CONDUCTORS & DIELECTRICS
Presented by
R.Lenin Raja, B.E.,M.E.,MIEEE.,(PhD)
Assistant Professor/ECE,
Sri Vidya College of Engineering & Technology.
1FDTP on Electromagnetic Fields [EC6403]4 July 2015
Definition of Current• Current is the flow of electrical
charge, i.e. amount of charge persecond moving through a wire, i =dq/dt.
• It is a scalar, not a vector, but it has adirection—positive in the direction offlow of positive charge carriers.
• Any way that you can get charges tomove will create a current, but atypical way is to attach a battery to awire loop.
• Charges will flow from the + terminalto the – terminal (again, it is reallyelectrons that flow in the oppositedirection, but current is defined asthe direction of positive chargecarriers).
Units: ampere 1 A = 1 C/s
Total current in
11 A
Total current out
- 3 A = 8 A
8 A
4 July 2015
Current in a Circuit
• What is the current in the wire marked i in the figure below?
Total current in
11 A
Total current out
- 3 A = 8 A
8 A
4FDTP on Electromagnetic Fields [EC6403]4 July 2015
Current At Junctions
• What is the current in all of the wire sections that are not marked?
8 A
2 A
5 A 3 A
6 A
5FDTP on Electromagnetic Fields [EC6403]4 July 2015
Current Density
• When we care only about the total current i in a conductor, we do not have to worry about its shape.
• However, sometimes we want to look in more detail at the current flow inside the conductor. Similar to what we did with Gauss’ Law (electric flux through a surface), we can consider the flow of charge through a surface. To do this, we consider (charge per unit time) per unit area, i.e. current per unit area, or current density. The units are amps/square meter (A/m2).
• Current density is a vector (since it has a flow magnitude and direction). We use the symbol . The relationship between current and current density is
J
AdJi
Small current densityin this region
High current densityin this region
6FDTP on Electromagnetic Fields [EC6403]4 July 2015
Resistance• Resistance is defined to be . That is, we apply a voltage V, and ask how
much current i results . This is called Ohm’s Law.
• If we apply the voltage to a conducting wire, the current will be very large so R is small.
i
VR
+
_
1.5 Vbatteryi
i large, soR small
wire
If we apply the voltage to a less conducting material, such as glass, the current will be tiny so R is very large.
i
i small, soR large glass
filament
The unit of resistance is the ohm, W. (Greek letter omega.)
1 ohm = 1 W = 1 volt per ampere = 1 V/A
RVCircuitDiagram
Resistor
7FDTP on Electromagnetic Fields [EC6403]4 July 2015
Current Through a Resistor
3. What is the current through the resistor in the following circuit, if V = 20 V and R = 100 W?
A. 20 mA.
B. 5 mA.
C. 0.2 A.
D. 200 A.
E. 5 A.
RVCircuitDiagram
4 July 2015
Current Through a Resistor
4. If the current is doubled, what changes?
A. The voltage across the resistor doubles.
B. The resistance of the resistor doubles.
C. The voltage in the wire between the battery and the resistor doubles.
D. The voltage across the resistor drops by a factor of 2.
E. The resistance of the resistor drops by a factor of 2.
RVCircuitDiagram
4 July 2015
Resistivity and Conductivity• Rather than consider the overall resistance of an object, we can discuss the
property of a material to resist the flow of electric current.
• This is called the resistivity. The text uses (re-uses) the symbol r for resistivity. Note that this IS NOT related to the charge density, which we discussed earlier.
• The resistivity is related not to potential difference V and current i, but to electric field E and current density J.
J
Er
Units V/m over A/m2 = Vm/A = ohm-meter =W m
Definition of resistivity
Note that the ability for current to flow in a material depends not only on the material, but on the electrical connection to it.
High resistance
Low resistance
r
1 Definition of conductivity
Note use (re-use) of for conductivity. NOT surface charge density.
10FDTP on Electromagnetic Fields [EC6403]4 July 2015
More on Resistivity• Since resistivity has units of ohm-meter, you might think that you can just divide by
the length of a material to find its resistance in ohms.
?/ LR r
AiJLVE / and / since
LRAAi
LV
J
E/
/
/rresistivity is
A
LR r Resistance from resistivity
Dependence on temperature: you can imagine that a higher temperature of a material causes greater thermal agitation, and impedes the orderly flow of electricity. We consider a temperature coefficient a: )( 000 TT -- arrr
11FDTP on Electromagnetic Fields [EC6403]4 July 2015
Resistivity of a Resistor
5. Three resistors are made of the same material, with sizes in mm shown below. Rank them in order of total resistance, greatest first.
A. I, II, III.
B. I, III, II.
C. II, III, I.
D. II, I, III.
E. III, II, I.
4
4
52
6
3
I.
II.
III.
Each hassquarecross-section
4 July 2015
Electric Power
• Recall that power is energy per unit time, (watts). Recall also that for an arrangement of charge, dq, there is an associated potential energy dU = dqV.
• Thus,
• In a resistor that obeys Ohm’s Law, we can use the relation between R and i, or Rand V, to obtain two equivalent expressions:
• In this case, the power is dissipated as heat in the resistor.
dt
dUP
iVVdt
dqP Rate of electrical energy transfer
Units: 1 VA = (1 J/C)(1 C/s) = 1 J/s = 1 W
RiP 2
R
VP
2
Resistive dissipation
13FDTP on Electromagnetic Fields [EC6403]4 July 2015
Superconductivity
• In normal materials, there is always some resistance, even if low, to current flow. This seems to make sense—start current flowing in a loop (using a battery, say), and if you remove the battery the current will eventually slow and stop.
• Remarkably, at very low temperatures (~4 K) some conductors lose all resistance. Such materials are said to be superconductors. In such a material, once you start current flowing, it will continue to flow “forever,” like some sort of perpetual motion machine.
• Nowadays, “high-temperature” superconductors have been discovered that work at up to 150 K, which is high enough to be interesting for technological applications such as giant magnets that take no power, perhaps for levitating trains and so on.
14FDTP on Electromagnetic Fields [EC6403]4 July 2015
Electrostatics
(Free Space With Charges & Conductors)
Outline
Maxwell’s Equations (In Free Space)
Gauss’ Law & Faraday’s Law
Applications of Gauss’ Law
Electrostatic Boundary Conditions
Electrostatic Energy Storage
15FDTP on Electromagnetic Fields [EC6403]4 July 2015
E-Gauss:
Faraday:
H-Gauss:
Ampere:
Maxwell’s Equations (in Free Space with Electric Charges present)
DIFFERENTIAL FORM INTEGRAL FORM
Static arise when , and Maxwell’s Equations split into decoupled
electrostatic and magnetostatic eqns.
Electro-quasistatic and magneto-quasitatic systems arise when one (but not
both) time derivative becomes important.
Note that the Differential and Integral forms of Maxwell’s Equations are related through
Stoke’s Theorem and Gauss’ Theorem 16FDTP on Electromagnetic Fields [EC6403]4 July 2015
Charges and Currents
There can be a nonzero charge density in the absence of a current density .
There can be a nonzero current density in the absence of a charge density .
Charge
conservation
and KCL for
ideal nodes
17FDTP on Electromagnetic Fields [EC6403]4 July 2015
Flux of through closed surface S = net charge inside V
Gauss’ Law
18FDTP on Electromagnetic Fields [EC6403]4 July 2015
Point Charge Example
Apply Gauss’ Law in integral form
making use of symmetry to find
• Assume that the image charge is
uniformly distributed at .
Why is this important ?
• Symmetry
19FDTP on Electromagnetic Fields [EC6403]4 July 2015
Gauss’ Law Tells Us …
… the electric charge can reside only on the surface of the conductor.
[If charge was present inside a conductor, we can draw a Gaussian surface around
that charge and the electric field in vicinity of that charge would be non-zero !
A non-zero field implies current flow through the conductor, which will transport the
charge to the surface.]
… there is no charge at all on the inner surface of a hollow conductor.
… that, if a charge carrying body has a sharp point, then the electric field at that
point is much stronger than the electric field over the smoother part of the body.
Lets show this by considering
two spheres of different size,
connected by a long, thin wire …20FDTP on Electromagnetic Fields [EC6403]4 July 2015
ba
SPHERE B
SPHERE A
WIRE
Because the two spheres are far apart, we can
assume that charges are uniformly distributed
across the surfaces of the two spheres,
with charge qa on the surface of sphere A
and qb on the surface of sphere B
… and the E-field on the surface of the spheres is:
Note that Ea >> Eb if b >> a 21FDTP on Electromagnetic Fields [EC6403]4 July 2015
Lighting Rods are connected to the ground. When a cloud
carrying electric charges approaches, the rod attracts opposite
charges from the ground. The Electric field at the tip of the
rod is much stronger than anywhere else. When the E-field
exceeds the
air breakdown strength (of 33 kV/cm), charges start to travel
to ground.
Lightning Rod
When a conductive body
contains sharp points,
the electric field on
these points is much
stronger than that on the
smooth part of the
conducting body.
22FDTP on Electromagnetic Fields [EC6403]4 July 2015
Faraday’s Law
Path I Path II
A unique path-independent potential
may be defined if and only if
Path I Path II
and
Dynamic form:
Static form:
23FDTP on Electromagnetic Fields [EC6403]4 July 2015
Normal is discontinuous at a surface charge.
Tangential is continuous at a surface.
A static field terminates
perpendicularly on a conductor
Boundary Conditions
24FDTP on Electromagnetic Fields [EC6403]4 July 2015
Point Charges Near Perfect Conductors
+ + + + + + ++ + + + + + ++ + + + + + +
-
- - - -
---
--
-
-
--
----
-
-
-
--
Time t = 0
Time t >> 0
+ + + + + + ++ + + + + + ++ + + + + + +
--
- - -
---
--
-- -
--
---- -
-- -
25FDTP on Electromagnetic Fields [EC6403]4 July 2015
Point Charges Near Perfect Conductors
- --- -- -- --- -- - -- + + + ++ + + ++ + + ++ + + +
+ + + +
Negative charge on top
surface of conductorPositive charge on
top and bottom
surface of
conductor
+
26FDTP on Electromagnetic Fields [EC6403]4 July 2015
Uniqueness and Equivalent Image Charges
- --- -- -- --- -- - -- + + + ++ + + +
Equivalent Image Charge
27FDTP on Electromagnetic Fields [EC6403]4 July 2015
Electrostatic Boundary Conditions
Tangential field is continuous
There is a jump in the normal electric field as
one passes through a surface charge
28FDTP on Electromagnetic Fields [EC6403]4 July 2015
Energy Stored in
Electric Fields
1. Begin with a neutral
reference conductor, the
charge reservoir. Its
potential is zero, by
definition.
2. Move charges from the
reference conductor into
free space, thereby creating
an electric field and doing
work in the process. The
work is stored as potential
energy in the electric fields.
3. Account for all the work
done, and thereby derive
the energy stored in the
electric fields.
4. The argument directly
extends to systems with
multiple conductors
(and dielectrics).
• The work done by moving charge to a location
with potential is . More generally, the work
done to make an incremental charge change to a
charge density is
• Gauss’ Law
ZERO ! WHY ? ENERGY DENSITY [J/m3]
29FDTP on Electromagnetic Fields [EC6403]4 July 2015
Energy Stored in Electric Fields
The energy stored in an electric field is ½ ε0 E2.
The maximum achievable field strength is
typically limited by electric breakdown
Note: Dielectric constant is
30FDTP on Electromagnetic Fields [EC6403]4 July 2015
Energ
y [
eV]
+
_
+
_Remember this
unit of energy:
1 eV = 1.6 x 10-19 J
0 V
ENERGY OF AN ELECTRON IN AN
ALKANEMOLECULE IN
GASOLINE
ENERGY OF AN ELECTRON IN H2O or CO2
ENERGY GIVEN OFF AS HEAT IN THE PROCESS OF GASOLINE COMBUSTION
O = C = OO
H H
H H
H - C – C – H
H H
--
--
1.16 Å1.54 Å
1.10 Å
Hydrogen ground state
energy is -13.6 eV
If the hydrogen radius was
twice as long, what would be
the ground state energy ?
Modeling Atoms and Molecules
as Capacitors that Store Energy
These product molecules have shorter bond lengths than the initial reactant molecule, hence the charge in them
sits closer together. This can be modeled as a higher capacitance.
Since voltage V = Q/C = E d is reduced,stored energy W/Volume = ½ εoE
2
is reduced in these molecules
Burn this molecule by reacting it with oxygen
31FDTP on Electromagnetic Fields [EC6403]4 July 2015
KEY TAKEAWAYS
• Maxwell’s Equations (in Free Space with Electric Charges present):
E-Gauss:
Faraday:
H-Gauss:
Ampere:
DIFFERENTIAL FORM INTEGRAL FORM
AIR BREAKDOWN STRENGTH is 33 kV/cm
• Boundary conditions for E-field:
. Normal E-field – discontinuous
. Tangential E-Field - continuous
• Energy stored in the
electric field per unit volume is:
• Dielectric constant in free space is
NEW UNIT OF ENERGY:
• Energy released when fuel molecules are oxidized since the charges in the products
are positioned closer together than in reactants (hence in a lower energy state)
GOOD FACTS TO REMEMBER:
34FDTP on Electromagnetic Fields [EC6403]4 July 2015
Propagation and polarizationIn isotropic media(e.g. free space, amorphous glass, etc.)
More generally,
(reminder :Anisotropic in media, e.g. crystals, one could E not have parallel to D)
planar wavefront
electric field vector E
wave-vector k
36FDTP on Electromagnetic Fields [EC6403]4 July 2015
/4 plate
Linear polarization
birefringentl/4 plate
Circular polarization
42FDTP on Electromagnetic Fields [EC6403]4 July 2015
λ/2 plate
Linear (90o-rotated)
polarization
Linear polarization
birefringentλ/2 plate
43FDTP on Electromagnetic Fields [EC6403]4 July 2015
Think about that
mirror
birefringent
λ/4 plate
Linear polarization
44FDTP on Electromagnetic Fields [EC6403]4 July 2015
Relationship between E and B
Vectors k, E, B form a
right-handed triad.
Note: free space or isotropic media only
45FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
46FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
47FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
Continuity of tangential electric fieldat the interface:
Since the exponents must be equalfor all x, we obtain
48FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
Continuity of tangential electric fieldat the interface:
law of reflection
Snell’s law of refraction
so wave description is equivalent to Fermat’s principle!! ☺
49FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
Incident electric field:
Reflected electric field:
Transmitted electric field:
Need to calculate the reflected and transmitted amplitudes E0r, E0t
i.e. need two equations
50FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
Continuity of tangential electric fieldat the interface gives us one equation:
which after satisfying Snell’s lawbecomes
51FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
The second equation comes from continuity of tangential magnetic field
at the interface:
52FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
So continuity of tangential magnetic field Bx at the interface y=0 becomes:
53FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
I. Polarization normal to plane of incidence
54FDTP on Electromagnetic Fields [EC6403]4 July 2015
Reflection & transmission@ dielectric interface
II. Polarization parallel to plane of incidence
Following a similar procedure ...
55FDTP on Electromagnetic Fields [EC6403]4 July 2015
1-D Finite-Element Methods
with Poisson’s Equation
56FDTP on Electromagnetic Fields [EC6403]4 July 2015
Outcomes Based Learning
Objectives
By the end of this laboratory, you will:
– Understand how to approximate the heat-
conduction/diffusion and wave equations in
two and three dimensions
– You will understand the differences between
insulated and Dirichlet boundary conditions
57FDTP on Electromagnetic Fields [EC6403]4 July 2015
The Target Equation
Recall the first of Maxwell’s equations (Gauss’s equation):
If we are attempting to solve for the underlying potential function, under the assumption that it is a conservative field, we have
This is in the form of Poisson’s equation
0
r
E
2
0
ur
58FDTP on Electromagnetic Fields [EC6403]4 July 2015
The Target Equation
In one dimension, this simplifies to:
Define:
and thus we are solving for
2
2
0
xdu x
dx
r
2
2
0
def xdV x u x
dx
r
-
0V x
59FDTP on Electromagnetic Fields [EC6403]4 July 2015
The Integral
If , it follows that
for any test function f(x) and therefore
Substituting the alterative definition of V(x)into this equation, we get
0x V xf
0V x
0
b
a
x V x dxf
2
2
0
0
b
a
xdx u x dx
dx
rf
-
60FDTP on Electromagnetic Fields [EC6403]4 July 2015
The Integral
Consider the first test function f1(x) :
3
1
2 2
2 210
0
xb
a x
xd dx u x dx u x dx
dx dx
rf
-
61FDTP on Electromagnetic Fields [EC6403]4 July 2015
Integration by Parts
Again, take
but before we apply integration by parts, expand the integral:
Everything in the second integral is known: bring it to the right:
2
2
0
0
b
a
xdx u x dx
dx
rf
-
2
2
0
0
b b
a a
xdx u x dx x dx
dx
rf f
-
2
2
0
b b
a a
xdx u x dx x dx
dx
rf f
62FDTP on Electromagnetic Fields [EC6403]4 July 2015
Integration by Parts
The left-hand integral is no different from before,
and performing integration by parts, we have
0
0
b b b
a aa
b b
a ax b x a
xd d dx u x x u x dx x dx
dx dx dx
xd d d db u x a u x x u x dx x dx
dx dx dx dx
rf f f
rf f f f
-
- -
2
2
0
b b
a a
xdx u x dx x dx
dx
rf f
63FDTP on Electromagnetic Fields [EC6403]4 July 2015
Integration by Parts
First, substituting in the first test function:
which yields
3 3
1 13 1
2 3 2 1 2 2
0
x x
x xx x x x
xd d d dx u x x u x x u x dx x dx
dx dx dx dx
rf f f f
- -
2
2 22
0
b b
a a
xdx u x dx x dx
d x
rf f
1 1 1
1 3x x x x - - -
64FDTP on Electromagnetic Fields [EC6403]4 July 2015
The System of Linear Equations
Again, recall that we approximated the solution by unknown piecewise linear functions:
where we define on
2 11 2 1 2
1 2 2 1
3 22 3 2 3
2 3 3 2
11 1
1 1
n nn n n n
n n n n
x x x xu u x x x
x x x x
x x x xu u x x x
x x x xu x
x x x xu u x x x
x x x x
-- -
- -
- - - -
- -
- -
- - - -
11
1 1
defk k
k k k
k k k k
x x x xu x u u
x x x x
- -
- -
1k kx x x
65FDTP on Electromagnetic Fields [EC6403]4 July 2015
Unequally Spaced Points
Recall that we approximated the left-hand integral by substituting the piecewise linear functions to get:
1 1
1 1
1
1
1
2
2
0
2
2
0
1 1
1 1 0
1 1 1
1 1 1 1 0
2 2
1 1 1 12 2 2
k k
k k
k
k
k
b b
k k
a a
x x
x x
x
k k k k
k k k k x
k k k
k k k k k k k k x
xdx u x dx x dx
d x
xdu x dx dx
d x
xu u u udx
x x x x
xu u u dx
x x x x x x x x
rf f
r
r
r
- -
-
-
-
-
- -
- -
- --
- -
-
- - - -
1kx
66FDTP on Electromagnetic Fields [EC6403]4 July 2015
67
LAPLACE’S EQUATION, POISSON’S EQUATION AND UNIQUENESS THEOREM
1 LAPLACE’S AND POISSON’S EQUATIONS
2 UNIQUENESS THEOREM
3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE
4 SOLUTION FOR POISSON’S EQUATION
FDTP on Electromagnetic Fields [EC6403]4 July 2015
LAPLACE’S AND POISSON’S EQUATIONS AND UNIQUENESS THEOREM
- In realistic electrostatic problems, one seldom knows the charge distribution –thus all the solution methods introduced up to this point have a limited use.
- These solution methods will not require the knowledge of the distribution of charge.
68FDTP on Electromagnetic Fields [EC6403]4 July 2015
1 LAPLACE’S AND POISSON’S EQUATIONS
To derive Laplace’s and Poisson’s equations , we start with Gauss’s law in point form :
vED r
VE -Use gradient concept :
r
r
v
v
V
V
-
-
2Operator :
Hence :
(1)
(2)
(3)
(4)
(5) => Poisson’s equation
is called Poisson’s equation applies to a homogeneous media.
22 / mVV v
r-
69FDTP on Electromagnetic Fields [EC6403]4 July 2015
0vrWhen the free charge density
=> Laplace’s equation(6)22 / 0 mVV
2
2
2
2
2
22
z
V
y
V
x
VV
In rectangular coordinate :
70FDTP on Electromagnetic Fields [EC6403]4 July 2015
2 UNIQUENESS THEOREM
Uniqueness theorem states that for a V solution of a particular electrostatic problem to be unique, it must satisfy two criterion :
(i) Laplace’s equation
(ii) Potential on the boundaries
Example : In a problem containing two infinite and parallel conductors, oneconductor in z = 0 plane at V = 0 Volt and the other in the z = d plane at V = V0 Volt,we will see later that the V field solution between the conductors is V = V0z / dVolt.
This solution will satisfy Laplace’s equation and the known boundary potentials at z= 0 and z = d.
Now, the V field solution V = V0(z + 1) / d will satisfy Laplace’s equation but will notgive the known boundary potentials and thus is not a solution of our particularelectrostatic problem.
Thus, V = V0z / d Volt is the only solution (UNIQUE SOLUTION) of our particularproblem. 71FDTP on Electromagnetic Fields [EC6403]4 July 2015
3 SOLUTION OF LAPLACE’S EQUATION IN ONE VARIABLE
Ex.1: Two infinite and parallel conducting planes are separated d meter, with one of the conductor in the z = 0 plane at V = 0 Volt and the other in the z = d plane at V = V0 Volt. Assume and between the conductors.0
2 0v
r
Find : (a) V in the range 0 < z < d ; (b) between the conductors ;
(c) between the conductors ; (d) Dn on the conductors ; (e) on the conductors ; (f) capacitance per square meter.
E
D sr
Solution :
0v
rSince and the problem is in rectangular form, thus
02
2
2
2
2
22
z
V
y
V
x
VV (1)
(a)
72FDTP on Electromagnetic Fields [EC6403]4 July 2015
0
0
2
2
2
2
22
dz
V
dz
d
dz
Vd
z
VV
We note that V will be a function of z only V = V(z) ; thus :
BAzV
Adz
dV
Integrating twice :
where A and B are constants and must be evaluated using given potential values at the boundaries :
00
BVz
dVA
VAdVdz
/0
0
(2)
(3)
(4)
(5)
(6)
(7)
73FDTP on Electromagnetic Fields [EC6403]4 July 2015
)(0 Vzd
VV
Substitute (6) and (7) into general equation (5) :
dz 0
)/(ˆˆ
ˆˆˆ
0 mVd
Vz
z
Vz
z
Vz
y
Vy
x
VxVE
-
-
--
(b)
)/(2ˆ
200 mCd
VzED
-
(c)
74FDTP on Electromagnetic Fields [EC6403]4 July 2015
)/(2
)ˆ(2
ˆˆ
2
ˆ2
ˆˆ
200
00
00
00
0
mCd
V
zd
VznD
d
V
zd
VznD
dzs
zs
r
r
--
-
-
(d) Surface charge :
0
/
V
ds
VQC
s
ab
r
)/(/2
/
2
0
00
0
2
mFdV
dV
VmC s
r
(e) Capacitance :
z = 0
z = d
V = 0 V
V = V0 V
75FDTP on Electromagnetic Fields [EC6403]4 July 2015
Ex.2: Two infinite length, concentric and conducting cylinders of radii a and b are located on the z axis. If the region between cylinders are charged free and , V = V0 (V) at a, V = 0 (V) at b and b > a. Find the capacitance per meter length.
03
Solution : Use Laplace’s equation in cylindrical coordinate :
and V = f(r) only :
011
2
2
2
2
2
2
z
VV
rr
Vr
rrV
f
76FDTP on Electromagnetic Fields [EC6403]4 July 2015
BrAV
r
A
r
V
Ar
Vr
r
Vr
r
r
Vr
rrV
ln
0
012
and V = f(r) only :
(1)
77FDTP on Electromagnetic Fields [EC6403]4 July 2015
BrAV ln
BbAV
BaAVV
br
ar
ln0
ln0
Boundary condition :
ba
bVB
ba
VA
/ln
ln;
/ln
00 -
Solving for A and B :
ab
rbVV
/ln
/ln0
Substitute A and B in (1) :
(1)
bra ;
78FDTP on Electromagnetic Fields [EC6403]4 July 2015
r
abr
VED
rabr
V
r
VrVE
ˆ/ln
ˆ/ln
ˆ
0
0
--
ab
rbVV
/ln
/ln0
abb
VrD
aba
VrD
brs
ars
/lnˆ
/lnˆ
0
0
r
r
--
Surface charge densities:
ab
Vb
ab
Va
brsbr
arsar
/ln
22
/ln
22
0
0
rr
rr
-
Line charge densities :
79FDTP on Electromagnetic Fields [EC6403]4 July 2015
oab V
d
V
QC
r
Capacitance per unit length:
)/(
/ln
2/
0
mFabV
mCr
80FDTP on Electromagnetic Fields [EC6403]4 July 2015
6/ and 0 ff 0f
VV 100 6/f
EV and
Ex.3: Two infinite conductors form a wedge located at
is as shown in the figure below. If this region is characterized by charged free. Find . Assume V = 0 V at
and at .
z
x f = 0
f = /6
V = 100V
81FDTP on Electromagnetic Fields [EC6403]4 July 2015
Solution : V = f ( f ) in cylindrical coordinate :
01
2
2
2
2 fd
Vd
rV
BAV
Ad
dV
d
Vd
f
f
f0
2
2
f
f
/600
)6/(100
0
6/
0
A
AV
BV
Boundary condition :
Hence :
f
ff
ˆ600
ˆ1
r
d
dV
rVE
-
--
f
600V
6/0 f
for region :
82FDTP on Electromagnetic Fields [EC6403]4 July 2015
BAV
A
d
dV
Ad
dV
d
dV
d
d
d
dV
d
d
rV
2/tanln
sin
sin
0sin
0sinsin
12
2
= /10
= /6
V = 50 V
x
y
z
6/ and 10/
E
Ex.4: Two infinite concentric conducting cone located at
10/ . The potential V = 0 V at
6/ and V = 50 V at . Find V and between the two conductors.
Solution : V = f ( ) in spherical coordinate :
2/tanlnsin
dUsing : 83FDTP on Electromagnetic Fields [EC6403]4 July 2015
BAV 2/tanln
BAV
BAV
12/tanln50
20/tanln0
6/
10/
Boundary condition :
Solving for A and B :
-
20/tan
12/tanln
20/tanln50 ;
20/tan
12/tanln
50
BA
1584.0
2/tanln1.95
20/tan
2/tanln
20/tan
12/tanln
50
V
ˆsin
1.95
ˆ1
r
d
dV
rVE
-
-
6/10/ Hence at region :
and
84FDTP on Electromagnetic Fields [EC6403]4 July 2015
4 SOLUTION FOR POISSON’S EQUATION
0vrWhen the free charge density
Ex.5: Two infinite and parallel conducting planes are separated d meter, with one of the conductor in the x = 0 plane at V = 0 Volt and the other in the x = d plane at V = V0 Volt. Assume and between the conductors.04 0
vr
Find : (a) V in the range 0 < x < d ; (b) between the conductors E
Solution :
BAxx
V
Axdx
dV
dx
Vd
V v
-
-
-
-
2
2
0
0
0
2
2
2
r
r
r
rV = f(x) :
85FDTP on Electromagnetic Fields [EC6403]4 July 2015
2
2
0
00
2
0
0
0
d
d
VA
Add
VV
BV
dx
x
r
r
-
Boundary condition :
BAxx
V -2
2
0
r
dx 0In region :
xd
Vxd
xV 00
2-
r
xxd
d
V
xdx
dVE
ˆ2
ˆ
00
--
-
r
;
86FDTP on Electromagnetic Fields [EC6403]4 July 2015
xrv 1 and 0 rEx.6: Repeat Ex.6.5 with
BxAV
x
A
dx
dV
Adx
dVx
Vxdx
d
Exdx
d
E
D v
-
-
-
-
)1ln(
1
1
01
01
0
0
rSolution :
87FDTP on Electromagnetic Fields [EC6403]4 July 2015
)1ln(
)1ln(
0
0
0
0
d
VA
dAVV
BV
dx
x
-
-
Boundary condition :
xdx
Vx
dx
dVE
d
xVV
ˆ)1ln()1(
ˆ
)1ln(
)1ln(
0
0
--
dx 0In region :BxAV - )1ln(
88FDTP on Electromagnetic Fields [EC6403]4 July 2015
?Contact : R.Lenin raja,
Contact: 088700-82081 ; 08807082081Email: [email protected] ; [email protected]
89FDTP on Electromagnetic Fields [EC6403]4 July 2015
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