CONDUCTORS, DIELECTRICS, AND CAPACITANCE.pdf
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Transcript of CONDUCTORS, DIELECTRICS, AND CAPACITANCE.pdf
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CONDUCTORS, DIELECTRICS,
AND CAPACITANCE
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CURRENT AND CURRENT
DENSITY
Current - is defined as electric charges in motion or defined as a rate of movement
of a charge passing in a given reference
point (or crossing in a given reference
plane) of one coulomb per second
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CURRENT AND CURRENT
DENSITY
Current - is defined as the motion of positive charges (even though conduction
in metals takes place through the motion
of electrons). The unit of current is ampere
(A) and current is symbolized by I,
dt
dQI ( convention current)
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CURRENT AND CURRENT
DENSITY
Consider the field theory, the events occurring at a point was the main concerned rather than within the large region, and in this case the concept of current density were important matter.
Current density - is defined as a vector represented by J and measured in amperes per square meter (A/m2).
The increment of current I crossing an incremental surface, S, is given as;
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CURRENT AND CURRENT
DENSITY
( normal to the current
density)
or,
( current density is not
perpendicular to the
surface)
SJI N
SJI
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CURRENT AND CURRENT
DENSITY
And the total current is obtained by integrating,
( convention current) S dSJI
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CURRENT AND CURRENT
DENSITY
Current density can be related to the velocity of volume charge density at a given point. Consider the element of charge Q = vv = vSL, as shown in the figure .
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CURRENT AND CURRENT
DENSITY
Assume that the charge element is
oriented within the
edges and parallel to
the coordinate axes,
and it shall only
posses an x
component of
velocity.
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CURRENT AND CURRENT
DENSITY
With respect to the time interval t, the element of charge has moved a distance x, as shown in the figure.
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CURRENT AND CURRENT
DENSITY
Therefore a charge Q = vSx has moved through a reference plane perpendicular to the direction of motion in a time increment t, and the resultant current is
t
xS
t
QI
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CURRENT AND CURRENT
DENSITY
And taking the limit with respect to time,
where: vx represents the x component of the velocity, v.
And in terms of current density,
xSI
xx vJ
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CURRENT AND CURRENT
DENSITY
and in general
where: J or vv is the convention current density
vJ
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Sample Problem 1
The vector current density is given as J =
(4/r2)cos ar + 20 e-2r sin a r sin cos
a A/m2. (a) Find J at r = 3, = 0, = .
(b) Find the total current passing through
the spherical cap r = 3, 0 < < 20O, 0 < < 2, in the ar direction.
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CONTINUITY OF CURRENT
The principle of conservation of charge states simply that charges can be neither created nor destroyed, although equal amounts of positive and negative charge may be simultaneously created, obtained by separation, destroyed, or lost by recombination.
The continuity equation, follow this principle when considering of any region bounded by a closed surface. And the current through the closed surface is
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CONTINUITY OF CURRENT
the outward flow of positive charge must
be balanced by a decrease of positive
charge (or perhaps an increase of
negative charge) within the closed surface.
S dSJI
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CONTINUITY OF CURRENT
The current at closed surface, however, is an outward-flowing current and it is the integral form of the continuity equation, and the differential, or point, form is obtained by using the divergence theorem to change the surface integral into a volume integral:
volS dJdSJ )(
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CONTINUITY OF CURRENT
Next represent the enclosed charge Qi by the volume integral of the charge density,
volvol ddt
ddJ )(
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CONTINUITY OF CURRENT
Keeping the surface constant, the derivative becomes a partial derivative and
may appear within the integral,
For an incremental volume,
volvol
dt
dJ
)(
dt
J
)(
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CONTINUITY OF CURRENT
And the point from the continuity equation,
tJ
)( (the current, or charge
per second, diverging
from a small volume per
unit volume is equal to
the time rate of
decrease of charge per
unit volume at every
point)
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Sample Problem 2
Assume that an electron beam carries a total current of 500 A in the az direction, and has a current density Jz that
is not a function of or in the region 0 < < 10-4 m and is zero for < 10-4 m. If the electron velocities are given by vz = 8 x
107 z m/s, calculate v at = 0 and z =:(a) 1 mm; (b) 2 cm: (c) 1 m.
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Seatwork
1. Given the vector current density J =
102za - 4cos2a A/m
2: (a) find the
current density at P( = 3, = 30O, z = 2); (b) determine the total current flowing
outward through the circular band = 3, 0 < < 2, 2 < z < 2.8.
Answer 180a 9a A/m2; 518 A
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Seatwork
2. Current density is given in cylindrical coordinates as J = - 106z1.5az A/m
2 in the region 0 20m; for 20 m, J = 0. (a) Find the total current crossing the surface z = 0.1m in the az direction. (b) If the charge velocity is 2 x 10
6 m/s at z = 0.1 m, find v there. (c) If the volume charge density at z = 0.15 m is -2000 C/m3, find the charge velocity there.
Answer -39.7 mA; - 15.81 kC/m3; -2900 m/s
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METALLIC CONDUCTOR
Metallic Conductor permit a higher-energy level in the valence band to merges smoothly, to a conduction band by the help of kinetic energy produce by an external field that will result in an electron flow.
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METALLIC CONDUCTOR
The valence electrons, or conduction, or free, electrons, having a charge Q = -e will move under the influence of an electric field, E, and will experience a force:
F = - eE
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METALLIC CONDUCTOR
And the valence electron velocity (drift velocity) is linearly related to the electric field intensity by the mobility of the electron in a given materials (i.e. free space or crystalline)
vd = - eE (electron velocity is opposite in
direction to E)
where is the mobility of an electron (positive)
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METALLIC CONDUCTOR
In terms of current density, J:
where:
Therefore;
EJ ee
ee
EJ
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METALLIC CONDUCTOR
Consider a uniform current density J and electric field intensity E in a cylindrical region of length L and cross-sectional area S.
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METALLIC CONDUCTOR
Therefore;
or
where:
or
but
JSdSJIS
ba
a
b
a
bab LEdLEdLEV
baab LEV IRV
ELV
L
VE
S
IJ S
LR
IS
LV
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METALLIC CONDUCTOR
Insulator did not permit any electron flow due to an existing gap between the valence band and the conduction band. In this case the electron cannot accept any additional amounts of energy.
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METALLIC CONDUCTOR
Semiconductors had a small forbidden region that separates the valence band and the conduction band. A small amounts of energy in the form of heat, light, or an electric field may raise the energy of the electrons and provide a conduction.
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Sample Problem 3
Find the magnitude of the electric field intensity in a sample of silver having = 6.17 x 107 mho/m and e = 0.0056 m
2/V.s if: (a) the drift velocity is 1 mm/s; (b) the current density is 107 A/m2; (c) the sample is a cube, 3 mm on a side, carrying a total current of 80 A; (d) the sample is a cube, 3 mm on a side, having a potential difference of 0.5 mV between opposite faces.
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Sample Problem 4
An aluminum conductor is 1000 ft long and
has a circular cross section with a
diameter of 0.8 in. If there is a dc voltage
of 1.2 V between the ends, find: (a) the
current density; (b) the current; (c) the
power dissipated, using your vast
knowledge of circuit theory.
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Conductivity of Metallic Conductor
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Seatwork
1. Find the magnitude of the current density in a
sample of silver for which = 6.17 x 107 S/m and e = 0.0056 m2/Vs if: (a) the drift velocity is 1.5 m/s; (b) the electric field intensity is 1 mV/m; (c) the sample is a cube 2.5 mm on a
side having a voltage of 0.4 mV between
opposite faces; (d) the sample is a cube 2.5 mm
on a side carrying a total current of 0.5A.
Answer: 16.53 kA/m2; 61.7 kA/m2; 9.87MA/m2;
80.0 kA/m2
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Seatwork
2. A copper conductor has a diameter of 0.6 in and
it is 1200 ft long. Assume that it carries a total dc
current of 50 A. (a) Find the total resistance of
the conductor. (b) What current density exists in
it? (c) What is the dc voltage between the
conductor ends? (d) How much power is
dissipated in the wire?
Answer: 0.0346; 2.74 x 105 A/m2; 1.729 V; 86.4 W
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
Conductor Characteristics:
1. It has a surface charge density that
resides on the exterior surface and within
has zero charge density.
2. In static conditions, in which no current
will flow, follows directly Ohms law: the electric field intensity within the conductor
is zero.
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
The principles applied to conductors in electrostatic fields:
1. The static electric field intensity inside a conductor is zero.
2. The static electric field intensity at the surface of a conductor is everywhere directed normal to the surface.
3. The conductor surface is an equipotential surface.
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
Prove:
Consider a closed surface conductor in a
free space boundary
( In static conditions, tangential
electric field intensity and electric
flux density are zero.)
0 dLE
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
For the tangential field around the small closed path at the surface:
and E = 0,
where: a to b = c to d = w and b to c = d to a = h
0 a
d
d
c
c
b
b
a
02
1
2
1,, hEhEwE ataNatbNt
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
( h = 0 and w 0(finite)) 0wEt0tE
0 tt ED
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
Consider the normal field in a small cylinder as the Gaussian surface;
(Using Gauss Law)
(Cylinder Sides)
where: bottom and
sides integral were equal to zero
QdSDS
Qsidesbottomtop
SQSD SN
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CONDUCTOR PROPERTIES AND
BOUNDARY CONDITIONS
or SND
SNN ED 0
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SAMPLE PROBLEM 5
A potential field is given as V = 100e-5x sin
3y cos 4z V. Let point P(0.1,/12,/24) be located at a conductor-free space
boundary. At point P, find the magnitude
of: (a) V; (b) E; (c) EN; (d) Et; (e) S
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Methods of Images
Dipole Characteristics infinite plane at zero potential that exists midway between the two charges.
Consider a vanishingly thin conducting plane that is infinite in extent and have an equipotential surface at a potential V = 0, and electric field intensity normal to the surface.
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Methods of Images
This can be represented by a single charge (image) above the plane and maintain the same fields, removing the plane and locating a negative charge at a symmetrical location below the plane.
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SAMPLE PROBLEM 6
A point charge of 25 nC is located in free
space at P(2,-3,5), and a perfectly
conducting plane is at z = 0. Find: (a) V at
(3,2,4); (b) E at (3,2,4); (c) S at (3,2,0).
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Solution to Problem 6
Given:
z = 0
V = 0 at z = 0 at (3,2,4)
R = (3,2,4) (2,-3,5) = (1,5,-1) R = (3,2,4) (2,-3,-5) = (1,5,9)
(a)
(b)
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Solution to Problem 6 (b)
(c)
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SEMICONDUCTORS
Intrinsic Semiconductor pure material (i.e. germanium or silicon)
Two types of current carriers:
1. Electrons are those from the top of the filled valence band which have received sufficient energy (usually thermal) to cross the relatively small forbidden (1 electron volt energy gap) band into the conduction band.
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SEMICONDUCTORS
2. Holes are vacancies left by the electrons represent unfilled energy states in the valence band which may also move from atom to atom in the crystal.
Both carriers move in an electric field, and they move in opposite directions; hence each contributes a component of the total current which is in the same direction as that provided by the other.
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SEMICONDUCTORS
The conductivity is a function of both hole and electron concentrations and mobilities,
where:
= conductivity
e and h = electron and hole charge densities
e and h = mobility of electrons and holes
hhee
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SEMICONDUCTORS
or
= e + h where: e = Neee h = Nheh Ne and Nh = electron and holes
concentrations
Electron charge, e = 1.602 x 10-19 C
e = mobility of electron (0.12, silicon; 0.36, germanium)
h = mobility of holes (0.025, silicon; 0.17, germanium )
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SEMICONDUCTORS
Doping the process of adding impurities to a pure semiconductor.
Acceptors furnish extra holes and form p-type materials.
Donor provide additional electrons and form n-type materials.
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SAMPLE PROBLEM 7
Using the values given in this section for
the electron and hole mobilities in
germanium at 300 K, and assuming hole
and electron concentrations of 2.7 x 1019
m-3, find: (a) the component of the
conductivity due to holes; (b) the
component of the conductivity due to
electrons; (c) the conductivity.
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Solution to Problem 7
1 = Neeh = (2.7 x 1019)( 1.602 x 10-19)(0.17)
= 0.735 mho/m
2 = Neee = (2.7 x 1019)( 1.602 x 10-19)(0.36)
= 1.557 mho/m
= 1 + 2 = 0.735 + 1.557 = 2.29 mho/m
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NATURE OF DIELECTRIC
MATERIALS
The characteristic which all dielectric materials have in common, whether they are solid, liquid, or gas, and whether or not they are crystalline in nature, is their ability to store electric energy.
This storage takes place by means of a shift in the relative positions of the internal, bound positive and negative charges against the normal molecular and atomic forces.
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NATURE OF DIELECTRIC
MATERIALS
A non-polar molecule does not have this dipole arrangement until after a field is applied. The negative and positive charges shift in opposite directions against their mutual attraction and produce a dipole which is aligned with the electric field.
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NATURE OF DIELECTRIC
MATERIALS
Qdp
n
i
itotal pp1
n
i
ipP1
0
1lim
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NATURE OF DIELECTRIC
MATERIALS
Consider a dielectric containing non-polar molecules, no dipole moment and P =0. In the interior, an incremental surface element , S and an electric field, E, produces a moment p = Qd in each molecule, such that p and d make an angle with S.
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NATURE OF DIELECTRIC
MATERIALS
Net total charge that crosses the elemental surface in an upward direction,
where: Qb = bound charge
Bound charge polarization,
Bound charge within the close surface,
SnQdQb
SPQb
Sb dSPQ
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NATURE OF DIELECTRIC
MATERIALS
Total enclosed charge,
where:
Q = total free charge enclosed by the surface S.
Enclosed free charge,
ST dSEQ 0
QQQ bT
SbT dSPEQQQ )( 0
PED N 0
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NATURE OF DIELECTRIC
MATERIALS
For Polarizable Material
where Q = free charge enclosed
In terms of volume charge densities
Equivalent Divergence relationships,
S dSDQ
vTT
vv
vbb
dvQ
dvQ
dvQ
T
b
E
P
0
vD
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NATURE OF DIELECTRIC
MATERIALS
In ferroelectric materials the relationship between P and E is not only nonlinear, but
also shows hysteresis effects.
The linear relationship between P and E is
where: e (chi) = a dimensionless quantity (electric susceptibility or dielectric
constant)
EP e 0
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NATURE OF DIELECTRIC
MATERIALS
Substituting to the value of D:
where:
Therefore
where: = permittivity
EEED ee 000 )1(
1 eR
R 0
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SAMPLE PROBLEM 8
A certain homogenous slab of lossless
dielectric material is characterized by an
electric susceptibility of 0.12 and carries a
uniform electric flux density within it of 1.6
nC/m2. Find: (a) E; (b) P;(c) the average
dipole moment if there are 2 x 1019 dipoles
per cubic meter; (d) the voltage between
two equipotentials 1 in. apart.
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Solution to Problem 8
Given:
Xe = electric susceptibility = 0.12
D = electric flux density = 1.6 nC/m2
dipoles/cu.m = 2 x 1019
a) D = 0E + Xe 0E = 0E(Xe + 1)
E = D/ [0(Xe + 1)] = (1.6 x 10-9)/[(8.854 x 10-12)(0.12 +1)]
= 161.35 V/m
b) P = Xe 0E = (0.12)(8.854 x 10-12)(161.35)
= 171.4 x 10-12 C/m2 or 171.4 pC/m2
c) Average dipole moment = (171.4 x 10-12)/(2 x 1019)
= 8.57 x 10-30 C-m
d) V = ES = (161.35)(0.0254) = 4.09829 V or 4.10 V
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
Boundary conditions existing at the interface between a conductor and a
dielectric:
1. D and E are both zero inside the
conductor.
2. The tangential E and D field
components must both be zero.
0 tt ED
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
3. The normal electric flux density is equal
to the surface charge density on the
conductor.
SNN ED
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
Consider the interface between two dielectrics having permitivities 1 and 2 and occupying regions 1 and 2 and the tangential components are given by,
0S dLE
02tan1tan wEwE (around the close path)
2tan1tan EE (h = negligible)
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
If the tangential electric field intensity is continuous across the boundary, then the
tangential D is discontinuous, for
or
2
2tan2tan1tan
1
1tan
DEE
D
2
1
2tan
1tan
D
D
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
Applying the Gausss law,
QdSDS
SQSDSD SNN 21
SNN DD 21
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BOUNDARY CONDITIONS FOR
PERFECT DIELECTRIC MATERIALS
(normal E is discontinuous)
(S = 0, the normal component of D is
continuous) or
SNN DD 21
21 NN DD
2211 NN EE
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SAMPLE PROBLEM 9
The region y < 0 contains a dielectric
material for which R1 = 2.5, while the region y > 0 is characterized by R2 = 4. Let E1 = -30ax + 50ay + 70az V/m, and
find: (a) EN1; (b) Et1; (c)Et1; (d) E1; (e) 1
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Solution to Problem 9
Given:
R1 = 2.5
R2 = 4
E1 = -30ax + 50ay + 70az V/m; y = normal
a) EN1 = E1ay = (-30ax + 50ay + 70az) ay = 0 +
50 + 0 = 50 V/m
b) Et1 = E1 EN1 = -30ax + 50ay + 70az - 50ay = -30ax + 70az V/m
c) Et1 = |Et1| =
-
Solution to Problem 9
d) E1 = |E1| =
e)
1 = cos-1 (E1ay)/|E1| = 56.71
O
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SAMPLE PROBLEM 10
Continue problem 9 by finding: (a) DN2; (b)
Dt2; (c) D2; (d) P2; (e) 2
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Solution to Problem 10
E1 = -30ax + 50ay + 70az E2 = -30ax + Ey2ay + 70az D1 = 0R1E1 = (8.854 x 10-12)(2.5)(-30ax + 50ay + 70az) = - 0.664 x 10
-
9ax + 1.107 x 10-9ay + 1.55 x 10
-9az D2 = Dx2ax + 1.107 x 10
-9ay + Dz2az
D2 = 0R2E2 Dx2ax + 1.107 x 10
-9ay + Dz2az = (8.854 x 10-12)(4)( -30ax +
Ey2ay + 70az)
Dx2 = -1.062 x 10-9
Ey2 = 31.26
Dz2 = 2.48 x 10-9
D2 = -1.062 x 10-9ax + 1.107 x 10-9ay + 2.48 x 10
-9az D2 = -1.062ax + 1.107ay + 2.48az nC/m
2
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Solution to Problem 10
a) DN2 = D2ay = (-1.062ax + 1.107ay + 2.48az) ay = 0 + 1.107 + 0 = 1.107 nC/m2
b) Dt2 = Et20R2 = (76.2)(8.854 x 10-12)(4) = 2.70 x 10-9
C/m2 = 270 nC/m2
c) D2 = -1.062ax + 1.107ay + 2.48az nC/m2
d)
P2 = -0.797ax + 0.830ay + 1.86az C/m2
e) ;
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CAPACITANCE
Capacitance the ratio of the magnitudes of the total charge on either conductor to
the potential difference between
conductors. And it is measured in farads
(F), where a farad = 1 Coulomb/Volt
where: Q = surface integral over the
positive conductor.
0V
QC
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CAPACITANCE
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CAPACITANCE
Consider two conductors embedded in a homogenous dielectric. Conductor M2
carries a total positive charge Q, and M1
carries an equal negative charge. There
are no other charges present, and the total
charge of the system is zero.
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CAPACITANCE
Let Vo, the potential difference between M2 and M1, and carrying a unit charge
from the negative to the positive surface,
and the capacitance is given by,
dLE
dSEC S
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CAPACITANCE
-
CAPACITANCE
Consider a simple two-conductor system in which the conductors are identical,
infinite parallel planes with separation d.
Choosing the lower conducting plane at z
= 0 and the upper one at z = d, a uniform
sheet of surface charge S on each conductor leads to the uniform field.
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CAPACITANCE
( = permittivity of the homogenous dielectric)
(charge on the lower plane)
(charge on the upper plane)
Potential difference between the lower and upper plane,
zS aE
zSaD
SZN DD
ZN DD
ddzdLEV Sd
Slower
upper
0
0
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CAPACITANCE
where:
S = surface charge density
S = surface area
or
The total energy stored in the capacitor,
Therefore,
And the charge value,
SQ S
d
S
V
QC
0
2
222
0 0 2
22
2
1
2
1
2
1
2
1
d
d
SSddSdzdvEW
SSS d S
volE
C
QQVCVWE
2
0
2
02
1
2
1
2
1
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SAMPLE PROBLEM 11
Find the relative permittivity of the
dielectric material used in a parallel plate
capacitor if : (a) C = 40 nF, d = 0.1 mm,
and S = 0.15 m2, (b) d = 0.2 mm, E = 500
kV/m, and S = 10 C/m2; (c) D = 50 C/m2 and the energy density is 20 J/m3.
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Solution to Problem 11 (a)
(b)
-
Solution to Problem 11 (c)
Given:
D = 50 x 10-6 C/m2
W = 20 J/m3
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SEVERAL CAPACITANCE
Consider a Coaxial cable or coaxial capacitor that has inner radius a, outer
radius b, and length L. The value of the
capacitance is given by,
)/ln(
2
ab
LC
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SEVERAL CAPACITANCE
Consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b > a. The expression for the electric field,
( = dielectric permittivity
between the region of the
spheres)
24 r
QEr
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SEVERAL CAPACITANCE
Potential difference using the line integral,
(Q =represents the total charge o the inner sphere)
ba
QVab
11
4
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SEVERAL CAPACITANCE
The Capacitance is given by,
or
(allowing the outer sphere to become infinitely large, b = .)
ba
V
QC
ab11
4
aC 4
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SEVERAL CAPACITANCE
Coating the sphere with a different dielectric layer, for which = 1, extending from r = a to r = r1,
)(4
)(4
4
12
0
12
1
2
rrr
QE
rrar
QE
r
QD
r
r
r
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SEVERAL CAPACITANCE
The potential difference is,
1011
2
0
2
1
1111
4
44
1
1
rra
QVV
r
Qdr
r
QdrVV
a
ra
ra
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SEVERAL CAPACITANCE
Therefore,
1011
1111
4
rra
C
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SEVERAL CAPACITANCE
Consider a parallel-plate capacitor of area S and spacing d, where d is small
compared to the linear dimensions of the
plate. The capacitance is given by 1S/d, using a dielectric permittivity 1. Suppose another part of the capacitor dielectric has
a permittivity of 2, then the potential difference given by Vo is,
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SEVERAL CAPACITANCE
(E1 and E2 are
both uniform)
(E is normal)
or
22110 dEdEV
21 NN DD
2211 EE
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SEVERAL CAPACITANCE
Eliminating E2;
The surface charge density,
Since D1 = D2, the capacitance is given by,
21210
1/dd
VE
2
2
1
1
01111
dd
VEDS
212
2
1
110011
11
CCS
d
S
dV
S
V
QC S
-
SAMPLE PROBLEM 12
Find the capacitance of: (a) 20 cm of 58C/U coaxial cable having an inner conductor 0.0295 in. in diameter, an outer conductor having an inside diameter of 0.116 in., and a polyethylene dielectric; (b) a conducting sphere 1 cm in diameter, covered with a layer of polyethelene 1 cm thick, in free space; (c) a conducting sphere 1 cm in diameter, covered with a layer of polyethelene 1 cm thick, and surrounded by a concentric conducting sphere 1.5 cm in radius.
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Solution to Problem 12
a)
b)
c)
-
Solution to Problem 12
b)
-
SAMPLE PROBLEM 13
A parallel-plate capacitor contains three
dielectric layers, Let R1 = 1, d1 = 0.2 mm, R2 = 2, d2 = 0.3 mm, R3 = 3, d3 = 0.4 mm, and S = 20 cm2. (a) Find C. (b) Find the
percentage of the total stored energy
located in each of the three regions.
-
Solution to Problem 13
a)
b)
-
CAPACITANCE OF A TWO-WIRE
LINE
-
CAPACITANCE OF A TWO-WIRE
LINE
Consider the potential field of two infinite line charges. A positive line charge in the
xz plane at x = a and a negative line
charge at x = -a. The potential of a single
line charge with zero reference at a radius
of Ro is,
R
RV L 0ln
2
-
CAPACITANCE OF A TWO-WIRE
LINE
Combining the potential field,
Letting R10 = R20 (placing zero reference at equal distance from each line) and R1 = x and R2 =y,
120
210
2
20
1
10 ln2
lnln2 RR
RR
R
R
R
RV LL
22
22
22
22
ln4
ln2 yax
yax
yax
yaxV LL
-
CAPACITANCE OF A TWO-WIRE
LINE
Choosing an equipotential surface V = V1 and letting K1 = dimensionless parameter of potential V1
LVeK /4
11
-
CAPACITANCE OF A TWO-WIRE
LINE
Therefore, 22
22
1yax
yaxK
2
1
12
2
1
1
1
2
1
1
K
Kay
K
Kax
01
12 22
1
12
ay
K
Kaxx
-
CAPACITANCE OF A TWO-WIRE
LINE
In equipotential surface where V = V1, the surface is independent value of z ( a cylinder) and intersects the xy plane in a circle of radius b,
Centered at x = h, y = 0;
where: ( x = 0)
1
2
1
1
K
Kab
1
1
1
1
K
Kah
22 bha
b
bhhK
22
1
-
CAPACITANCE OF A TWO-WIRE
LINE
But
Therefore,
The value of the capacitance is given by,
or
LVeK /2
10
1
0
ln
4
K
VL
110 ln
2
ln
4
K
L
K
L
V
LC L
)/(cosh2
/ln
2122 bh
L
bbhh
LC
-
SAMPLE PROBLEM 14
Find the capacitance between a circular
conducting cylinder in air, 2 mm in radius,
and: (a) a conducting plane, 1 cm distant
from the cylinder axis; (b) a conducting
plane, 1 cm distant from the cylinder axis;
(c) a similar cylinder, axes separated by 1
cm.
-
Solution to Problem 14
a)
b)
c)
L = 1
= 0 = 8.854 x 10-12
r = b = 2mm = 2 x 10-3 m
d = h = 0.01 m
-
Solution to Problem 14
0.024 = 1cm + 1cm + 2mm + 2mm