Electrostatic Fields:

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CHAPTER 2
ELECTROMAGNETIC FIELDS THEORY

• Electrostatic Fields
• Electric flux and electric flux density
• Gauss's law

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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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Electric Flux And Electric Flux Density

• Consider an amount of charge Q is applied to a
  metallic sphere of radius a.
• Enclosed this charged sphere using a pair of
  connecting hemispheres with bigger radius.

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Electric Flux (? )and Electric Flux Density
The outer shell is grounded. Remove the ground
then we could find that Q of charge has
accumulated on the outer sphere, meaning the Q
charge of the inner sphere has induced the Q
charge on the outer sphere.
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Electric Flux And Electric Flux Density
Figure 1
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Electric Flux And Electric Flux Density

• From figure 1, it can be seen that electric flux
  ? from Q to Q is induced/displaced
• ? originate on Q and terminates on Q, but if
  there is no negative charge, ? terminates at
  infinity
• 1 coulomb of electric charge gives rise to one
  coulomb of ?. Therefore
• Electric flux ? is a scalar field

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Electric flux density

• D is defined as total flux ? per unit area.
• Electric flux density D is a vector field.
• In practice, one cannot measure D, but use D to
  obtain E

D
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Electric flux density

• From figure 1, it can be seen that
• Equation (1) is true even if the inner sphere
  becomes smaller and smaller until becomes a point
  charge Q

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Electric flux density

• In free space, E is related to D, the electric
  flux density only by a constant ?0. Therefore D
  is scaled version of E.
• For a general volume charge distribution in free
  space, we know that
• and

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Electric flux density
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Gausss Law

• The flux crossing an infinitesimal surface, ?S is
  defined as

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Gausss Law

• The total flux passing through the closed surface
  is obtained by

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Gausss Law

• Where Q can be

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EXAMPLE 7
Find the amount of electric flux through the
surface at z 0 with and
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SOLUTION
The differential surface vector is
We could have chosen
but the positive differential surface vector is
pointing in the same direction as the flux, which
give us a positive answer.
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SOLUTION
Therefore,
Why?!
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Why do we need to use Gausss law?

• An alternative statement of Coulombs law
• Easier to find E or D for symmetrical (uniform)
  charge distribution point charge, infinite line
  of charge, infinite cylindrical surface charge
  and spherical distribution of charge

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Gaussians surface

• The first step in solving a Gauss law problem is
  to create an imaginary surface that encloses the
  charge called the Gaussians surface
• Gaussians surface is a closed two-dimensional
  surface through which a flux or electric field is
  to be calculated.
• A Gaussian surface has the property that the
  normal component of the field is constant over
  the enclosed surface .

Gaussian surface
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Gaussians surface

• Gaussian surfaces are usually carefully chosen to
  exploit symmetries of a situation to simplify the
  calculation of the surface integral.
• If the Gaussian surface is chosen such that for
  every point on the surface the component of the
  electric field along the normal vector is
  constant, then the calculation will not require
  calculus as the constant can be pulled out of the
  integration sign.

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Gaussians surface

• Spherical Surface
• A spherical Gaussian surface is used when
  finding the electric field or the flux produced
  by any of the following
• a point charge
• a uniformly distributed spherical shell of charge
  
• any other charge distribution with spherical
  symmetry

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Gaussians surface

• Cylindrical Surface
• A cylindrical Gaussian surface is commonly used
  to calculate the electric field produced by any
  of the following
• a long, straight wire with uniform charge
  distribution
• any long, straight cylinder or cylindrical shell
  with uniform charge distribution

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Gaussians surface

• Gaussian Pillbox
• This surface is most often used to determine the
  electric field due to an infinite sheet of charge
  with uniform charge density, or a slab of charge
  with some finite thickness.

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Application of Gausss Law

• To apply Gausss law, these conditions must first
  be satisfied
• 1. Ds is everywhere normal or tangential to the
  closed surface, so that becomes either
  or zero respectively.
• 2. On that portion of the closed surface for
  which is not zero, Ds constant.

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Point Charge

• Consider point charge q, located at the origin.
• To determine D at a general point P ? choose a
  spherical surface containing P.

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Point Charge

• Next step is to determine the area of the
  Gaussian surface.
• Since the surface chosen is a sphere with radius
  r, the area would be

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Point Charge

• D is everywhere normal to the Gaussian surface,
  DDrar. Thus from Gausss Law,

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Infinite line Charge
Because of symmetry, D is only in the a?
direction, so
and
Thus
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Infinite line Charge
Because of symmetry, D is only in the a?
direction, so
and
Thus
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Infinite Sheet of Charge
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Infinite Sheet of Charge

• As D is normal to the sheet, DDzaz

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Example Uniformly Charged Sphere

• Consider a sphere with radius a with uniform
  charge ?v C/m3. To determine D everywhere, we
  must
• Determine the appropriate Gaussian surface
• Construct Gaussian surfaces for case r a and r
  a

r
r
a
a
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Example Uniformly Charged Sphere

• For r a

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Example Uniformly Charged Sphere

• For r a

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Gausss Law and Divergence Theorem

• Gausss Law can be written as
• If we apply divergence theorem,
• Thus

1st Maxwells equation The volume charge density
is the same as the divergence of the electric
flux density
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Gauss or Coulomb ?

• To calculate E of a charge distribution, it is
  simplest to apply Gausss Law if a symmetrical
  Gaussian surface enclosing the charge can be
  found.
• If no Gaussian surface exist, then you would have
  to use Coulombs Law formula to calculate E.
• Besides finding E, Gausss Law can also be used
  to find Q for a given volume. By using Gausss
  Law, we can avoid calculating triple integration
• Gausss law in point form (Maxwells equation) is
  useful in solving problem where ?v is unknown.
  The next example demonstrates this

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Example 4.8 (from Sadikus book)

• D is given as C/m2
• Calculate the charge density at (1, p/4,3) and
  the total charge enclosed by the cylinder of
  radius 1m with
• -2 z 2m

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Solution

• we can find ?v using the Maxwells equation

At point (1,p/4,3),
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Solution for Q Method 1

• There are 2 methods to find the total charge
  enclosed, Q. The first method is by using volume
  charge distribution

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Solution for Q Method 2

• The second method is by using Gausss law
• Where are flux through top, bottom and side
  of the surface

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Solution for Q Method 2

• Since D is in az direction only, there is no flux
  in a? direction, ?s 0
• So only ?t and ?b exist

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Solution for Q Method 2

• And
• Therefore Q is

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Example

• Given prove that
• For the cube given below

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Example
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Example

• So how do we prove that
• We must solve the equation, part by part, and
  prove that the left-hand side of the equation
  will produce the same answer as the right-hand
  side

?
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Solution Left-hand side

• We begin with Left-hand side of the equation

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Solution Left-hand side

• By examining the figure, and from the equation of
  D,
• It can be clearly seen that D and dS only exist
  in ax (front) direction. x1 for this surface.
  Therefore

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Solution Right-hand side

• On the Right-hand side, we make use of the
  equations
• Where

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Solution Right-hand side

• Thus,
• Therefore Maxwell theorem is proven