Transcript EE3321 Electromagnetic Field Theory

Potential Energy, Energy Density
Capacitance, Polarization
Boundary Conditions
Potential Energy
 The 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)
Example
Potential Energy U = +qV
Charge Q
R
V(∞→R) =Vref = 0
Test Charge +q
Potential V = kQ
R
Observations
 The potential at infinity is zero
 A positive test charge +q gains potential as it gets
closer to the charge +Q
 A negative test charge –q loses potential as it gets
closer to the charge +Q.
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.
CRT Simplified Set Up
 Once 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.
Exercise
 An 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.
Energy Stored in an E Field
 Using 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.
Example
 Let 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 constant
 Calculate the energy density
 Calculate the volume between the
capacitor plates
Exercise
 Calculate the energy stored in the field produced by a
metal sphere of radius a holding a charge Q.
 Determine the electric field E
 Find the energy density u
 Set up the integral for U
 Integrate over space a<R< ∞
Capacitance
 As 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.
Capacitance
 An 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
Parallel Plate Capacitor
 From 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
Exercise
 Consider a parallel plate capacitor. Derive an
expression for the stored energy U in terms of the
capacitance C and the potential V.
Polarization
 Suppose 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
Polarization
 Next 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
Polarization
 The decrease of E is said to be due to the polarization
P of the dielectric molecules which opposes E:
 But the capacitance increases to
Exercise
 The 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.
Electric Boundary Conditions
 On 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 material
 The component of E parallel to the interface is
continuous
 The component of D normal to the interface is
continuous
Perfect Dielectric Medium
 Tangential components of E are continuous
E1t = E2t
Medium 1 (air)
Medium 2
WATER DROPPLET
Perfect Dielectric Medium
 Normal components of E are discontinuous
ε1E1n = ε2E2n
Medium 1 (air)
no free charges
Medium 2
WATER DROPPLET
Perfect Conductor Medium
 Tangential components of E are zero
E1t = E2t = 0
Medium 1 (air)
X
Short circuit
Medium 2
WATER DROPPLET
Perfect Conductor Medium
 Normal components of E are discontinuous
E1n≠ 0
Medium 1 (air)
E2n= 0
Medium 2
WATER DROPPLET
Examples of Field Lines
Exercise
 Consider a dielectric interface at z = constant.
 Let єr1 = 2, єr2 = 5, and E1 = 2ax + 3ay + 5az
 Find E2
єr1 = 2
єr2 = 5
Homework
 Read textbook sections 4-8, 4-9, 4-10, 4-11
 Solve problems 4.43, 4.45, 4.50, 4.51, 4.52, 4.54