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TRANSCRIPT
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
CHAPTER 4. ELECTROSTATICS
– 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
CHAPTER 4. ELECTROSTATICS
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
CHAPTER 4. ELECTROSTATICS
� 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
CHAPTER 4. ELECTROSTATICS
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
4.3. COULOMB’S LAW
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
CHAPTER 4. ELECTROSTATICS
� 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
4.3. COULOMB’S LAW
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
CHAPTER 4. ELECTROSTATICS
(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
4.3. COULOMB’S LAW
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
CHAPTER 4. ELECTROSTATICS
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
4.3. COULOMB’S LAW
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
CHAPTER 4. ELECTROSTATICS
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
4.3. COULOMB’S LAW
� 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
CHAPTER 4. ELECTROSTATICS
� 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
4.3. COULOMB’S LAW
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
CHAPTER 4. ELECTROSTATICS
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
CHAPTER 4. ELECTROSTATICS
� 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
CHAPTER 4. ELECTROSTATICS
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
CHAPTER 4. ELECTROSTATICS
� 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
CHAPTER 4. ELECTROSTATICS
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
𝐄𝐄 = �𝐑𝐑𝐸𝐸𝑅𝑅
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
CHAPTER 4. ELECTROSTATICS
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
CHAPTER 4. ELECTROSTATICS
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
4.5. ELECTRIC SCALAR POTENTIAL
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
CHAPTER 4. ELECTROSTATICS
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
4.5. ELECTRIC SCALAR POTENTIAL
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
CHAPTER 4. ELECTROSTATICS
– 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
4.5. ELECTRIC SCALAR POTENTIAL
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
CHAPTER 4. ELECTROSTATICS
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
𝐄𝐄 = �𝐑𝐑𝐸𝐸𝑅𝑅
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
CHAPTER 4. ELECTROSTATICS
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
CHAPTER 4. ELECTROSTATICS
� 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
CHAPTER 4. ELECTROSTATICS
� 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
CHAPTER 4. ELECTROSTATICS
– 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
CHAPTER 4. ELECTROSTATICS
� 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
4.8. ELECTRIC BOUNDARY CONDITIONS
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
CHAPTER 4. ELECTROSTATICS
� 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
4.8. ELECTRIC BOUNDARY CONDITIONS
� 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
CHAPTER 4. ELECTROSTATICS
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. ELECTRIC BOUNDARY CONDITIONS
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
CHAPTER 4. ELECTROSTATICS
� 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
CHAPTER 4. ELECTROSTATICS
� 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;
4-58
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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CHAPTER 4. ELECTROSTATICS
� 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/
4-60
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
CHAPTER 4. ELECTROSTATICS
� 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
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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
CHAPTER 4. ELECTROSTATICS
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