dielectric - Erwin Sitompul
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Transcript dielectric - Erwin Sitompul
Engineering Electromagnetics
Lecture 9
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
President University
Erwin Sitompul
EEM 9/1
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
Consider the interface between two dielectrics having
permittivities ε1 and ε2, as shown below.
We first examine the tangential components around the small
closed path on the left, with Δw<< :
E dL 0
Etan1w Etan 2w 0
Etan1 Etan 2
President University
Erwin Sitompul
EEM 9/2
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
The tangential electric flux density is discontinuous,
Dtan1
Dtan 2
Etan1 Etan 2
1
Dtan1 1
Dtan 2 2
2
The boundary conditions on the normal components are found
by applying Gauss’s law to the small cylinder shown at the right
of the previous figure (net tangential flux is zero).
DN1S DN 2S Q S S
DN1 DN 2 S
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• ρS cannot be a bound surface charge
density because the polarization
already counted in by using dielectric
constant different from unity
• ρS cannot be a free surface charge
density, for no free charge available in
the perfect dielectrics we are
considering
• ρS exists only in special cases where
it is deliberately placed there
Erwin Sitompul
EEM 9/3
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
Except for this special case, we may assume ρS is zero on the
interface:
DN 1 DN 2
The normal component of electric flux density is continuous.
It follows that:
1EN1 2 EN 2
President University
Erwin Sitompul
EEM 9/4
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
Combining the normal and the tangential
components of D,
DN1 D1 cos1 D2 cos2 DN 2
Dtan1 D1 sin 1 1
Dtan 2 D2 sin 2 2
2 D1 sin 1 1D2 sin 2
After one division,
tan 1 1
tan 2 2
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1 2 1 2
Erwin Sitompul
EEM 9/5
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
The direction of E on each side of
the boundary is identical with the
direction of D, because D = εE.
E1
1EN1 2 EN 2
Etan1 Etan 2
1 2 1 2
E2
President University
Erwin Sitompul
EEM 9/6
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
The relationship between D1 and D2 may be found from:
2
D2 D1 cos 1
1
2
2
sin 2 1
The relationship between E1 and E2 may be found from:
1
2
E2 E1 sin 1
2
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2
cos 2 1
Erwin Sitompul
EEM 9/7
Chapter 6
Dielectrics and Capacitance
Boundary Conditions for Perfect Dielectric Materials
Example
Complete the previous example by finding the fields within the
Teflon.
Eout E0ax
Dout 0 E0ax
Pout 0
• E only has normal
component
Din Dout 0 E0a x
0 E0a x
Din
Ein
0.476E0a x
r 0
r 0
0 E0a x
Pin 1.1 0Ein 1.1 0
0.524 0 E0a x
r 0
President University
Erwin Sitompul
EEM 9/8
Chapter 6
Dielectrics and Capacitance
Boundary Conditions Between a Conductor and a Dielectric
The boundary conditions existing at the interface between a
conductor and a dielectric are much simpler than those
previously discussed.
First, we know that D and E are both zero inside the conductor.
Second, the tangential E and D components must both be zero
to satisfy:
E dL 0
D E
Finally, the application of Gauss’s law shows once more that
both D and E are normal to the conductor surface and that
DN = ρS and EN = ρS/ε.
The boundary conditions for conductor–free space are valid
also for conductor–dielectric boundary, with ε0 replaced by ε.
Dt Et 0
DN EN S
President University
Erwin Sitompul
EEM 9/9
Chapter 6
Dielectrics and Capacitance
Boundary Conditions Between a Conductor and a Dielectric
We will now spend a moment to examine one phenomena:
“Any charge that is introduced internally within a conducting
material will arrive at the surface as a surface charge.”
Given Ohm’s law and the continuity equation (free charges
only):
J E
v
J
t
We have:
v
E
t
v
D
t
President University
Erwin Sitompul
EEM 9/10
Chapter 6
Dielectrics and Capacitance
Boundary Conditions Between a Conductor and a Dielectric
If we assume that the medium is homogenous, so that σ and ε
are not functions of position, we will have:
v
D
t
Using Maxwell’s first equation, we obtain;
v
v
t
Making the rough assumption that σ is not a function of ρv, it
leads to an easy solution that at least permits us to compare
different conductors.
The solution of the above equation is:
v 0e( )t
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• ρ0 is the charge density at t = 0
• Exponential decay with time constant of ε/σ
Erwin Sitompul
EEM 9/11
Chapter 6
Dielectrics and Capacitance
Boundary Conditions Between a Conductor and a Dielectric
Good conductors have low time constant. This means that the
charge density within a good conductors will decay rapidly.
We may then safely consider the charge density to be zero
within a good conductor.
In reality, no dielectric material is without some few free
electrons (the charge density is thus not completely zero), but
the charge introduced internally in any of them will eventually
reach the surface.
ρv
ρ0
v 0e( )t
ρ0/e
ε/σ
President University
Erwin Sitompul
t
EEM 9/12
Chapter 6
Dielectrics and Capacitance
Homework
No homework this week.
President University
Erwin Sitompul
EEM 9/13