Embed Size (px)
Citation preview
ECE 307: Electricity and Magnetism
Fall 2012
Instructor: J.D. Williams, Assistant Professor
Electrical and Computer Engineering
University of Alabama in Huntsville
406 Optics Building, Huntsville, Al 35899
Phone: (256) 824-2898, email: [email protected]
Course material posted on UAH Angel course management website
Textbook:
M.N.O. Sadiku, Elements of Electromagnetics 5th ed. Oxford University Press, 2009.
Optional Reading:
H.M. Shey, Div Grad Curl and all that: an informal text on vector calculus, 4th ed. Norton Press, 2005.
All figures taken from primary textbook unless otherwise cited.
8/17/2012 2
Electrostatic Fields
• Applications of electrostatics – Electric Power Transmission
– X-ray machines
– Lighting Protection
– Solid State Electronics
• Capacitors, Resistors, Field Effect Transistors
– Touch pads
– Keyboards
– Cathode Ray Tubes
– Electrocartiograms
– Particle separation
– Aspects of Electrochemistry
– Electrostatic charge transfer in living organisms
8/17/2012 3
Chapter 4: Electrostatic Fields
• Topics Covered
– Coulomb’s Law and Field
Intensity
– Electric Fields due to
Continuous Charge Distributions
– Electric Flux Density
– Gauss’s Law
– Applications of Gauss’s Law
– Electric Potential
– Relationship Between E and V
– An Electric Dipole and Flux
Lines
– Electric Density in Electrostatic
Fields
All figures taken from primary textbook unless otherwise cited.
Homework: 3,9,11,17,19,25,35,35,45,51, 49,54,55
8/17/2012 4
Fundamental Problem of
Electromagnetic Field Theory:
• If one has some number of source charges (Q1,Q2,Q3), then what force do they exert on another (test) charge?
– In general the positions of the source are given as a function of time.
– Also, as a general rule, both source and test charges are in motion.
– For such cases, our goal is to calc. the trajectory of the test charge.
– Solutions to these problems rely heavily on the principle of superposition:
• The interaction between 2 charges is completely independent of the presence of other neighboring charges. Thus, we can determine the force acting on a test charge by summing each force applied to the test charge by neighboring charges individually.
– Sound easy? It’s not!!!
• In general, the force acting on a charge is not simply dependent on the position vector, r, but also on the velocities and accelerations of the two charges. As you might imagine, such solutions are very complex.
• However we can reduce the problem to something more manageable by examining only cases in which charges are fixed in space over the time interval posed by the problem
• Such problems are referred to as Electrostatic Problems and are the subject of the next three chapters of this book.
8/17/2012 5
Two Methods Useful Methods for
Solving Electrostatic Problems • Coulomb’s Law
– Can be used for any spatial configuration
– Will be used to study point, line, surface and volume charges
• Gauss’ Law
– Often a much easier method for cases involving highly symmetric
geometries.
2
21
4 R
QQF
o
o
S
o
E
adEQ
Integral Form
Differential Form
8/17/2012
6
Summary Diagram of Electrostatics
VE
0, EldEV
0
E
Eo
v
E V
o
vV
2
d
rV v
o4
1
d
r
aE rv
o
24
1
Figure is recopied from Griffiths, Introduction to Electrodynamics,3rd ed., Benjamin Cummings, 1999.
Charge Density
Electric Field Potential Electric Field
8/17/2012 7
Point Charge Distributions and
Coulomb’s Law • The force, F, between two point charges Q1 and Q2 is:
– Along the line joining them
– Directly proportional to the product between them
– Inversely proportional to the square of the distance between them
– The equation above is easily calculated for a test charge, Q1, at the
origin and a source charge, Q2, at a distance R away.
– The solution is slightly more complicated as we move our reference
frame away from the two charge such that Q1 is referenced by the
vector r1 and Q2 is referenced by the vector r2.
2
21
4 R
QQF
o
where , 2212 /10854.8 NmCo
o
k4
1
JAVA Applet
8/17/2012 8 http://www.falstad.com/mathphysics.html
8/17/2012 9
Point Charge Distributions and
Coulomb’s Law (cont.)
12
1212
RR
rrR
• The solution is slightly more
complicated as we move the reference
frame to some location away from the
two charges
• The location of Q1 is referenced by the
vector r1
• The location of Q2 is referenced by the
vector r2
• The vector between Q1 and Q2 is R12
• Where the R12 normal vector is a12
R
Rar
1212
ˆ
8/17/2012 10
Point Charge Distributions and
Coulomb’s Law (cont.)
• Things to note: – .
– Like charges repel each other, opposites attract
– The distance |R12| between Q1 and Q2 must be large with respect to the diameter of the
charges
– Q1 and Q2 must be static for this solution to be valid
– The signs of Q1 and Q2 must be taken into account when solving for the force
3
12
1221
2
12
1221122
12
2112
)(
4
ˆ
44
rr
rrQQ
rr
aQQa
R
QQF
o
o
r
o
or )( 1212211221 rr aFaFF
1221 FF
8/17/2012 11
Multipole Solutions
• If, as stated before, there are more than two charges, then one can solve
for the same force acting on the test charge Q, as
k k
kk
o
ooo
Tot
rr
rrQ
Q
rr
rrQQ
rr
rrQQ
rr
rrQQF
3
3
3
33
3
2
22
3
1
11
)(
4
...)(
4
)(
4
)(
4
Insert Example in class
• Three equal point charges are separated as shown. Each of the charges is connected by a very thin string designed to break when a force of 0.1N is applied in tension.
– Calculate the charge required to break the strings, if a = 20mm
– What is the electric field intensity at the center of the string AB?
Multi-pole Force Example
8/17/2012 12
Problem and Figure from N. Ida, Engineering Electromagnetics, 2ed, Springer, 2003
8/17/2012 13
Electric Field
• The electric field intensity vector, E, is the force per unit charge when placed in an electric field
• The electric field is independent of the test charge defined within the space
• It makes no reference at all to a test charge, thus, one accurate understanding of the electric field is to asses the force acting on a point, P, and then divide out the test charge used to sum each of the component forces
k k
kk
o rr
rrQE
3
)(
4
1
8/17/2012 14
Continuous Charge Distributions
• Line, surface, and volume charge distributions are customarily solved as
continuous charge media. As such, these distributions of charge can be
evaluated by integration (of differentiation) of the “charge density” in space.
Line
Surface
Volume
Leading to E field solutions:
dvQ
dsQ
dlQ
VV
SS
LL
Lr
o
L aR
dlE
24
S
r
o
S aR
dsE
24
V
r
o
V aR
dvE
24
Ls v
Note: your text uses for density here instead of . Do not let yourself get
confused between surface charge density and the cylindrical axis unit vector
8/17/2012 15
Line Charges
• Consider a uniform charge density from A to B along the Z axis
zB
zAL
LL
dzQ
dzdldQ
dzdl
E(x,y,z) is found by integration
22
222
)'(
)'(
]',,[)',0,0(),,(
'
zz
zzyxRR
zzyxzzyxR
dzdl
Let,
Note: this form would be difficult to solve for
Then, Lr
o
L aR
dlE
24
2/32232)'(
)'(
zz
azza
R
R
R
a zR
And
8/17/2012 16
Line Charges (cont.)
sec
cos)'( 22 zzR
L
z
o
L
zz
dzazzaE
2/322 )'(
')'(
4
2sec'
tan'
dz
OTzFinally
A
B
R
hypadj /cos Using
O
T
2
1
22
2
sec
sincossec
4
daaE
z
o
L
22 )'(cos
zz
22 )'(
'sin
zz
zz
L
z
o
L
zz
dzaaE
22 )'(
'sincos
4
2
1
22 sec
'sincos
4
z
z
z
o
Ldzaa
E
Next,
8/17/2012 17
Line Charges (cont.)
• Consider a uniform charge density from A to B along the Z axis
z
o
L
z
o
L
z
o
L
aaE
daaE
daaE
1212
2
1
2
1
22
2
coscossinsin4
sincos4
sec
sincossec
4
Special Case: Infinite line charge
aaE
o
L
o
L
22
4
8/17/2012 18
Line Charge Example 2
Find the electric field a distance z above the midpoint of a straight line segment
of length 2L, which carries a uniform line charge density, .
Rz
dxdl
L
/cos
P
z
x
O
Note: symmetry provides a field propagating
in the z direction and Ez = E cos
L
z
o
aR
zdxE
0
32
4
L
zL
o
Lz
L
oL
r
o
L
aR
dl
aR
dla
R
dlE
0
2
22
cos24
1
cos4
1
4
22
22
coszx
z
zxR
L
z
o
azx
zdxE
0
2/322 )(2
4
8/17/2012 19
Line Charge Example 2 (cont.)
Find the electric field a distance z above the midpoint of a straight line segment
of length 2L, which carries a uniform line charge density, .
P
z
x
O
z
o
L
z
o
L
z
o
azLz
L
azxz
xz
azx
zdxE
22
0
222
0
2/322
2
4
1
4
2
)(2
4
Z >> L
z
o
z
o
az
E
az
LE
2
4
1
2
4
12
L
Note: symmetry provides a field propagating
in the z direction and Ez = E cos
8/17/2012 20
Surface Charge • Consider a sheet of charge in the xy-plane with uniform charge density, s.
Solve for the electric field at point P=(0,0,h)
RRa
hRR
ahaR
dddsdQ
R
z
SS
/
)(
22
S
z
o
S
h
ddahaE
2/3224
Sr
o
S aR
dsE
24
8/17/2012 21
Surface Charge (cont.) • Consider a sheet of charge in the xy-plane with uniform charge density, s.
Solve for the electric field at point P=(0,0,h)
z
o
Sz
o
S
z
o
S
z
o
S
aahh
ah
dh
ah
ddhE
22
2
)(
4
)2(
4
0
2/122
0 2/322
2
2
00 2/322
Electric Field is independent of the
distance away from the plane!!!!!!
Can you calculate the field between
two planes? What is the physical
significance of this question?
RRa
hRR
ahaR
dddsdQ
R
z
SS
/
)(
22
Note: The symmetry of this problem
requires that all solutions for E cancel
8/17/2012 22
More Examples of Distributed
Charge Problems • Circular ring of charge with radius
a, carries a uniform charge density
L, and is placed on the xy plane
and centered on the z axis – Find the electric field at P(0,0,h)
– What values of h give the max electric
field intensity?
– If the total charge on the ring is Q,
approximate the field intensity as the
radius goes to zero. _____________________________________________________________________
Find the electric field at P(0,0,h)
2/32232
22
)(
ha
ahaa
R
R
R
a
haR
ahaaR
addl
zr
z
2/322
2/322
2
2
4
4
ha
aha
adha
ahaa
dlR
aE
z
o
L
z
o
L
r
o
L
zero
8/17/2012 23
Ring of circular charge (cont.)
What values of h give the max electric field intensity?
The extremum of any function can be found by taking the differential with respect to the variable
of interest and setting it equal to zero.
If the total charge on the ring is Q, approximate the field intensity as the radius goes to zero.
Uniform charge density requirement implies that Q=L2a, thus
22/32
2/3222/322
44
42
h
aQ
h
ahQ
ha
ahQ
ha
ahaE
z
o
z
o
z
o
z
o
L
zero
2
02
02
22/3
20
22
222/122
322
2/1222/322
ah
ha
haha
ha
hahhhaa
dh
Ed
o
l
8/17/2012 24
Circular Disk of Charge
A circular disk of radius a, and s centered at z=0. Find E at P(0,0,h)
2/3222
22
)(
h
aha
R
a
hRR
ahaR
dddsdQ
zR
z
SS
z
o
S
z
a
o
S
az
o
S
Sr
o
S
ahha
h
ah
h
ddh
ahE
aR
dsE
11
4
)2(
1
4
)2(
4
4
2/122
0
2/122
2
00 2/322
2
z
o
S
z
o
S
z
o
S
aE
az
aha
h
ahah
hE
2
,0
12
11
4
)2(
2/122
2/122
8/17/2012 25
Rectangular Surface Charge
• Consider the following flux sheet
CdxdyyxxydsQs
S
9
1
0
1
0
2/322 10)25(
22/322 /)25(
10
10
mnCyxxy
y
x
s
x
y
1
1
s
Force expressed by Q=-1mC at P
So
S
Sr
o
S
rr
rrdsa
r
dsE
32'
)'(
44
Total enclosed charge on the surface
where
2/322325
)5,,(
'
'
)5,,(
)0,,()5,0,0(
''
yx
yx
rr
rr
yx
yx
rrrr p
EqF
mV
mVdxdyyxxy
Cdxdyyx
yx
mF
yxxyE
/)25.11,5.1,5.1(
/)5,,(9
1025
)5,,(
/36
104
)25(
1
0
1
0
9
1
0
1
0
2/3229
2/322
The field intensity at P(0,0,5)
mFo /36
10 9
Using the value
given for permittivity
of free space on
page 106 of your
text
8/17/2012 26
Solutions for Multiple Charge Densities
• What if we want to look at the field intensity due to multiple charge densities
of different shapes at different locations?
Well,
So for a series of different charge densities at:
s1=10 nC/m2 at x=2
s2=15 nC/m2 at y=-3
l3=10 nC/m2 along x=0, z=2
Find E at P(1,1,-1)
i
itot EE
yyy
o
s
xxx
o
s
amVa
mF
mCaE
amVa
mF
mCaE
/270
/36
102
)/1015()(
2
/180
/36
102
)/1010()(
2
9
29
22
9
29
11
8/17/2012 27
Solutions for Multiple
Charge Densities (2) l3=10 nC/m2 along x=0, z=2
mVE
mVaa
aa
mF
mC
aaE
aaa
R
LPR
R
aE
tot
zx
zx
zx
o
s
zx
o
l
/)54,270,162(
/)3(18
)3(
)10(/36
102
)/1010(
)3(10
1
10
1
2
)3(10
1
10
)3,0,1()2,0,0()1,0,1(
2
9
29
13
33
8/17/2012 28
Charge Volumes
• Assuming a uniform charge density, the total
charge on in a sphere of radius, a, is:
• What is the electric field on the surface?
• What about the electric field at a point P (0,0,z)
from the from the center of a sphere with radius
a, centered at the origin?
• where
v
v
v
RdvQ
3
4 3
R
o
v aR
dvEd
24
aaa zR
sincos
r
o
vrv
o
r
v o
v aa
aa
aa
a
dvE
3)4(3
4
4 2
3
2
8/17/2012 29
Charge Volumes (2)
• The symmetry of the charge distribution
requires that the sum of the contributions
from Ex and Ey equal zero.
• We are left only with Ez terms in the
solution
• where
2
cos
4cos
R
dvEdaEE
o
vzz
'''sin''2 dddrrdv
8/17/2012 30
Geometry of Oblique Triangles
• Consider
• The most convenient way
to express the interval is
in terms of R and r’,
cos2'
'cos'2
222
222
zRRzr
zrrzR
''sin'
'2
''cos
2
'cos
222
222
zr
RdRd
zr
Rrz
zR
rRz
Where
z
r’ R
’
Because the length of the
vector from the surface to the
point varies as a function of R
Law of
Cosines
8/17/2012
31
Charge Volumes (3)
• As ’ goes from 0 to , R varies from (z-r’) to (z+r’) if P is outside the sphere
• Direct substitution yields:
a
r
rz
rzRo
v
a
r
rz
rzRo
v
a
r
rz
rzRo
vz
o
v
o
vz
dRdrR
rzr
z
drrRzR
rRz
zr
RdR
ddrrR
dE
dddrrRR
dvEdE
0'
'
'
2
22
2
2
0' 0'
'
'
2
2
222
2
0' 0'
'
'
2
2
2
22
''
1'8
2
'''1
2
'
'4
'''1
cos''sin4
'''sincos
4
cos
4cos
• Note that this is identical to the electric field due to a point charge at a
distance z>>a .
8/17/2012
32
Charge Volumes (3)
• As ’ goes from 0 to , R varies from (z-r’) to (z+r’) if P is outside the sphere
• Direct substitution yields:
2
3
2
0'
2
2
0'
'
'
22
2
0'
'
'
2
22
2
43
4
4
1''4
4
''
'4
''
1'8
2
z
Qa
zdrr
z
drR
rzRr
z
dRdrR
rzr
zE
o
v
o
a
ro
v
a
r
rz
rzo
v
a
r
rz
rzRo
vz
• Note that this is identical to the electric field due to a point charge at a
distance z>>a .
8/17/2012 34
Electric Flux Density • The electric field intensity, E, is used to quantify a field in vacuum
where the dielectric of the space between charges is said to be 1
(equal to the permittivity of free space).
• This alone is not representative of fields in various media
• Thus we define the electric flux density (or electric displacement),
D, as D=oE+P where P is the polarization due to the media
– For the moment we will assume that no polarization occurs and that P is equal
to zero. A detailed discussion of polarized media will be presented in Chapter 5
• Using the displacement vector we can find the electric flux through
a surface as
• Examples: Infinite sheet charge Volume charge distribution
2
2
xs
o
xs
aD
aE
v
Rv
v o
Rv
R
advD
R
advE
2
2
4
4
s
SdD
8/17/2012 35
Gauss’ Law
• Gauss’ Law: The electric flux, , through any closed surface is equal to the
total charge enclosed by that surface, thus =Qenc
Integral Form
Applying the divergence theorem, we have
yielding the differential form
This is the first of the 4 Maxwell Equations which clearly states that the
volume charge density is equal to the divergence of the electric flux density
t
BE
D v
dvdvDSdDv
v
vs
vD
enc
v
v
s
QdvSdD
t
DJH
B
0Maxwell’s
Equations
In matter
8/17/2012 36
Gauss’ Law(2)
• Gauss’ law is simply an alternative statement of Coulomb’s law.
• Gauss’ law provides an easy means of finding E or D for symmetrical charge
distributions
• Applications:
rr
enc
rrenc
s
enc
r
r
ar
QEa
r
QD
rDddrDQ
ddraaDQ
SdDQ
ddraSd
aDD
2
0
2
22
2
0 0
2
2
0 0
2
4,
4
4sin
sin
sin
Point Charge
8/17/2012 37
Gauss’ Law:
Uniformly Charged Sphere
r
o
vr
v
r
s
r
s
v
v
v
venc
r
o
vr
v
r
s
r
s
v
v
v
venc
ar
aEa
r
aD
rDddrDSdD
addrdrdvQ
ar
ar
Ear
D
rDddrDSdD
rddrdrdvQ
ar
2
3
2
3
22
32
22
32
3,
3
4sin
3
4sin
3,
3
4sin
3
4sin
0
Outside Sphere Inside Sphere
8/17/2012 38
Gauss’ Law:
Uniformly Charged Sheet
zs
enc
s
zz
bottomstops
z
s
s
s
senc
z
zz
aD
QSdD
ADAAD
dsdsDSdD
AdsQ
dxdyaSd
aDD
2
2)(
8/17/2012 39
Gauss’ Law: Infinite Line Charge
aEaD
llD
Q
dzdDSdD
lQ
dzdaSd
aDD
LL
L
enc
l
s
Lenc
0
2
0 0
2,
2
2
Gauss’s Law Example:
Two Uniformly Charged Shells
(or Hollow Conducting Shell)
8/17/2012 40
Problem and Figure from N. Ida, Engineering Electromagnetics, 2ed, Springer, 2003
0
4
4
,ˆ4
4
0
0
2
0
2
0
2
0
0
2
2
E
Rb
b
QE
a
QE
bRarR
QE
QRESdE
bRa
E
aR
b
a
S
Gauss’s Law Example:
Point Charge Inside Two Uniformly Charged Shells
8/17/2012 41 r
R
QE
QRESdE
bR
E
bRa
rR
QE
QRESdE
aR
S
S
ˆ4
4
0
ˆ4
4
0
2
30
0
2
3
3
2
2
10
0
2
1
1
3
1
To simplify the problem, solve each
step for a different radial value, Ri
Problem and Figure from N. Ida, Engineering Electromagnetics, 2ed, Springer, 2003
8/17/2012 42
Solutions where D is
varied in space
222
22
22
22
/5.0/4
cos1
3,4
,1),,(
/cos
/cos
/cos
mCmC
PzP
mC
z
mCzD
mCazD
v
v
v
z
4m 1m
Z=0
D = zcos2 az C/m2
P(1,/4,3)
Now solve for the total volume charge:
Use Gauss’s Law Differential form:
Method 1: CdzdddvQ
vv
venc
3
4cos2
8/17/2012 43
Solutions where D
varies in space (2)
4m 1m
Z=0
D =zcos2az C/m2
P(1,/4,3)
Now solve for the total volume charge:
Use Gauss’s Law Differential form:
Method 2:
CQ
CCddz
CCddz
SdDSdDSdDSdDQ
i
i
zbottom
ztop
bottoms sidestopssi
i
3
4
3
2cos
3
2cos
2
22
2
22
Note: 0 aaz
222
22
22
22
/5.0/4
cos1
3,4
,1),,(
/cos
/cos
/cos
mCmC
PzP
mC
z
mCzD
mCazD
v
v
v
z
8/17/2012 44
Path Independence
of the Electric Field • Let us calculate E using the simplest possible configuration
– a point charge at the origin
• Now let us calculate the integral of this field between any points a and b
• This solution is clearly path independent and depends only on the distances of a and
b from the point charge. It cost nothing to move in the and directions, because
the field has no angular dependence. Thus, if I solve for any closed loop in which
ra=rb, then obviously
r
o
ar
qE
24
1
bao
b
ao
b
ao
b
arr
q
r
qdr
r
qldE
11
44
1
4
12
b
a
ldE
adrardadrld r
sin
0 ldE
8/17/2012 45
Path Independence
of the Electric Field (2) • Next, recall Stokes Theorem:
• Thus the Curl of E must also equal zero for any single point charge at the
origin. Note however that these results hold no matter where the charge is
located or if we have many different charges. Why? Because the fields of
individual charges are linearly independent, allowing us to sum any field in
its entirety using the principle of superposition.
• Also recall that If Curl E = 0, then the field is said to be an irrotational or a
potential field. These fields are described using scalars (magnitude but no
direction). This is why we use electrostatic potential, V, so often to
describe systems in which only electrostatic fields are of concern.
LS
ldESdE
0 E
0...)()(...)(
...
2121
21
EEEEE
EEE
Maxwell’s 2nd equation
for electrostatics
8/17/2012 46
Electric Potential
• We define the electric potential (or the irrotational scalar field), V, describing any electrostatic vector field, E, as the magnitude of the difference between E at two
points a and b and some standard (common) reference point o.
• Now, the Gradient theorem states that
b
a
o
a
b
o
p
o
ldEldEldEaVbV
ldEpV
)()(
)(
Edl
dV
Edl
dV
ldEdV
max
cos
Note the crucial role that independence
of path plays: If E was dependent on
path, then the definition of V would be
nonsense because the path would alter
the value of V(p)
negative by convention
VE
8/17/2012 47
• The word potential is a hideous misnomer
– Potential is connected to potential energy
– Electrostatic potential is NOT the same as potential energy
• Advantages of the potential formulation
– If you know V, then it is easy to get E at any point in space by simply taking the gradient
– This is because the three components of E are not as independent as the they look. Instead,
Electric field components are explicitly interrelated by the requirement that xE is zero.
– Thus, one can simply reduce the problem to a scalar one and ignore the fuss of vector
addition when so desired
• The reference point, 0
– Note that the reference point chosen for the potential is arbitrary and thus the value of the
potential can be altered simply by moving the reference point (or the position of the line
integral in space) to any different location.
– Because any change in reference is simply an additive constant to the field, the gradient
does not change, yielding the same electric field from both locations.
– Thus the potential contains no physical significance because we can adjust its value at any
given point without altering our perception of the field at all.
• The potential obeys the principle of superposition
• Units of the potential: Joules/Coulomb = Volt
Comments on the Electric Potential
8/17/2012 48
Summary Diagram of Electrostatics
VE
ldEV
0
E
Eo
v
E V
o
vV
2
d
rV v
o4
1
d
r
aE rv
o
24
1
Figure is recopied from Griffiths, Introduction to Electrodynamics,3rd ed., Benjamin Cummings, 1999.
Potential from a Point Charge in
a Spherical Conductor
8/17/2012 49
1000
2
0
2
10
1
111
444
4
ˆ4
0
1
1
Rab
Q
R
Q
b
Q
drr
QVldEVV
rR
QE
aR
R
a
R
a
b
L
b
b
QVVE
bRa
R
Qdr
r
QldEV
rR
QE
bR
b
R
L
0
2
30
2
0
0
2
3
3
4,0
44
ˆ4
3
Note: we are starting
backwards from and
move in toward the origin
Problem and Figure from N. Ida, Engineering Electromagnetics, 2ed, Springer, 2003
8/17/2012 50
Using Gauss’ Law With Dielectrics
A
Qdld
DV
QDAdaD
QSdD
aED
aEEE
enc
S
zstot
z
o
st
21
z
o
s aE
21
z
o
s aE
22
Concentric Conducting Spheres
with radius a, b (b>a) with a
dielectric fill
Two flat conductive plates of
area, A, filled with dielectric
d
z
a
b
o
o
ba
Q
drr
Q
ldD
V
R
aQE
o
b
a o
o
R
11
4
4
4
2
2
If capacitance, C=Q/V, then what is the value for each example?
- + +
-
8/17/2012 51
Using Gauss’ Law With Dielectrics: Electric Field and
Voltage Everywhere for a Charged Dielectric Cylindrical
Shell over a Grounded Cylinder
0,0,0
VDE
a
• Electric field and voltage everywhere for a charged dielectric cylindrical
shell over a grounded inner conductor
aaaa
da
ldD
V
aD
aE
LaLE
LadvSdE
ba
r
a
v
v
S
ln24
)ln(22
2
ˆ2
,ˆ2
2
1
222022
0
22
0
22
0
22
0
220
22011
11
Problem and Figure from N. Ida, Engineering Electromagnetics, 2ed, Springer, 2003
Let r in the diagram equal in our coordinate system
8/17/2012 52
Using Gauss’ Law With Dielectrics: Electric Field and
Voltage Everywhere for a Charged Dielectric Cylindrical
Shell over a Grounded Cylinder
• Electric field and voltage everywhere for a charged dielectric cylindrical shell over a grounded inner
conductor
b
ab
a
baa
dab
a
baa
ldD
VV
abD
abE
b
r
b
b
ln2
ln24
2ln2
4
ˆ2
,ˆ2
0
22
02220
0
22
02220
22
0
0
22
0
Problem and Figure from N. Ida, Engineering Electromagnetics, 2ed, Springer, 2003
Let r =
8/17/2012 53
Work Done by the
Electrostatic Field • It is often useful to characterize any system that can impose forces on an object by
the work it does to that object
• Suppose a charge q1 is located near another charge, q. The force acting on q causes work to be done by displacing q a distance dl.
• The negative sign indicates that the work done by an external agent, q1. Thus the total work done, or potential energy required) to move q a distance dl from a to b
is:
• Notice the useful relation presented: The work done to move a charge from one location to the next is the charge times the difference in potential required to move from location a to location b in the vector field. In this case, the scalar is a much simpler means of calculating work done without having to sum individual vector components
ldEqldFdW
qaVbVldEqW
b
a
)()(
8/17/2012 54
Work and Energy
in Electrostatic Fields • To determine the energy present in an assembly of charges, we must first determine the
amount of work necessary to assemble them.
• Suppose we position 3 charges, q1, q2, and q3 in an initially empty space. Initially there is no
work done to transfer charge q1 from infinity to our work space, because the space is initially
charge free with no electric field in the region.
• However, there is a field present from q1 when we move q2 into position. The work done by
transferring q2 into our workspace is the product of q2 times the potential difference between q1
and q2.
• The same is true as we position charge,q3, with respect to charges q1 and q2
• Note that if the charges were positioned in reverse order then:
• Thus by adding all of the work possibly performed we obtain
)(0 32313211
321
VVqVqW
WWWW
)(0 13123232
321
VVqVqW
WWWW
k
n
k
kVqW
VqVqVq
VVqVqVVqVqW
1
332211
3231321113123232
2
1
)(0)(02
• Note that there is no reason preventing us from evaluating distributed charges in
the same manner. Therefore one can write that the energy, W, present in an
electrostatic field is:
• And since we can show by Gauss’ Law that
8/17/2012 55
Energy and Energy Density in
Electrostatic Fields
L
LVdlW 2
1
L
LVdlW 2
1
S
SVdSW 2
1V
VVdvW 2
1
Dv
v
v
dvVDDVW
VdvDW
)()(2
1
)(2
1
vs
dvVDSdDVW
2
1
2
13
1
r
2r
first integral 0 as S becomes large
o
o DEwdW
22
22
Energy Density
v
o
v
dvEdvEDW 2
2
1
2
1
v
dvVD
2
1
Energy
8/17/2012 56
Electric Field Dipoles
• An electric dipole created when two point charges of equal magnitude but opposite sign
are separated by a small distance
21
12
21
1
2
2
4
11
4
1
4
4
1)(
rr
rrq
rr
q
r
qV
ldar
qldEpV
ooo
p
o
r
o
p
o
r
o
ar
qE
24
1
2
cos
4 r
dqV
o
32'
)'(
44 rr
rrp
r
apV
oo
r
• Using the geometric symmetry present in the
system, one can find the potential at a point P as
• Furthermore, we can define the dipole moment as
the charge times the displacement vector d = dcos,
such that the potential can now be written as
8/17/2012 57
Electric Field Dipoles (2)
• The electric field due to the dipole centered at the origin may now be determined
exactly by calculating the gradient of the potential field
)sincos2(4
)sincos2(4
1
cos
4'
)'(
44
3
3
2322
aar
p
aar
qd
aV
ra
r
VVE
r
dq
rr
rr
r
p
r
apV
r
o
r
o
r
ooo
r
