ELECTROMAGNETICS THEORY

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ELECTROMAGNETICS THEORY (SEE 2523) ELECTROSTATIC
FIELD
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CHAPTER 2 COULOMBS LAW ELECTRIC FIELD
INTENSITY
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INTRODUCTION
An electrostatic field is produced by a static
charge distribution.
A magnetostatic field is produced by the moving
charges or a constant current flow (DC current)

AC current produced the electromagnetic field.
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2.1 ELECTRIC CHARGES

• Charges are measured in Coulombs (C).
• The smallest unit of electric charge is found on
  the negatively charged electron and positively
  charged proton.
• The electron mass, qe -1.602x10-19 (C).
• So, 1 C would represent about 6x1018 electrons.
• In real world, electric charges
• can be found

at a point
along a line
on a surface
in a volume or
combination the above distributions
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2.1.1 A POINT CHARGE
The concept of a point charge is used when the
dimensions of an electric charge distribution are
very small compared to the distance to
neighboring electric charges.
Fig. 2.1 A point charge
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2.1.2 A LINE CHARGE
A line charge denotes as the electric charge
distribution along a thin line.

• A line charge density
• Total charge along the line

?Q
?l
Fig 2.2 A line charge
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2.1.3 A SURFACE CHARGE
A surface charge is defined as charges
distribution on a thin sheet.
A surface charge density
?Q
?S
Total charge on the surface
Fig. 2.3 A surface charge
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2.1.4 A VOLUME CHARGE
A volume charge means electric charges in a
volume. This volume charge can be viewed as a
cloud of charged particles.

• A volume charge density
• Total charge in the volume

?Q
?V
Fig. 2.4 A volume charge
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2.2 COULOMBS LAW

• Coulombs law states that the force F between two
  point charges Q1 and Q2 is Directly
  proportional to the magnitude of the both
  products of the charges and inversely
  proportional to the square of the distance R
  between them.

? Expressed mathematically,
and
or
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Fig. 2.5 Coulombs law
where
is known as the permittivity of free space
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• A unit vector, then the
  force
• The force, on Q1 due to Q2

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• Things need to considered in Coulombs law
• gtgt The force expressed by Coulombs law is a
  mutual force, for each of the two charges
  experiences a force of the same magnitude,
  although of opposite direction.
• gtgt Coulombs law is linear, if the charge
  multiply by n, the force is also multiplied by
  the same factor, n.
• gtgt If there are more than two point charges, the
  principle of superposition is used. The force on
  a charge in the presence of several other charges
  is the sum of the forces on that charges due to
  each of the other charges acting alone.

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2.3 ELECTRIC FIELD INTENSITY DUE TO POINT CHARGE

• Consider one charge fixed in position, Q1 and
  test charge, Qt that exists everywhere around Q1
  , shown in Fig. 2.6.
• From the definition of Coulombs law, there will
  exist force on Qt cause by Q1.

Fig. 2.6
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The electric field intensity E is the force per
unit charge when placed in the electric field.
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• For n point charges, the electric field
  intensity at a point in space is equal to the
  vector sum of the electric field intensities due
  to each charge acting alone. (Following
  superposition theorem). Shown in Fig. 2.7.

Fig. 2.7
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? The total field at point P
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2.4 ELECTRIC FIELD INTENSITY DUE TO THE LINE
CHARGE

• We have
• Thus, the electric field for a line charge

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Ex4. Find the electric field intensity about the
finite line charge of uniform distribution
along the z axis as shown in Fig. 2.8.
Fig. 2.8
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Solution
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• Using

• Hence

• Hence

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• If the line is infinite

• We obtain

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2.5 ELECTRIC FIELD INTENSITY DUE TO
THE SURFACE CHARGE

• We have
• Thus, the electric field for a surface charge

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Ex6. Find the electric field intensity along z
axis on a disc with a radius a(m) which is
located on xy plane with uniform charge
distribution as shown in Fig.
2.9.
Fig. 2.9
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Solution

• Using cylindrical coordinate

The component is cancel because the charge
distribution is symmetry. Only component
exists.
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For an infinite sheet of charge
In general, it can be summarized to be
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• The electric field is normal to the sheet and
  independent of the distance between the sheet and
  the point.
• In the parallel plate capacitor, the electric
  field existing between the two plates having
  equal and opposite charge is given by
• The electric field is zero for both above and
  below plates.

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2.6 ELECTRIC FIELD INTENSITY DUE TO THE VOLUME
CHARGE

• We have
• Thus, the electric field intensity for a volume
  charge is

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Ex8. Find the electric field intensity for a
sphere with a radius a(m) and with uniform charge
density, shown in Fig. 2.10.

Fig. 2.10
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Solution

• The electric field at P due to elementary
  volume charge is
• where

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• Due to the symmetry of the charge distribution,
  the Ex and Ey becomes zero.
• Only Ez exists. Thus,
• Using Cosine law,

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• It is convenient to evaluate the integral in
  terms
• of R and r, so
• Differentiating (keeping z and r
  fixed) then

where
and
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? Substituting all the equations into electric
field equation yields
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• Because and
  then electric field at P(0,0,z) is given by
• Due to the symmetry of the charge distribution,
  the electric field at any point
  can be written as

Which is identical to the electric field at the
same point due to a point charge Q located at the
origin.