ee3321 electromagnetic field theory
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EE3321 Electromagnetic Field Theory. Potential Energy, Energy Density Capacitance, Polarization Boundary Conditions. Potential Energy. The electric field and potential energy are directly related: - PowerPoint PPT PresentationTRANSCRIPT
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EE3321 Electromagnetic Field Theory
Potential Energy, Energy DensityCapacitance, Polarization
Boundary Conditions
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Potential EnergyThe electric field and potential energy are directly
related: As a test charge +q moves in the direction that the field
opposed it, its potential energy increases.
The electrostatic potential energy is the energy of an electrically charged particle (at rest) in an electric field. The energy difference between two potentials is given by
U = q(V2 – Vref) Joules (VAs)
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Example
Test Charge +q
Charge Q
Potential V = kQ R
Potential Energy U = +qV
RV(∞→R) =Vref = 0
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ObservationsThe potential at infinity is zeroA positive test charge +q gains potential as it
gets closer to the charge +QA negative test charge –q loses potential as it
gets closer to the charge +Q.
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Cathode Ray Tube The CRT is a vacuum tube
containing an electron gun (a source of electrons) and a fluorescent screen used to create images in the form of light emitted from the fluorescent screen.
The image may represent electrical waveforms (oscilloscope), pictures (television, computer monitor), radar targets and others.
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CRT Simplified Set UpOnce the electrons leave the cathode, they
accelerate toward the grid.Electrons entering the deflecting plate region
and change directions depending on the voltage between the plates.
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ExerciseAn electron moves at a constant velocity v =
vo ax. Assume that the electron enters in a field E = - 1 a z (V/m) at x =0.
Compute the potential energy U the electron loses as it moves from A (at z = 0 cm) to B (at z = 0.5 cm). Recall e = 1.602 x 10 – 19 As.
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Energy Stored in an E FieldUsing Gauss’ Law in differential form and the
Divergence theorem and it can be shown that the energy density or energy per unit volume (J/m3) of the electric field is:
u = ½ є |E|2 (Joules/m3)
The total energy stored in the electrostatic field is
U = ∭ u dV (Joules)where dV is the volume differential.
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ExampleLet E = 9 V/mm in between the plates.
Suppose that the area A = 1 cm2 and the dielectric thickness is d = 1 mm. Find the energy stored by the capacitor for a relative permittivity of 2.8. Neglect (field) fringing effects.Notice that the field is constantCalculate the energy densityCalculate the volume between the
capacitor plates
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ExerciseCalculate the energy stored in the field
produced by a metal sphere of radius a holding a charge Q.Determine the electric field EFind the energy density uSet up the integral for UIntegrate over space a<R< ∞
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CapacitanceAs shown above a capacitor consists of two
conductors separated by a non-conductive region. The non-conductive substance is called the
dielectric medium. The conductors contain equal and opposite
charges on their facing surfaces, and the dielectric contains an electric field.
A capacitor is assumed to be self-contained and isolated, with no net electric charge and no influence from an external electric field.
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CapacitanceAn ideal capacitor is wholly
characterized by its capacitance C (in Farads), defined as the ratio of charge ±Q on each conductor to the voltage V between them
C = Q/V
More generally, the capacitance is defined in terms of incremental changes
C = dq/dv
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Parallel Plate CapacitorFrom Gauss’ Law the charge and the electric
field between the plates is related by
Likewise, the line integral relating the potential and the electric field simplifies to
Thus the capacitance is given by
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ExerciseConsider a parallel plate capacitor. Derive an
expression for the stored energy U in terms of the capacitance C and the potential V.
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PolarizationSuppose that a capacitor is charged up by
connecting it to a voltage source V which is then removed.
A fixed charge Q is placed on its upper plate and –Q on the lower plate.
Suppose the capacitor is air filled. In this case,
The capacitance is
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PolarizationNext assume that the
capacitor is filled with dielectric material as illustrated here.
Since the charge does not change, the electric flux D is the same as before.
However, the electric field E changes to
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PolarizationThe decrease of E is said to be due to the
polarization P of the dielectric molecules which opposes E:
But the capacitance increases to
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ExerciseThe relative permittivity of air is 1.0 and that
of quartz is about 4.5. Calculate the difference in capacitance for two capacitors with identical geometry using these two dielectric materials.
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Electric Boundary ConditionsOn a perfect conductor
The component of E parallel to the conducting surface is zero
The component of D normal to the conducting surface is numerically equal to the charge density
On a perfect dielectric materialThe component of E parallel to the interface is
continuousThe component of D normal to the interface is
continuous
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Perfect Dielectric Medium
WATER DROPPLET
Medium 1 (air)
Medium 2
Tangential components of E are continuousE1t = E2t
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Perfect Dielectric Medium
WATER DROPPLET
Medium 1 (air)
Medium 2
Normal components of E are discontinuousε1E1n = ε2E2n
no free charges
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Perfect Conductor Medium
WATER DROPPLET
Medium 1 (air)
Medium 2
Tangential components of E are zeroE1t = E2t = 0
Short circuitX
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Perfect Conductor Medium
WATER DROPPLET
Medium 1 (air)
Medium 2
Normal components of E are discontinuousE1n≠ 0 E2n= 0
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Examples of Field Lines
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ExerciseConsider a dielectric interface at z =
constant. Let єr1 = 2, єr2 = 5, and E1 = 2ax + 3ay + 5az
Find E2
єr2 = 5
єr1 = 2
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HomeworkRead textbook sections 4-8, 4-9, 4-10, 4-11Solve problems 4.43, 4.45, 4.50, 4.51, 4.52,
4.54