Let us Review the topics we have learnt earlier

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Let us Review the topics we have learnt
earlier In the last class
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Review

• Conductors
• Dielectrics
• Boundary Conditions

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Review

• Conductors
• Dielectrics
• Boundary Conditions

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Conduction Current
v drift velocity , suffixes e for electron
and h for holes
v µ E, µ is the mobility in m2/(v-sec)
s ? µ
s
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Example The current flowing through a 300 m long
conductivity wire of uniformCross section has a
denstiy of 3x105 A / m2. Find the voltage drop
across the length of the wire if the wire
material has a conductivity of 2x107 S/m..
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Solution
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Example A 25 cm long copper wire of circular
cross section has a radius 2 cm. For copper, s
5.8 x 107 S/m, determine The resistance of the
wire and The power dissipated in the wire if the
voltage across the length is 1.5mV
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Aliter to find P
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Dielectrics

• Polar dielectrics
• Non polar dielectrics

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Relative Dielectric constants of some dielectric
materials
For most conductors, e e0 , er 1 , er 1
for free space as well
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With reference to the figure, find E1 if E2 is
Given, e1 3 e2 8 e0 and the boundary is of
charge free We have,
Let x-y plane represent the boundary between the
two media.. The x and y components of E2 are
parallel to the boundary. Therefore they are the
same across the two sides of the boundary.
Therefore E1x E2x 4, E1y E2y - 3.
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From the conditions on the normal components of D
across the boundary, we get, D1nor D2nor ?S
0. i.e., e1 E1nor e2 E2nor or e1 E1z e2 E2z
and therefore E1z ( e2/ e1) E2z (8 e0/ 2 e0 )
3 12 . Therefore .
In the previous example if the charge density at
the boundary is non-zero, i.e., ?S 3.54X 10-11
C/m2 , find E1. e1 E1nor - e2 E2nor ?S
E1nor E1z (e2 E2nor ?S )/e1
14.Therefore...
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Dielectric Conductor Boundary
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Conductor Conductor Boundary
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Electrostatic condition
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• Obtain an expression for capacitance in terms of
  the applied voltage and charge stored.
• Obtain the expression for the capacitance of a
  parallel plate capacitor
• Obtain an expression for the capacitance of a
  coaxial cable
• Obtain an expression for the capacitance of a two
  wire transmission line.
• Obtain an expression for the capacitance of a
  spherical capacitor
• Work out all the Drill problems and examples from
  Hayt book and Schaum Series book.

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Dielectrics
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Dipoles
Dielectrics have very no DC conductivity. Their
charges are not mobile they are strongly bound to
the atoms. Such charges are called bound Charge(
as opposed to free charges in conductors)
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• External Electric fields influence the
• dielectric atoms and molecules despite
• the fact that their charges are more or
• less fixed.
• Macroscopic Displacement of the center
• of the electron cloud makes the atom look
• like a dipole

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• The example shown is called electronic
• polarisation.
• It occurs in dielectrics whose atoms
• and molecules are originally neutral.
• Such matrials are known as non
• polar dielctrics.
• The process of polarisation is balanced
• by the Coulombs force of attraction.

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Ionic polarisation occurs in molecules
consisting of positively and negatively charged
ions which are originally mixed, randomly
oriented, and may have zero net charge
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Oriental polarisation occurs in materials which
have permanent Microscopic separation of charges
(dipoles), like electrets, polar liquids,Water
etc. Such materials are called polar materials.
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Each microscopic polarised region
is characterised by the dipole moment p
To quantify the polarisation effect on a
macroscopic level, We define polarisation
vector P(defined as dipole moment Per unit
volume,
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Electric Dipoles

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The density of uncompensated surface charges (see
fig) gives the magnitude of the polarisation
vector P inside the dielectric
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Here, N is the number of uncompensated
microscopic charges, Qb on each surface. Clearly
we observe that the Polarisatrion is zero
outside the dielctric i.e., in the air.
We have assumed that throught the volume, the
microscopic dipole Moment p were assumed to be
the same (isotropic). This means That The
positive and negative charges perfectly cancel
resulting in zero net Charge.
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However, in some cases, the polarisation is not
homogeneous throught the volume. Then Bound
volumetric charges appear whose density is given
by the divergence of the polarisation vector.
i.e., ?b - div P
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Thus there are two types of charges. 1. The free
charges in conductors and 2. The bound charges
in dielectrics The bound charges represent the
behaviour of the dielectric atoms/molecules in
vacuum. Applying the Gauss law for the electric
flux density in vacuum in the presence of both
types of charges gives
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Now we can define a vector which depends on the
free sources only and has nothing to do with the
properties of the medium
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Note the similatity between the D vector and
Free charge density with the relation between
The P vector and the bound charge density
Thus we get
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From the experimental results we find that the
Polarisation vector P is strongly related to the
Electric field E . For most dielectrics Commonly
used in Engineering Applications, E and P are
linearly related over a wide range of E, as
where ?e is the electric susceptibility of the
material.
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The dielctric constant er is not really a
constant. It may depend on Frequency or on the
field intensity. It also called the relative
permittivity
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E
S
?
d/2 cos ?
d/2 cos ?
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?Qb n q ?S d cos? P. dS
D
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er (1 ?e)
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Relative Dielectric constants of some dielectric
materials
For most conductors, e e0 , er 1 , er 1
for free space as well
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Format-1(a) Example
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