Electromagnetic Engineering:
• Electromagnetics deals with space concepts and requires thinking
in the three dimensions of the real world. Hence the three
dimensional coordinate systems are essential.
• Rectangular Coordinate System
• Cylindrical Coordinate System
• Spherical Coordinate System
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Electrostatics:
• The fundamental quantity of interest in electromagnetics is
‘charge’. Charges produce forces that are related to the
fundamental electromagnetic field quantities of an electric filed
and magnetic field.
• The term ‘field’ is used to indicate that these quantities have
values that vary throughout space and perhaps with time. They
also have direction of effect at each point, which is quantified by
describing the field quantities as vectors.
• Stationary charges produce the electrostatic field, and dc or
steady currents (flow of charges) produce magnetic field
(magnetostatic field).
• Charges moving at a velocity that is not constant produce both
an electric and a magnetic field that are not independent of each
other, and these effects are described by the fundamental
Maxwell’s Equations.
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Electrostatics:
• Static electric fields in vaccum or free space – such fields are found
in the focussing and deflection systems of electrostatic cathoderay tubes.
Coulomb’s Law: The force between two very small objects
separated in a vacuum or free space by a distance, which is large
compared to their size, is proportional to the charge on each and
inversely proportional to the square of the distance between them.
where Q1 and Q2 are the positive or negative quantities of charge, R
is the separation, and k is a proportionality constant.
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Electrostatics: (Coulomb’s Law)
• The new constant 0is called the permittivity of free space and has
magnitude, measured in farads per meter (F/m).
The quantity 0 is not dimensionless.
Coulomb’s law is now,
• The coulomb is an extremely large unit of charge. Charge of
electron (negative) or proton (positive is 1.602 × 10−19 C.
• Hence a negative charge of one coulomb represents about
6 × 1018 electrons.
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Electrostatics: (Coulomb’s Law)
• Coulomb’s law shows that the force between two charges of one
coulomb each, separated by one meter, is 9 × 109 N, or about one
million tons.
• The electron has a rest mass of 9.109 × 10−31 kg and has a radius of
the order of magnitude of 3.8×10−15 m.
• This does not mean that the electron is spherical in shape, but
merely describes the size of the region in which a slowly moving
electron has the greatest probability of being found.
• All other known charged particles, including the proton, have
larger masses and larger radii, and occupy a probabilistic volume
larger than does the electron.
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Electrostatics:(Coulomb’s Law)
• Vector Form of Coulomb’s Law: The force acts along the line
joining the two charges, and is repulsive if the charges are alike in
sign or attractive if they are of opposite sign.
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Electrostatics: (Coulomb’s Law)
• Let the vector r1 locate Q1, whereas r2 locates Q2. Then the
vector R12 = r2 − r1 represents the directed line segment from
Q1 to Q2. The vector F2 is the force on Q2 (where Q1 and Q2
have the same sign),
where a12 = a unit vector in the direction of R12,
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Electrostatics: (Coulomb’s Law)
• The force expressed by Coulomb’s law is a mutual force, for each
of the two charges experiences a force of the magnitude, although
of opposite direction.
• Coulomb’s law is linear, for if we multiply Q1 by a factor n, the
force on Q2 is also multiplied by the same factor n. It is also true
that the force on a charge in the presence of several other charges
is the sum of the forces on that charge due to each of the other
charges acting alone.
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Electric Field Intensity:
• If we now consider one charge fixed in position, say Q1, and move
a second charge slowly around, we note that there exists
everywhere a force on this second charge; in other words, this
second charge is displaying the existence of a force field that is
associated with charge, Q1. Call this second charge a test charge
Qt . The force on it is given by Coulomb’s law,
• Force per unit charge gives the electric field intensity, E1 arising
from Q1:
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• E1 is interpreted as the vector force, arising from charge Q1, that
acts on a unit positive test charge.
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Electric Field Intensity:
• In general, we write the expression for Electric Field Intensity as,
(V/m) i.e. [ N/C = (N.m)/(C.m) = J/(C.m) = V/m]
• The electric field intensity is evaluated at the test charge location
that arises from all other charges in the vicinity—meaning the
electric field arising from the test charge itself is not included in E.
• The electric field of a single point charge becomes:
• The vector R is the directed line segment from the point at which
the point charge Q is located to the point at which E is desired, and
aR is a unit vector in the R direction.
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Electric Field Intensity:
• If we arbitrarily locate Q1 at the center of a spherical coordinate
system. The unit vector aR then becomes the radial unit vector ar ,
and R is r . Hence
The field has a single radial component.
• If we consider a charge that is not at the origin of our coordinate
system, the field no longer possesses spherical symmetry, and we
might as well use rectangular coordinates. For a charge Q located
at the source point r’ = x’ax + y’ay + z’az, as illustrated in Figure on
next slide, we find the field at a general field point r = xax+ yay +zaz
by expressing R as r − r’, and then
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Electric Field Intensity:
The vector r’ locates the point charge Q, the vector r identifies the
general point in space P(x, y, z), and the vector R from Q to P(x, y, z)
is then R = r − r’.
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Electric Field Intensity:
• The coulomb forces are linear, the electric field intensity arising
from two point charges, Q1 at r1 and Q2 at r2, is the sum of the
forces on Qt caused by Q1 and Q2 acting alone, or
• where a1 and a2 are unit vectors in the direction of (r−r1) and (r−r2),
respectively.
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Electric Field Intensity:
• If we add more charges at other positions, the field due to n point
charges is,
Example:- Find Eat P(1, 1, 1) caused by four identical 3-nC charges
located at P1(1, 1, 0), P2(−1, 1, 0), P3(−1,−1, 0), and P4(1,−1, 0), as
shown in Figure on next slide.
Solution: We find that r = ax + ay + az , r1 = ax + ay , and thus r − r1 = az.
The magnitudes are: |r − r1| = 1, |r − r2| = √5, |r − r3| = 3, and |r − r4|
= √5. Because Q/4π0 = 3 × 10−9/(4π × 8.854 × 10−12) = 26.96V.m,
or,
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Electric Field Intensity:
• A symmetrical distribution of four identical 3-nC point charges
produces a field at P, E = 6.82ax + 6.82ay + 32.8az V/m.
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Problem:• A charge of −0.3μC is located at A(25,−30, 15) (in cm), and a second
charge of 0.5μC is at B(−10, 8, 12) cm. Find E at: (a) the origin; (b)
P(15, 20, 50) cm.
Ans. 92.3ax − 77.6ay − 94.2az kV/m; 11.9ax − 0.519ay + 12.4az kV/m
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• D3.3: Given the electric flux density, D = 0.3r2ar nC/m2 in free
space: (a) find E at point P(r = 2,  = 250, = 900); (b) find the
total charge within the sphere r = 3; (c) find the total electric
flux leaving the sphere r = 4.
• Ans. 135.5ar V/m; 305 nC; 965 nC
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Application of Gauss’s Law to Differential Volume Element
(Rectangular Coordinates)
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Application of Gauss’s Law to Differential Volume Element
• Let us consider any point P, shown in Figure on previous slide,
located by a rectangular coordinate system. The value of D at the
point P may be expressed in rectangular components,
D0 = Dx0ax + Dy0ay + Dz0az.
• We choose as our closed surface the small rectangular box,
centered at P, having sides of lengths x, y, and z, and apply
Gauss’s law,
• In order to evaluate the integral over the closed surface, the
integral must be broken up into six integrals, one over each face,
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Divergence Theorem:
• The divergence of the vector flux density D is the outflow of flux
from a small closed surface per unit volume as the volume shrinks
to zero.
• div D =D = v
…. Point form of Gauss’s Law or Maxwell’s
First Equation in point form.
• We have obtained from applying Gauss’s Law to differential volume
element,
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Divergence Theorem:
• The divergence theorem relates the closed surface integration to
volume integration.
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Application of Gauss’s Law to Differential Volume Element:
(Example)
• D3.6: In free space, let D = 8xyz4ax+4x2z4ay+16x2yz3az pC/m2. (a) Find
the total electric flux passing through the rectangular surface z = 2, 0
<x < 2, 1 < y < 3, in the az direction. (b) Find E at P(2,−1, 3). (c) Find an
approximate value for the total charge contained in an incremental
sphere located at P(2,−1, 3) and having a volume of 10−12 m3.
• Ans. 1365 pC; −146.4ax + 146.4ay − 195.2az V/m; −2.38 × 10−21 C
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Energy Expended in Moving a Point Charge in an Electric Field:
• The electric field intensity was defined as the force on a unit test
charge at that point at which we wish to find the value of this
vector field.
• If we attempt to move the test charge against the electric field,
we have to exert a force equal and opposite to that exerted by the
field, and this requires us to expend energy or do work.
• Suppose we wish to move a charge Q a distance dL in an electric
field E. The force on Q arising from the electric field is,
• The component of this force in the direction dL which we must
overcome is
where aL = a unit vector in the direction of dL.
• The force that we must apply is equal and opposite to the force
associated with the field,
and the expenditure
of energy is the product of the force and distance.
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Energy Expended in Moving a Point Charge in an Electric Field:
• The differential work done by an external source moving charge Q
is dW = −QE · aLdL, or,
where we have replaced aLdL by the simpler expression dL.
• The work required to move the charge through a finite distance
must be determined by integrating,
• If we wish to move the charge in the direction of the field, our
energy expenditure turns out to be negative; we do not do the
work, the field does.
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Potential Difference and Potential:
• Potential difference V is defined as the work done (by an external
source) in moving a unit positive charge from one point to another
in an electric field,
• Potential difference is measured in joules per coulomb, for which
the volt is defined as a more common unit, abbreviated as V.
Hence the potential difference between points A and B is,
and VAB is positive if work is done in carrying the positive charge
from B to A.
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Potential difference between two points in the field of a
point charge:
• The potential difference between points A and B at radial
distances rA and rB from a point charge Q. Choosing an origin at Q,
and dL = dr ar
• If rB > rA, the potential difference VAB is positive, indicating that
energy is expended by the external source in bringing the positive
charge from rB to rA.
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Potential Gradient
• E = –V
• No work is done in carrying the unit charge around any closed
path,
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