Electrostatic Fields:

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CHAPTER 2
ELECTROMAGNETIC FIELDS THEORY

• Electrostatic Fields
• Electric Fields in Material Space

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In this chapter you will learn

• Electrostatic Fields
• -Charge, charge density
• -Coulomb's law
• -Electric field intensity
• - Electric flux and electric flux density
• -Gauss's law
• -Divergence and Divergence Theorem
• -Energy exchange, Potential difference, gradient
• -Ohm's law
• -Conductor, resistance, dielectric and
  capacitance
• - Uniqueness theorem, solution of Laplace and
  Poisson equation

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Classification of Materials

• What is conductivity?
• What is conductor?
• What is metal?
• What is insulator?
• What is semiconductor?
• What is superconductor?

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Classification of Materials

• Materials can be classified in terms of
  conductivity, s in mhos per meter ( /m), or
  by siemens per meter (S/m)
• Classification of materials are
• Conductors Nonconductors
• Or technically as metals and insulators (or
  dielectric)
• Metal has high conductivity (s gtgt1)
• Insulators has low conductivity (s ltlt1)

O
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Current

• Current through a given area is defined as the
  electric charge passing through the area per unit
  time
• Thus 1 A means charge is being transferred at a
  rate of one coulomb per second

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Current Density

• Current density J is defined by current ?I
  flowing through surface ?S
• or
• assuming that the current density is
  perpendicular to the surface.

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Current Density for J not normal to the surface

• If J is not normal to the surface,
• Thus
• Compare with the general definition of flux
• I through S is the flux of current density J

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Current Density

• Depending on how I is produced, there are three
  types of current densities
• Convection current densities
• Conduction current densities
• Displacement current densities ? will be covered
  in chapter 3 (magnetic)

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Conductor

• inside of a conductor, under static condition
• E0
• ?v0
• Vab0

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Non-isolated Conductor
A conductor whose ends are maintained at a
potential difference V E ? 0 inside the conductor

• Conductor is not isolated it is wired to a
  source of electromotive force. So
• Free charge moves
• Prevent the establishment of electrostatic
  equilibrium
• Electric field must exist inside the conductor to
  sustain the flow of current
• The moving electrons encounter damping forces
  called resistance

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Example

• If
• calculate the current passing through
• A hemispherical shell of radius 20cm
• A spherical shell of radius 10cm

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Dielectrics (Insulators)

• Q Why insulators are bad conductors??

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Dielectrics (Insulators) In Static Electric Field

• The electrons in the atoms of insulators or
  dielectrics, are confined to their orbits
• They cannot be liberated in normal circumstances,
  even by the application of an external field.
• Insulator or dielectrics are materials with a
  completely filled upper band
• So conduction could not normally occur because of
  the existence of a large energy gap to the next
  higher band.

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Dielectrics (Insulators) In Static Electric Field

• ve charge is displaced from its equilbrium
  position in the direction of E by the force F
  QE
• -ve charge is displaced in the opposite direction
  of E by the force F- QE
• ?a dipole exist? the dielectric is polarized

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Dielectrics (Insulators) In Static Electric Field

• The distorted charge distribution is equal to the
  original distribution plus a dipole whose moment
  is
• p Qd
• Where d is the distance vector from Q to Q of
  the dipole
• If there are N dipoles in a volume ?v of the
  dielectric, the total dipole moment due to the
  electric field is

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Polarization in Dielectrics

• As a measure of intensity of the polarization,
  we define a polarization vector P as
• where N is the number of molecules per unit
  volume and the numerator represents the vector
  sum of the induced dipole moments contained in a
  very small volume ?v.
• The vector P, a smoothed point function, is the
  volume density of electric dipole moment. P is a
  measure of the intensity of polarization.

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Polarization in Dielectrics
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Polarization in Dielectrics

• Q concentrate on the upper surface S, while Q
  concentrate on the lower surface S

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Polarization in Dielectrics

• Since and give the
  same effects, the nett bounded charge to surface
  dS is

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Polarization in Dielectrics

• If the dielectric material is linear and
  isotropic, the polarization ,P vary directly as
  the applied electric field, E

Electric susceptibility of the material ? A
measure of how susceptible (or sensitive) a given
dielectric is to electric field
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Dielectric Constant Strength
and
e is the permittivity of dielectric
and
er is the dielectric constant or relative
permittivity the ratio of the permittivity of
the dielectric to that of free space
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Linear, Isotropic Homogeneous Dielectrics

• A dielectric material (D eE) is
• linear if e does not change with the applied E
  field
• Homogeneous if e does not change from point to
  point
• Isotropic if e does not change with direction

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Boundary Conditions

• If electric field exists in a region consisting
  of two different media, the conditions that the
  field must satisfy at the interface separating
  the media are called boundary conditions
• Dielectric and dielectric
• Conductor and dielectric
• Conductor and freespace

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Boundary Conditions

• E can be decomposed into two orthogonal
  components
• E Et En
• Where Et and En are the tangential and normal
  components of E respectively.

Et
En
E
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Boundary Conditions of Dielectric/Dielectric
First boundary condition can be determined by
performing a line integral of E around a closed
rectangular path,
Figure 1
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Boundary Conditions of Dielectric/Dielectric

• From figure 1,
• E1E1t E1n
• E2E2t E2n

In figure 1, consider closed path abcda. Assuming
the path is very small with respect to the
variation of E. thus
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Boundary Conditions of Dielectric/Dielectric
What Happen if abcd0 (h0)?
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Boundary Conditions of Dielectric/Dielectric
Second boundary condition can be determined by
applying Gausss Law over a small pillbox shaped
Gaussian surface
Figure 2
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Boundary Conditions of Dielectric/Dielectric
From figure 2 D1D1t D1n D2D2t D2n
In Figure 2, Consider Gaussian surface, ?S.
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Boundary Conditions of Dielectric/Dielectric
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Boundary Conditions of Dielectric/Dielectric
If no free charges exist at the interface Then
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Boundary Conditions of Conductor/Dielectric
Dielectric
Conductor
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Boundary Conditions of Conductor/Dielectric
Dielectric
Conductor
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Boundary Conditions of Conductor/Dielectric
For conductor, so
For perfect conductor, no electric field may
exist within a conductor, that is ?v 0, E 0
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Boundary Conditions of Conductor/Dielectric

• Since E - V 0, there can be no potential
  difference between any two points in the
  conductor
• E can be external to the conductor and normal to
  its surface, that is

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Boundary Conditions of Conductor/Free Space
Dielectric
Conductor
Dielectric
Conductor
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EXAMPLE 5.10

• REGION y 0 is perfect conductor
• y 0 dielectric with e1r2 and ?s 2nC/m2
• Find E and D at points
• A (3, -2, 2)
• B(-4, 1, 5)

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Solution

• 1st step is to sketch the problem

z
conductor
dielectric
A
y
0
B
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Solution

• (a) A (3, -2, 2) is in the conductor region,
• Therefore E D 0
• (b) B(-4, 1, 5) is in the dielectric medium, so

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Exercise 5.9

• REGION 1, x 0 is homogeneous dielectric with
  e1r2.5
• REGION 2 x 0 is free space
• If D1 12ax-10ay4az nC/m2 Find D2 and ?2

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Solution
Free-space
dielectric
E2
E2t
E1n
?2
x
0
E2n
E1t
E1
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Solution (cont)

• Given D1 12ax-10ay4az
• Since boundary is at x 0 surface, an ax.
• Therefore D1n 12ax
• And D1t -10ay4az
• For dielectric-free space condition,
• D2n D1n 12ax

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Solution (cont)
Thus
and
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Solution (cont)
E2
E2t
?2
E2n
To find ?, we use trigonometry
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Exercise 1
Assume the boundary to be charge free, find E1
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Solution
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Exercise 2

• Repeat ex. 1 for a boundary with charge density

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Solution Ex.2

• Repeat the same step as in ex. 1, in finding E1t

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Solution Ex.2

• To find E1n, we must include the surface charge
  density given

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Problem

• A xerographic copying machine is an important
  application of electrostatics. The surface of the
  photoconductor is initially charged uniformly.
• When light from the document to be copied is
  focused on the photoconductor, the charges on the
  surface combine with those on the upper surface
  to neutralize each other.

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Problem

• The image is developed by pouring a charged black
  powder over the surface of the photoconductor.
  The electric field attracts the charged powder,
  which is later transferred to paper and melted to
  form a permanent image.
• How do we determine the electric field below and
  above the surface of the photoconductor??

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Problem
light
air


E2
- - - - - - - - - - - -
recombination
photoconductor
The charge distribution is unknown, so how do we
solve this problem???
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Poissons and Laplace Equations

• The two fundamentals are
• ??D ? ?? E ?v
• And
• E -? V
• Thus
• ??D ?? (- ??V)
• ?(?2V) - ? v

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Poissons and Laplace Equations

• So for an inhomogeneous medium,
• For a homogeneous medium,(? is a constant)
• Poisson's Equation

?(?2V) - ? v
?2V - ? v /?
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Poissons and Laplace Equations

• ?2 is the Laplacian Operator
• ?2V is the divergence of the gradient of V
• If there are no free charges in the region, then
• ?v 0
• ?2V 0 Laplace's Equation

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How to interpret the 2 equations of Laplaces and
Poisson?

• If there are no charges in the region of
  interest (in the case of a parallel plate
  capacitor, where no charges cannot be found in
  the perfect dielectric, then you would use
  Laplace Equation.
• If there are charges in the region of interest,
  in the case of the depletion layer of a diode
  where there exist positive/negative ions
  (charges), then you would apply Poissons
  Equation.

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Uniqueness Theorem

• Any solution of Laplace's Equation which satisfy
  a given set of boundary conditions must be the
  only solution regardless of the method used.
• General procedure For Solving a Boundary value
  Problem
• Solve Laplace Equation by using direct
  integration
• Apply Boundary Conditions to determine a unique
  solution for V
• From V, obtain E -?V

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Uniqueness Theorem

• (iv) Obtain D from ?E
• (vi) Obtain surface charge density ?s D
• (vii) Obtain free charge
• (viii) Obtain capacitance C Q/V
• The solution is always V, the voltage everywhere
  in the region of interest i.e the values of V in
  all points in the dielectric region.

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Example
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Solution
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Capacitance

• Capacitor consists of two conductors (may or may
  not be parallel plates) separated by free space
  or a dielectric medium. The conductors may be of
  arbitrary shape.

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Capacitance

• The capacitance C of the capacitor is the ratio
  of the magnitude of the charge on one of the
  plates to the potential difference between them

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Capacitance

• C can be obtained from two methods
• Q method Assuming Q and determining V in terms
  of Q (involving Gauss Law)
• V method Assuming V and determining Q in terms
  of V (involving solving Laplaces equation)

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Q Method

• Assume a charge Q on plate a and a charge Q on
  plate b.
• Solve for E using the appropriate method
  (Coulombs Law, Gausss Law, boundary conditions)
• Solve for the potential difference Vab between
  the plates (The assumed Q will divide out)

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Steps to find Capacitance

• Choose a suitable coordinate system
• Let the two conducting plates carry charges Q
  and Q
• Determine E using Coulombs or Gauss Law and
  find V from.
• The negative sign may be ignored in this case
  because we are interested in the absolute value
  of V
• Obtain C from

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V Method

• Assume Vab between the plates.
• Find E , then D using Laplaces equation.
• Find ?S, and then Q at each plate using conductor
  dielectric boundary condition (DN ?S )
• C Q/Vab (the assumed Vab will divide out)

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Physical Interpretation of Capacitance

• Capacitance is a property of a device or system
  with the ability to store charge or energy. It is
  a constant for the device and do not depend on
  the charge or voltage applied to the device.
• The capacitance depends only on the physical
  structure such as the area of the conductor (a),
  separation distance (d) of the conductors and
  also the type of material (permittivity ?) of the
  device.

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Common type of capacitors

• Parallel Plate capacitor
• Coaxial capacitor
• Spherical Capacitor

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Parallel Plate capacitor
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Parallel Plate capacitor
Place charge Q on the inner surface of the top
plate, and Q charge on the upper surface of the
bottom plate, where the charge density,
from
Use conductor dielectric boundary, to obtain
from
_______ (1)
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Parallel Plate capacitor
From eq. (1), we could find the electric field
intensity, E
The potential difference across the plates is
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Parallel Plate capacitor
Finally, to get the capacitance
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A Coaxial Capacitor
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A Coaxial Cable
Thus
To find V, we use
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A Coaxial Cable
Eventually, C is found by combining the previous
findings
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Spherical Capacitor
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Spherical Capacitor
Repeat as before. The first step is to find E
Then find V
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Spherical Capacitor
Then find C
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Example 1

• I given d 5 mm, S 30 cm2
• Determine the capacitance in the figure above.

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Solution

• Since D and E are normal to the dielectric
  interface, the capacitors can be treated as
  consisting of 2 capacitors C1 and C2 in series.
  So total capacitance is

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Example 2
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Solution
D and E are parallel to the dielectric interface,
the capacitors can be treated as consisting of 2
capacitors C1 and C2 in parallel. So total
capacitance is
C1
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Example 3

• A coaxial capacitor consists of two concentric,
  conducting cylindrical surfaces, one of radius a
  and another of radius b, as shown in the figure.
  The insulating layer separating the two
  conducting surfaces is divided equally into two
  semi-cylindrical sections, one filled with
  dielectric e1 eand the other filled with
  dielectric e2

• Develop an expression for C in terms of the
  length l and the given quantities
• Evaluate the value of C for a 2 mm, b 6 mm,
  e1 2, e2 4 and l 4 cm

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Solution
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Solution
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Solution