� An important special case to consider is when E travels from amaterial having permittivity �1 to �2

� From physics you may recall that when light rays pass fromone medium to the next, there is a direction change that takesplace due to the change in the index of refraction

4-51


� The same concept applies here and is in fact related

� Consider now the scenario of Figure 4.30

E1z

E1t

E2t

E2z

E1

E2 θ2

ε1

ε2

θ1

z

x-y plane

Figure 4-19 Application of boundary conditions at theinterface between two dielectric media (Example 4-10).

Figure 4.30: Electric field angle change at an �1 to �2 interface.

� Assume that E1 D OxE1xC OyE1yC OzE1z and find E2 D OxE2xCOyE2yC OzE2z in terms of E1, also find the relationship betweenthe angles �1 and �2 (assume �s D 0 at the interface)

� The continuity of Et means that

E1t D E2t ) E1x D E2x and E1y D E2y

� Similarly for the normal components, which by constructionof the problem lie along the z axis,

�1E1z D �2E2z

� So we can write thatE2 D OxE2x C OyE2y C OzE2zD OxE1x C OyE1y C

�1

�2OzE1z

4-52


� The angle reationships are

tan �1 DE1t

E1nDE1t

E1zD

qE21x CE

21y

E1z

tan �2 DE2t

E2nDE2t

E2zD

qE22x CE

22y

E2z

alsoD

qE22x CE

22y

.�1=�2/E1z

� In particulartan �2tan �1

D�2

�1

4.8.1 Dielectric-Conductor Boundary

� Consider the special case of medium 1 a dielectric and medium2 a conductor

� In a perfect conductor there are no fields or fluxes, i.e., E D

D D 0, so for the tangential and normal boundary conditionswe have

E1t D D1t D 0

D1n D �1E1n D �s

� The key result from the above is that a charge density at theconductor surface is induced by the normal component of theelectric field

D1 D �1E1 D On�s

4-53


E1 E1 E1

E1 E1 E1Ei Ei Ei

+ + + + +

– – – – – – – – – – –– – – – –

+ + + + + + + ++ + +

ρs = ε1E1

−ρs

Conducting slab

ε1

ε1

Figure 4-20 When a conducting slab is placed in an external electric field E1, charges that accumulate on the conductorsurfaces induce an internal electric field Ei = −E1. Consequently, the total field inside the conductor is zero.

Figure 4.31: The interaction of an electric field in a sandwidge ofdielectric-conductor-dielectric.

� Also, the flux lines at the conductor interface are always nor-mal ( On) with a positive charge density when E1 is away fromthe interface and a negative charge density when E1 is towardthe interface (see Figure 4.31)

Example 4.15: Metallic Sphere in a Uniform E Field

� Consider a metallic sphere placed in a uniform electric field

–

E0

metalsphere

+ + ++

++

++

++

– –––

––

–––

Figure 4-21 Metal sphere placed in an external electricfield E0.

Figure 4.32: The flux lines bend at the surface of a metallic sphereto insure they remain normal everywhere.

4-54


4.8.2 Conductor-Conductor Boundary

� Taking the special case one step further, suppose we have twoconductors of different conductivity interfaced

– Note, neither conductor is perfect in this scenario

Medium 1ε1, σ1

Medium 2ε2, σ2

J1n

J2n

J1t

J2t

J1

J2

n̂

Figure 4-22 Boundary between two conducting media.Figure 4.33: A finite conductivity conductor–conductor interfacescenario.

� The boundary conditions require that

E1t D E2t and �1E1n � �2E2n D �s

� Since for conductors Ji D �iEi we can also write that

J1t

�1D

J2t

�2and �1

E1n

�1� �2

E2n

�2D �s

� The tangential components can coexist as parallel current flow

� The normal components cannot be different, since this requiresa �s that is not constant with time and hence not a static condi-tion

4-55


� To resolve this dilemma, we force J1n D J2n, so

J1n

��1

�1��2

�2

�D �s

4.9 Capacitance

� Any two conductors in space, separated by a dielectric (air isvalid), form a capacitor

� A voltage placed across the two conductors allows CQ and�Q charges to accumulate

� The ratio of charge to voltage defines the capacitance in farads

C DQ

V.C/V or F/

–– – –

–

–––––

–

E

Surface S

V+

–

ρs

+ + + ++

+

++++

+ +Q

Conductor 1

−QConductor 2

Figure 4-23 A dc voltage source connected to acapacitor composed of two conducting bodies.

Figure 4.34: Establishing the capacitance between two conductorsby applying voltage V .

4-56

4.9. CAPACITANCE

� Note: The charge on each conductor is distributed to insurethat E D 0 within the conductor and the potential is the sameat all points

� We know from an earlier discussion that only the normal com-ponent of E exists, so

En D On � E D�s

�

� The total charge over the positively charged conductor is

Q D

ZS

�sds D

ZS

�E � ds

� The voltage V is formally

V D V12 D �

Z Cond. 1

Cond. 2E � d l

Putting the pieces together we have,

C D

RS�E � ds

�RlE � d l

.F/

Note: C is positive and a function of only the geometry andthe permittivity

� If the material has loss, i.e., a small conductivity, then there isalso a resistance between the two conductors given by

R D�RlE � d lR

S�E � ds

as we have seen earlier

4-57


� Interesting Observation: For materials having uniform � and�, it follows that

RC D�

�;

so given C then R is known and likewise given R, C is known

Example 4.16: Classical Parallel Plate Capacitor

� From physics you likely recall the parallel plate capacitor and�A=d , where A is the plate area and d is the plate separation

++

++

++

+

––

––

––

–ρs

–ρs

–Q

EE

Area A

+Q

z = d

z

+ + + + + + + + + +

–

+

–– – – – – – – – –

ds E EEV

Conducting plate

Conducting plate

Dielectric ε

z = 0

Fringing field lines

Figure 4-24 A dc voltage source connected to a parallel-plate capacitor (Example 4-11).Figure 4.35: Parallel plate capacitor analysis model using appliedvoltage V .

� With respect to the top plate E D �OzE and from the boundaryconditions E D �s=�

� Also the neglecting fringing fields, the charge density �s is uni-form, so Q D �sA! E D Q=.�A/

� The voltage across the plates is

V D �

Z d

0

E � d l D �

Z d

0

.�OzE/ � Oz dz D Ed;

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4.9. CAPACITANCE

so

C DQ

VDE�A

EdD�A

d.F/

Example 4.17: Coax Capacitance

� Another classical structure to analyze is the coax capacitor

+ + + + + + + + + + + +

+ + + + + + + + + + + +

– – – – – – – – – – –

+

– –

– – – – – – – – – – – –

+

–

V+

– b

a

ρl–ρl

l

E E E

E E EDielectric material ε

Outer conductor

Inner conductor

Figure 4-25 Coaxial capacitor filled with insulating material of permittivity ε (Example 4-12).Figure 4.36: Coax capacitor analysis model using applied voltage V .

� Using Gauss’s law and knowing the form of the E field for aninfinite line charge (pure radial and inverse proportional to r),we have

E D �OrQ

2��rl

� The potential can be calculated as

V D �

Z b

a

E � d l D �

Z b

a

��Or

Q

2��rl

�� Or dr

DQ

2��lln.r/

ˇ̌̌̌ba

DQ

2��lln�b

a

�� Finally,

C DQ

VD

2��l

ln.b=a/.F/

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� In terms of capacitance per unit length we have

C 0 DC

lD

2��

ln.b=a/.F/m/

4.10 Electrostatic Potential Energy

� When a voltage is applied to a lossless capacitor energy goesinto the structure and is stored in the electric field

� What work is done in charging up this capacitor?

� When the voltage is applied we are moving charge from oneplate to another

� A voltage increment � corresponds to charge q=C , so the dif-ferential electrostatic work, We, is

dWe D � dq Dq

Cdq

� Building up a total charge Q accumulates total work

We D

Z Q

0

q

Cdq D

1

2

Q2

C.J/;

but since C D Q=V , Q2 D C 2V 2 and substituting yields

We D1

2CV 2 .J/

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4.10. ELECTROSTATIC POTENTIAL ENERGY

� When an electric field is present in a region it can be viewed asan electrostatic energy density via

we DWe

V D1

2�E2 .J/m3/

� From the energy density it follows that the potential energystored in volume V is,

We D1

2

ZV�E2 dV .J/

Example 4.18: Energy Stored in Coax

� In the coax capacitance example we found that

C D2��l

ln.b=a/

and

V DQ

2��lln.b=a/ D

l�`

2��lln.b=a/;

where Q D l�`

� Using We D .1=2/CV2 we have

We D1

2

�2��l

ln.b=a/

�� `2��

ln.b=a/�2

D1

2

l�`

2��ln�b

a

�.J/

4-61


� Starting from energy density and integrating over the volumewe should get the same answer

We D1

2

Z 2�

0

Z l

0

Z b

a

�� `

2��r

�2rdr dz d�

D2��l

2

�2`.2�/2�2

ln�b

a

�D1

2

�2`2��

ln�b

a

�.J/

The same result!

� Suppose that a D 2 cm, b D 5 cm, �r D 4, �` D 10�4 C/m,and l D 20 cm

� Plugging the numbers into either of the above equations yields

We D 4:12 .J/

Page 2 of 2ECE3110_Chapter4a

xxxxxxxxx

yyyyyyyyy

6.56.56.56.56.56.56.56.56.5−0.037−0.037−0.037−0.037−0.037−0.037−0.037−0.037−0.037

8.188.188.188.188.188.188.188.188.18

0.50.50.50.50.50.50.50.50.5

f1 x =e x,lsf2 x =

1

x 2

ls =1.91. 10.

Coax Stored Energy

W> = 12

·0.2· 10−4 2

2·π·4·8.85·10−12· ln

52

▸ 4.11955 (J)

On the nspire you can enter an array of three element lists usingFigure 4.37: TI Nspire calculation of We.

4.11 Image Method

� When a charge distribution is placed over an infinite groundplane Coulomb’s law and Gauss’s law cannot be readily ap-plied

4-62

4.11. IMAGE METHOD

� Solving Poisson’s or Laplace’s equation is an option, but this isalso mathematically challenging, likely to require a numericalsolution

� It turns out that an electrically equivalent problem can be cre-ated using the image distribution with the ground plane re-moved

� A simple example of a single point chargeQ distance d abovea ground plane is shown if Figure 4.38

(a) Charge Q above grounded plane (b) Equivalent configuration

Electric fieldlines

V = 0d

d

Q

−Q

ε

ε

+

–

V = 0 d

z

Q

σ = ∞

ε+

n̂

Figure 4-26 By image theory, a chargeQ above a grounded perfectly conducting plane is equivalent to Q and its image−Qwith the ground plane removed.

Figure 4.38: The image theory concept in solving electrostaticsproblems with charge over a uniform ground plane.

4-63


Example 4.19: Ulaby 4.71 – A Corner Reflector Charge

� Construct image distribution for a corner reflector charge

d

d

z

y

P = (0, y, z)

Q = (0, d, d)

Figure P4.61 Charge Q next to two perpendicular,grounded, conducting half-planes.

Figure 4.39: Ulaby 4.61.

4-64

4.11. IMAGE METHOD

Example 4.20: Ulaby 4.63 – Conducting Cylinder Over a GroundPlane

� An infinite length charged cylinder over a ground plane is anassigned homework problem

V = 0

a

d

Figure P4.63 Conducting cylinder above a conductingplane (Problem 4.63).

Figure 4.40: Capacitance of an Infinite Cylinder Over a GroundPlane.

4-65

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