Download Lecture Notes 10: Electrostatic Fields Inside A Dielectric, Boundary Conditions on E, D and P

UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
LECTURE NOTES 10
The Macroscopic Electric Field Inside a Dielectric
When we discuss electric (and/or magnetic) fields, whether they are outside of/exterior to
matter, or inside the matter itself, implicitly, we physically interpret these field quantities to be
associated with macroscopic averages over (vast) numbers of electromagnetic quanta (i.e. virtual
photons), atoms, molecules, electric charges (both +ve and –ve) etc. The “true” E & B-fields
inside of matter - at the atomic scale - are wildly varying from point to point (and also wildly
varying in time, e.g. on short/atomic time-scales due to fluctuation(s) in thermal energies at finite
temperature). For almost all applications that we are interested in, we are not concerned with these
wild spatial (and temporal) fluctuations on the atomic scale; we are primarily concerned with the
average / mean fields extant in these media, suitably averaged over large numbers of constituent
particles involved. These (space and time-averaged) fluctuations die out as 1 N where N is the
number of constituents involved. If N
1023 , then since σ N = N , then for random fluctuations
(i.e. Gaussian-distributed) the fractional fluctuations, σ N N = N N = 1 N 3.2 × 10−12 are
extremely small – essentially negligible! Hence the macroscopic (i.e. microscopically averagedover) E-field can be seen as being truly electrostatic, for so-called time-independent situations.
Suppose we want to calculate the macroscopic electric field E ( r ) at some point, r inside a
solid dielectric sphere of radius, R as shown in the figure below.
ẑ
Field point, P
R
ŷ
r
O
x̂
Small imaginary sphere of radius δ
centered on the field point, P @
| r | < R (for averaging purposes)
The macroscopic electric field at the field point P @ r inside the sphere consists of two parts:
– A contribution from the average electric field Eout ( r ) due to electric charges outside / external
to a small imaginary sphere (of radius δ R ) centered on the point P, and:
– A contribution from the average electric field Ein ( r ) due to electric charges inside this small
conceptual sphere.
In other words, the macroscopic electric field at the field point P located at r
(inside the dielectric sphere, i.e. r < R ), using the Principle of Linear Superposition is:
E ( r ) = Eout ( r ) + Ein ( r )
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
1
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
In Griffith’s problem 3.41(d), we learned that the electric field averaged over an imaginary
sphere due to a single charge q outside of/exterior to the imaginary sphere was the same as the
electric field due to the charge q, as observed at the center of that imaginary sphere. By the
principle of superposition, this result then holds for any collection of exterior charges.
Thus, here for our dielectric sphere of radius R, Eout ( r ) (with r < R ) is the electric field at
r due to the electric dipoles contained within the dielectric sphere of radius R that are outside
of/exterior to the imaginary/conceptual sphere of infinitesimal radius δ centered on r .
Outside of/excluding the region of this small imaginary sphere of radius δ centered on the field
point P @ | r | < R, the atomic/molecular electric dipoles are far enough away from the field point
P that we may safely write the potential Vout ( r ) corresponding to Eout ( r ) (with r < R ) as:
Vout ( r ) =
rˆ ⋅ Ρ ( r ′ )
1
∫ r 2 dτ ′, r = r − r ′, r = r − r ′
4πε o outside
where the integral is over the volume of the dielectric sphere, but excluding the small volume
associated with the small imaginary sphere of radius δ centered on the field point P @ | r | < R.
The electric dipoles inside the small conceptual/imaginary sphere of radius δ centered on the
field-point P @ r are too close to treat in this fashion.
However, in Griffith’s problem 3.41(a-c), we also learned that the average electric field inside a
sphere of radius δ due to all of the electric charge contained within the sphere of radius
δ (regardless of the details of the charge distribution within that sphere) is:
1 p
Eave = −
4πε 0 δ 3
where p is the total electric dipole moment of that sphere.
Thus, we know that we know that the average electric field @ r within the small conceptual /
imaginary sphere of radius δ centered on the field-point P @ r must be:
1 p
Ein ( r ) = −
4πε 0 δ 3
where p ( r ) is the total/net macroscopic electric dipole moment associated with the (microscopic)
electric dipoles contained within this conceptual/imaginary sphere centered on the field point P @
r:
⎛ Volume of conceptual / ⎞
= Ρ ( r ) 43 πδ 3 = 43 πδ 3 Ρ ( r )
p (r ) = Ρ (r ) *⎜
3⎟
4
⎝ imaginary sphere, 3 πδ ⎠
where Ρ ( r ) = macroscopic electric polarization = electric dipole moment per unit volume (@ r ).
p ( r ) = Ρ ( r ) ( 43 πδ 3 )
Thus:
And thus:
2
Ein ( r ) = −
1
p (r )
4πε 0 δ
3
=−
1
4/ πε
/ 0
(
4
3
)
π/ δ 3 Ρ ( r )
δ3
=−
1
Ρ (r )
3ε 0
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Thus we obtain:
Ein ( r ) = −
1
Ρ (r )
3ε 0
Now because of the (infinitesimal) size of the conceptual/imaginary sphere of radius δ R
centered on the field point P @ r , then the (macroscopic) electric polarization Ρ ( r ) should not
vary (on average) significantly over this small volume, thus the term/contribution that was left out
of the integral for the outside potential :
rˆ ⋅ Ρ ( r ′ )
1
Vout ( r ) =
dτ ′, r = r − r ′, r = r − r ′
∫
4πε o outside r 2
actually corresponds to that associated with the electric field at the center of a uniformly polarized
dielectric sphere of radius δ , which is − 1 3ε 0 Ρ ( r ) !!!
{see/read Griffiths Example 4.2 and/or Prof. S. Errede’s P435 Lect. Notes 9, p. 25-26}.
But this is precisely what the electric field Ein ( r ) = −
1
Ρ ( r ) puts back in!!!
3ε 0
In other words, using the principle of superposition: VToT ( r ) = Vout ( r ) + Vin ( r )
and thus:
Ein ( r ) = −∇Vin ( r ) = −
Thus we see that VTot ( r ) =
∫
whole
volume v′
1
Ρ (r )
3ε 0
rˆ ⋅ Ρ ( r ′ )
dτ ′ works fine for the entire dielectric!!!
r2
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
3
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
The Macroscopic Electric Field Due to Near Dipoles in a Polarized Dielectric
Consider a very large block of polarized dielectric (e.g. polarized by a uniform external E field,
e.g. Eext = Eoext xˆ Imagine a small spherical volume of radius δ ~ 1 cm deep within the polarized
dielectric. The electric polarization Ρ inside the dielectric will then be uniform e.g. Ρ = Ρ o xˆ and
Eint inside the dielectric will also uniform, Eint = Eoint xˆ
Imagine “excising” this small spherical volume from the polarized dielectric –
but still having it precisely/magically retain all of its EM properties as they were when it was part
of the polarized dielectric. By itself, it will appear as shown below:
Mathematically & physically, note that this situation here is equivalent to two overlapping spheres,
one with uniform volume charge density ρ + = + Q 43 πδ 3 and another sphere with uniform volume
charge density ρ − = − Q
d
4
3
πδ 3 whose centers are offset from each other by a distance
δ ( d 1Å = 10−10 m ) . Thus equivalently, this sphere now has only a bound surface charge
density σ B (ξ ) = σ o cos (ξ ) where the angle ξ is measured with respect to the + x̂ axis.
Thus a uniformly polarized dielectric sphere of radius δ with uniform polarization Ρ = Ρ o xˆ is
equivalent to two uniformly oppositely charged spheres whose centroids are displaced from each
other by a distance d δ . See figures on the immediately following page:
4
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
GREATLY EXAGGERATED PIX:
⎛ What is the E -field @ the center of this polarized dielectric sphere? ⎞
⎜⎜
⎟⎟
⎝ = E -field due to the near dipoles inside the polarized dielectric!!! ⎠
We know that for a single, uniformly electrically charged sphere (volume charge density ρ =
constant), that the electric field inside such a single sphere is given (from Gauss’ Law) by
1 Qencl
Einside ( r < δ ) =
rˆ
4πε 0 r 2
where r is defined from center of that sphere.
But the charge enclosed by the Gaussian surface of radius r ( r < δ ) is Qencl = ρ *V = ρ 43 π r 3 .
Noting that the total charge contained in a single uniformly charged sphere is
QTot1 = ρ *VTot1 = ρ 43 πδ 3 , or ρ = QTot1 43 πδ 3 , then we can rewrite Einside ( r < δ ) as:
1 Qencl
1 ρ 43 π r 3
1
1 QTot1 ⎛ r ⎞
ˆ
ˆ
ˆ
Einside ( r < δ ) =
r
=
=
r
rr
=
ρ
⎜ ⎟ rˆ
4πε 0 r 2
3ε 0
4πε 0 δ 2 ⎝ δ ⎠
4 π ε0
r2
Radius δ of uniformly charged sphere
Gaussian surface
of radius r.
ρ = QTot
1
4
3
πδ 3
Now for two oppositely-charged spheres of uniform charge density ρ ± whose centroids are
laterally displaced from each other by an infinitesimal distance d 10−10 m δ ~ 1 cm the net /
total E -field at the center of the two overlapping spheres (by the principle of linear superposition)
is:
1
1
ρ+
ρ−
Tot
Einside
= Einside
( r ) + Einside
( r ) = ρ + r+ + ρ− r−
3ε 0
3ε 0
where ρ ± = ±QTot1
4
3
πδ 3 and where the vectors r+ and r− are defined in the figures shown below:
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
5
UIUC Physics 435 EM Fields & Sources I
Define: ρ + ≡ ρ , then ρ − = − ρ
Fall Semester, 2007
Tot
Thus: Einside
=
Lecture Notes 10
Prof. Steven Errede
1
1
ρ ( r+ − r− ) = −
ρd
3ε 0
3ε 0
=− d
Thus the E -field at center of two overlapping oppositely-but-uniformly charged spheres whose
centroids are laterally displaced from each other by an infinitesimally small distance
δ 10−10 m δ ∼ 1 cm and d = d xˆ, is:
Tot
Einside
=−
But ρ = QTot1
Tot
Einside
=−
4
3
1
1
ρd = −
ρ d xˆ for d = d xˆ.
3ε 0
3ε 0
πδ 3 and p = QTot d = total dipole moment of polarized dielectric sphere.
1
(
)
1
1 QTot1
1 QToT1 d
ˆ
d
x
xˆ
ρ d xˆ = −
=
−
3
3ε 0
4πε 0 δ 3
3ε 0 43 πδ
Tot
∴ For ( r < δ ) : Einside
=−
1
but p = QToT1 dxˆ
p
.
4πε 0 δ 3
We now drop the “Tot” superscript, since this simply referred to our (equivalent) model of the
polarized dielectric sphere as the superposition of two uniformly-but-oppositely-electricallycharged spheres displaced by an infinitesimal distance δ 10−10 m δ ∼ 1 cm .
The electric polarization Ρ = electric dipole moment per unit volume = p
4
3
πδ 3
⇒ p = Ρ 43 πδ 3 = 43 πδ 3Ρ
∴ For ( r < δ ) : Einside = −
1
p
4πε 0 δ
3
=−
1
4 π ε0
4
3
π δ3Ρ
1
=−
Ρ ⇒
3
3ε 0
δ
Einside = −
1
Ρ
3ε 0
This is the macroscopically averaged E -field at the center of/inside an imaginary/conceptual small
diameter sphere of radius δ (somewhere) deep inside of a uniformly polarized dielectric.
Note that this E -field arises solely from the contributions of the near dipoles in the dielectric
within this sphere of radius δ . Note further that it explicitly does NOT include the externally
applied electric field (that was used to polarize the block of dielectric in the first place).
This E -field DOES NOT include ANY contributions from electric dipoles (or anything else)
EXTERIOR to this imaginary/conceptual sphere!
6
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
THE MACROSCOPIC ELECTRIC SUSCEPTIBILITY χ e OF A DIELECTRIC
(Lossless) (in Eext) (uniform, no voids) (rotationally invariant → e.g. not a crystalline material)
(i.e. amorphous)
a.k.a. "Class A" Dielectric
For an "ideal", linear, homogeneous & isotropic dielectric the electric polarization (a.k.a. the
electric dipole moment per unit volume) Ρ is simply related to the internal electric field, Eint of the
dielectric, by a simple proportionality constant, i.e.
Ρ ( r ) = m Eint ( r )
Ρ (r )
m = slope of straight line
Eint ( r )
m = simple constant
(i.e. m = scalar quantity)
n.b. This relation is ONLY true for CLASS A dielectrics - i.e. ones which are linear, homogenous,
ideal and isotropic. (We will discuss modifications to this relation shortly…)
Now: SI units of Ρ ( r ) : Coulombs
SI units of E ( r ) : Newtons
meter 2
Coulomb
Ρ (r )
Coulombs meter 2
Coulombs 2
⇒ m has SI units of m =
Define: m ≡ ε 0 χ e
=
=
Eint ( r ) Newton Coulombs Newton-m 2
where ε 0 = (macroscopic) electric permittivity of free space (vacuum) = 8.85 ×10
−12
Coulombs 2
Newton-m 2
= Farads/m
and the (macroscopic) electric susceptibility of the dielectric material, χ e is a pure number
(i.e. χ e is a scalar quantity – it is dimensionless).
⎛ Coulombs ⎞
= ε 0 χ e Eint ( r )
Then: Ρ ( r ) ⎜
2 ⎟
⎝ meter ⎠
⎛ Coulombs 2 ⎞ ⎛ Newtons ⎞
⎜⎜
⎟
2 ⎟
⎟*⎜
⎝ Newton -m ⎠ ⎝ Coulomb ⎠
= Coulombs m 2
For class-A dielectrics: Ρ ( r ) = ε 0 χ e Eint ( r )
For free space (“empty” vacuum), the (macroscopic) electric susceptibility χ e = 0 because free
space/vacuum has no MATTER in it.
The electric susceptibility χ e and electric polarization Ρ ( r ) explicitly refer to the dielectric
properties of matter (and not the underlying/inter-penetrating vacuum). By the principle of linear
superposition, the dielectric properties of matter and vacuum are additive to / independent of each
other, thus we can define the (total) electric permittivity associated with a block of “Class-A” type
dielectric as a scalar point function, defined at each point r in space as:
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
7
UIUC Physics 435 EM Fields & Sources I
ε
=
total electric
permittivity
of dielectric
Fall Semester, 2007
Lecture Notes 10
+ ε o χ e = ε o (1 + χ e ) = (1 + χ e ) ε o ⇐
εo
electric
permittivity
of vacuum
electric
permittivity
of dielectric
Prof. Steven Errede
SI Units same as
for ε o (Farads/m)
In some dielectrics, under certain conditions χ e → ∞ . In plasmas (i.e. ionized gases), χ e < 0 .
In most typical/garden-variety dielectric materials, 0 ≤ χ e ≤ 10.
We can also define a relative electric permittivity (a.k.a. dielectric “constant”) which is
(obviously) dimensionless:
K e ( = " ε r ") ≡
⎛ε ⎞
ε (1 + χ e ) ε o
=
= (1 + χ e ) and/or: χ e = K e − 1 = ⎜ ⎟ − 1
εo
εo
⎝ εo ⎠
Consider a “real life” situation (i.e. an actual physics experiment): A Class-A dielectric block of
insulator-type material is inserted between two parallel plates, which have a potential difference
ΔV across the parallel plates of the capacitor, as shown in the figure below:
We know that: Eext =
σ free
ΔV
xˆ =
xˆ
ε0
a+b+c
( Volts/m )
from the (empty) parallel plate capacitor
If the Class-A dielectric is in a uniform/constant Eext (i.e. the gap of the parallel-plate capacitor
is small relative to size (length/width dimensions of the parallel plates), then the electric
polarization Ρ ( r ) = Ρ o xˆ is must also be uniform/constant inside the gap of the parallel-plate
capacitor, and thus no bound volume charge density exists inside the dielectric material:
ρ Bound ( r ) = −∇iΡ ( r ) = 0
However, on the RHS and LHS surfaces of the dielectric (see above figure, with
nˆ1 = + xˆ , nˆ2 = − xˆ ), that σ Bound+ = Ρ ( r )inˆ1 RHS = +Ρ o and σ Bound− = Ρ ( r )inˆ2 LHS = −Ρ o ,
surface
respectively, or, expressing this more compactly:
8
surface
σ Bound = Ρ ( r )inˆ
±
surface
= ±Ρ o
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Thus, here again we see that we can replace the polarization Ρ of the dielectric altogether, here
simply by the (equivalent) bound surface charge distributions σ Bound± (since ρ Bound ( r ) = 0 inside
the dielectric). Then we see that:
Ρ = Ρ o xˆ = σ Bound xˆ = ε 0 χ e E int
What is Eint ( r ) ???
macroscopic
Eint ( r ) = vector sum of Eext ( r ) and Emolecular
(r )
dipoles
using the principle of linear superposition!!!
macroscopic
Eint ( r ) = Eext ( r ) + Emolecular
(r )
Thus:
dipoles
macroscopic
What is: Emolecular
(r ) ?
dipoles
Note that the situation (here) with bound surface charges of σ Bound+ = +Ρ o on the RHS surface
and σ Bound− = −Ρ o on the LHS surface is analogous to that for the free surface charge densities
σ free = +σ o (RHS) and σ free − = −σ o (LHS) associated with the parallel plate capacitor itself, except
+
macroscopic
note the direction of Emolecular
( r ) relative to Eext (and hence note the sign change below)!!! We
dipoles
can thus easily see that:
macroscopic
Emolecular
(r ) = −
dipoles
σ
σ Bound
xˆ c.f. with the external field of ||-plate capacitor: Eext = + free xˆ
ε0
ε0
Thus we also see that:
σ free
σ
xˆ − bound xˆ
ε0
ε0
dipoles
σ
Ρ
1
Ρ
1
macroscopic
xˆ = o xˆ = Ρ , Thus: Emolecular
( r ) = − bound xˆ = − o xˆ = − Ρ ( r )
ε0
ε0
ε0
ε0
ε0
dipoles
macroscopic
Eint ( r ) = Eext ( r ) + Emolecular
(r ) = +
But:
σ bound
ε0
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
9
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
macroscopic
Therefore: Eint ( r ) = Eext ( r ) + Emolecular
( r ) = Eext ( r ) −
dipoles
Lecture Notes 10
1
ε0
Prof. Steven Errede
Ρ (r )
Rearranging this relation:
1
Eext ( r ) = Eint ( r ) +
∴
ε0
1
Eext ( r ) = Eint ( r ) +
ε0
Ρ (r )
But: Ρ ( r ) = ε 0 χ e ( r ) Eint ( r )
ε 0 χ e Eint ( r ) = Eint ( r ) + χ e Eint ( r ) = (1 + χ e ) Eint ( r )
Thus: Eext ( r ) = (1 + χ e ) Eint ( r )
or: Eint ( r ) = Eext ( r ) (1 + χ e )
We see that the macroscopic/averaged-over internal electric field inside the dielectric Eint ( r ) is
reduced by a factor of 1 (1 + χ e ) relative to the external/applied electric field Eext ( r ) , because the
macroscopic
electric field associated with the (now polarized) molecular dipoles, Emolecular
( r ) opposes the
dipoles
external applied electric field! Using the dielectric constant, K e ≡ ε ε o = (1 + χ e ) we see the same
thing, namely that Eint ( r ) = Eext ( r ) (1 + χ e ) = Eext ( r ) K e i.e. the internal electric field is
“screened” / reduced from the Eext value by the dielectric constant K of the dielectric material.
We can also show that, since: Ρ ( r ) = ε 0 χ e Eint ( r ) Then: Eint ( r ) = Ρ ( r ) ε 0 χ e and Eext = (σ free ε 0 ) xˆ
Ρ ( r ) ⋅ nˆ
Thus: Eint ( r ) =
Ρ (r )
ε 0 χe
surface
=
= Ρ o = σ Bound
⎛σ
⎞
= ⎜ Bound ⎟ xˆ
ε 0 χe ⎝ ε 0 χe ⎠
Ρ o xˆ
{Ρ ( r ) = Ρ xˆ}
o
⎛ 1 ⎞
⎛ 1 ⎞ ⎛ σ free ⎞
and: Eint ( r ) = ⎜
⎟ Eext ( r ) = ⎜
⎟⎜
⎟ xˆ
⎝ 1 + χe ⎠
⎝ 1 + χe ⎠ ⎝ ε o ⎠
Then we see that:
⎛ σ Bound
⎛ χe ⎞
⎞ ⎛ 1 ⎞⎛ σ free ⎞
⎟ σ free or: ⎜⎜
⎟=⎜
⎟⎜
⎟ or: σ Bound = ⎜
⎝ 1 + χe ⎠
⎠ ⎝ 1 + χ e ⎠⎝ ε o ⎠
⎝ σ free
⎛ε ⎞
But: χ e = K e − 1, or: 1 + χ e = K e = ⎜ ⎟
⎝ ε0 ⎠
⎛ K −1 ⎞
⎛ χe ⎞
⎛ ε − ε0 ⎞
∴ σ Bound = ⎜ e ⎟ σ free = ⎜
⎟ σ free
⎟ σ free = ⎜
⎝ ε ⎠
⎝ Ke ⎠
⎝ 1 + χe ⎠
⎛ σ Bound
⎜
⎝ ε 0 χe
⎞ ⎛ χe ⎞
⎟⎟ = ⎜
⎟
⎠ ⎝ 1 + χe ⎠
i.e. The bound surface charge density σ Bound on the surface of a dielectric is directly related to the
free surface charge density σ free on the surface of the conducting plates of the parallel plate
capacitor!!!
IMPORTANT NOTE: This relation between bound surface charge density σ Bound and surface
charge density σ free is NOT a universal one!!! It is specific only to the case of the parallel-plate
capacitor!!!
10
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
The potential difference ΔV between the two capacitor plates of the parallel plate capacitor is:
ΔV = − ∫ E id = aEext + bEint + cEext
C
If a = c (i.e. the air gaps in the parallel plate capacitor the same dimension)
⎛
b ⎞
1
Then: ΔV = 2aEext + bEint But: Eint =
Eext ∴ ΔV = ⎜ 2a −
⎟ Eext
Ke ⎠
Ke
⎝
Define: d ≡ ( 2a + b ) = total gap between parallel plates of capacitor.
Now: Eext =
∴
σ free
xˆ
ε0
⎛
b ⎞ σ free
ΔV = ⎜ 2a +
⎟
Ke ⎠ ε 0
⎝
Capacitance of parallel plate capacitor: C ≡
Q (σ free A )
=
ΔV
ΔV
A = surface area of one of the
plates of the ||-plate capacitor
Capacitance of the ||-plate capacitor (including the dielectric):
σ free A
σ A
ε0 A
=
C = free =
ΔV
⎛
b ⎞ σ free ⎛⎜ 2a + b ⎞⎟
2
+
a
Ke ⎠
⎜
⎟
⎝
Ke ⎠ ε 0
⎝
If there is no dielectric, then K e = 1 =
Then: Cno dielectric =
ε0 A
2a
=
ε
ε0
( ε =ε 0 = vacuum ) and b = 0, d = 2a
ε0 A
d
If there are no air gaps, then a = c = 0 and d = b
Then: Cdielectric =
ε0 A
d
Ke
⎛ε A⎞
= K e ⎜ 0 ⎟ = K eCno dielectric
⎝ d ⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
11
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
THE MACROSCOPIC ELECTRIC FIELD INSIDE A DIELECTRIC
BOUNDARY CONDITIONS ON E
Suppose we have a linear, homogeneous, isotropic (Class-A) dielectric. We concern ourselves
with the macroscopic E -field inside the dielectric (the microscopic/atomic scale
E -field is wildly fluctuating, both in space (position) and time due to thermal fluctuations).
We note that the basic properties of macroscopic E -field in a dielectric material should not
change wildly/dramatically from those of the vacuum (free/empty) space. i.e. imagine we
adiabatically change ε 0 → ε dielectric no fundamental changes will occur during this process.
In particular, we must keep in mind that the macroscopic E ( r ) is a conservative field, whether
inside the dielectric or the vacuum (microscopically, virtual photons are all the same kind/type in a
dielectric vs. the vacuum) and since the macroscopic force (e.g. on a test charge,
QT ) F ( r ) = QT E ( r ) is also conservative in a dielectric medium or vacuum ⇒ hence both
F ( r ) and E ( r ) are derivable from a scalar potential, V ( r ) .
⇒ ∇ × E ( r ) = 0 (always), since ∇ × ( ∇V ( r ) ) = 0 (always) for a conservative force.
Equivalently: ∇ × E ( r ) = 0 ⇒
∫ E ( r )i d
c
= 0 (Stoke’s Theorem) and vice-versa.
Consider a parallel plate capacitor with a dielectric between the plates of the capacitor.
The dielectric has a very thin hole drilled through it, parallel to the electric field(s), as shown in the
figure below:
Now
∫ E ( r )id
= 0 . Take contour C as shown in picture above, but shrink the contour C down
C
to just ε (i.e. infinitesimally) inside/outside the hole drilled in the dielectric:
∫ E ( r )id
C
(
+ (E
= 0; Evac ⊥ε 2
= Evac i
12
i
34
diel
(
= Evac i
12
) + (E
)+ (E
= 0; Ediel ⊥ε 3
) (
ε ) + (E
i
1
diel 2
4
= 0; Ediel ⊥ε 4
12
) + (E
diel
) (
ε ) (
i ε 2 + Ediel i 12 ε 3
1
vac 2
i
34
i
1
vac 2
1
)
)
1
2
ε 2 = 1st half of
23
, 12 ε 3 = 2nd half of
23
1
2
ε 4 = 1st half of
41
, 12 ε1 = 2nd half of
41
= 0; Evac ⊥ε1
)=0
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
But:
12
=−
34
≡
⇒ ∴
(E
vac
Fall Semester, 2007
)
Lecture Notes 10
− Edie i = 0 But: Evac ||
Prof. Steven Errede
and Ediel ||
∴ Evac = Edie at the surface/boundary of the dielectric.
tangent
tangent
Specifically: Evac
= Ediel
@ the interface/boundary of the dielectric.
More generally:
The tangential components of E are equal @ a dielectric interface
i.e. E1t = E2t @ the interface of dielectric.
The tangential component of E is continuous across a dielectric interface.
Note that this result is valid regardless of the orientation of cavity/hole, provided (if and only if)
the dielectric is Class-A (i.e. linear, homogeneous isotropic) – it is not necessarily true otherwise.
SOME EXAMPLES OF DIELECTRICS
∃ all kinds of dielectric materials - some are gases, some are liquids and some are solids.
⎛ε ⎞
Dielectric “constant” K ≡ ⎜ ⎟ = (1 + χ e )
⎝ ε0 ⎠
−12
ε 0 = 8.85 ×10 Farads/m
= electric permittivity of free space/vacuum
= macroscopic constant/scalar quantity
= constant @ all frequencies (Lorentz invariant quantity)
ε = electric permittivity of dielectric
= macroscopic constant/scalar quantity
for Class-A dielectrics
SI Units: Farads/m
χ e = electric susceptibility of dielectric
= macroscopic constant/scalar quantity
for Class-A dielectrics
SI Units: Dimensionless
n.b. The macroscopic parameters ε , χ e (and thus K e ) have/exhibit frequency dependence because
microscopically, the induced and/or permanent electric dipole moments in atoms/molecules in the
dielectric (in general) are frequency dependent over the frequency range 0 ≤ f ≤ ∞ Hz !!!
Dielectric “Constants”
of various materials at
STP and f = 0 Hz.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
13
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
THE MACROSCOPIC ELECTRIC DISPLACEMENT FIELD, D ( r )
GAUSS’ LAW IN THE PRESENCE OF DIELECTRICS
We have seen that the effect of polarization of a dielectric is to produce bound surface and
volume charge densities within and/or on the surface(s) of the dielectric:
Bound volume charge density: ρ Bound ( r ) = −∇iΡ ( r )
( Coulombs meter 3 )
Bound surface charge density: σ Bound ( r ) = Ρ ( r )inˆ
( Coulombs meter )
2
surface
We have also shown that the E -field inside a dielectric medium due to the electric polarization,
Ρ ( r ) is simply (equivalently) due to the bound charge distributions ρ Bound ( r ) and/or σ Bound ( r ) .
Suppose now that this dielectric also had embedded in it free electric charges – e.g. either
embedded electrons or positive ions (e.g. by irradiating it with an e− beam or proton/ion beam).
Within the dielectric, since the electric charge density distributions (obviously) obey the principle
of linear superposition (i.e. due to charge conservation!), then the TOTAL volume electric charge
density can be written as:
ρTot ( r ) = ρ Bound ( r ) + ρ free ( r )
Then Gauss’ Law (in differential form) becomes:
ε 0∇i ETot ( r ) = ρTot ( r ) = ρ Bound ( r ) + ρ free ( r )
where: ETot ( r ) = total electric field = "Ebound ( r ) " + " E free ( r ) and ρ Bound ( r ) = −∇iΡ ( r )
We can rearrange Gauss’ Law Law (in differential form) as follows (dropping the “Tot” subscript
on the E-field – but please keep this in mind!!!):
ε 0∇i E ( r ) − ρ Bound ( r ) = ∇i ε 0 E ( r ) + Ρ ( r ) = ρ free ( r )
(
)
≡ D ( r ) = Electric Displacement
The (macroscopic) Electric Displacement Field: D ( r ) ≡ ε 0 E ( r ) + Ρ ( r )
SI units of D ( r ) are the same as that for Ρ ( r ) (same as that for σ Bound & σ free !! ): Coulombs m 2
Then we realize that Gauss’ Law (for dielectrics) becomes: ∇i D ( r ) = ρ free ( r )
i.e. the divergence of the (macroscopic) D -field at the point ( r ) is due to (i.e. equal to) the
free volume charge density, ρ free that is present at the point ( r ) !
In integral form, Gauss’ Law (for dielectrics) becomes:
∫ D ( r ′)idA′ = Q
encl
free
S′
Gauss’ Law for D physically tells us that the electric displacement field, D ( r ) is sensitive to the
free charge that is present in a given situation, whereas Gauss’ Law for E tells us that the electric
field intensity E ( r ) is sensitive to the total charge that is present in this same situation. Gauss’ Law
for Ρ tells us that Ρ ( r ) is sensitive to the bound charge that is present in this same situation.
14
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Thusfar, we have obtained several useful relations for Class-A dielectrics, summarized here:
D ( r ) ≡ ε 0 E ( r ) + Ρ ( r ) and Ρ ( r ) = ε 0 χ e E ( r )
SI units of D ( r ) are the same as that for Ρ ( r ) (same as that for σ Bound & σ free !! ): Coulombs m 2
∇i E ( r ) = ρTot ( r ) ε 0
∫ E ( r′)idA′ = Q
encl
Tot
and
ε o with ρTot ( r ) = ρ Bound ( r ) + ρ free ( r )
S′
∇i D ( r ) = ρ free ( r )
∫ D ( r ′)idA′ = Q
and
encl
free
encl
encl
= QBound
+ Q encl
with QTot
free
S′
∇iΡ ( r ) = − ρ Bound ( r ) and
∫ Ρ ( r′)idA′ = −Q
encl
Bound
S′
and σ Bound ( r ) = Ρ ( r )inˆ surface
⎛ε ⎞
and: K e ≡ ⎜ ⎟ = (1 + χ e )
⎝ ε0 ⎠
The tangential components of E are continuous across a dielectric interface
i.e. E1t = E2t @ the interface of a dielectric.
For Class-A dielectric materials (i.e. linear, ideal, homogeneous isotropic materials):
D ( r ) = ε 0 E ( r ) + Ρ ( r ) , but Ρ ( r ) = ε 0 χ e E ( r ) inside the dielectric
∴ D ( r ) = ε 0 E ( r ) + χ eε 0 E ( r ) = ε 0 (1 + χ e ) E ( r )
but ε = ε 0 (1 + χ e ) and K e ≡ ε ε o = (1 + χ e )
∴
D ( r ) = ε E ( r ) = ε 0 (1 + χ e ) E ( r ) = ε 0 E ( r ) + Ρ ( r ) in a Class-A dielectric material.
Griffiths Example 4.4:
A (very) long, straight conducting wire carries a uniform, free line electric charge λ which is
surrounded by rubber insulation out to radius, a. Find the electric displacement D ( r ) .
L
s
Take a cylindrical Gaussian surface of radius,
λ Coulombs/meter
free line charge
a
ẑ
s and length, L:
∫
S′
D ( r ′ )idA = Q enclosed
free
From the intrinsic symmetry of this problem, we realize that D ( r ) will be radial
(n.b. The E -field associated with the free line charge λ (alone) is radial)
The only contribution to surface integral is from the cylindrical portion of the Gaussian-surface,
i.e. D ( r ) || rˆ , and the end caps of the Gaussian surface (|| zˆ ) do not contribute since D ( r ) ⊥ zˆ .
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
15
UIUC Physics 435 EM Fields & Sources I
(
)
D 2π s L = λ L = Q
Thus: D ( r ) =
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
In Cylindrical Coordinates:
s =r s =r
sˆ = rˆ
s= s = r =r
enclosed
free
λ
rˆ (Coulombs/m2)
2π r
Note that this formula holds inside the rubber dielectric ( r < a ) as well as outside the rubber
dielectric ( r > a ) , i.e. this formula is valid for any r.
However, since Ρ ( r > a ) = 0 (i.e. no rubber dielectric for r > a )
Then: E ( r ) =
1
ε0
D (r ) =
λ ⎛1⎞
⎜ ⎟ rˆ for r > a
2πε 0 ⎝ r ⎠
Inside the rubber dielectric ( r < a ) , since we do not explicitly know the analytic form of
Ρ ( r < a ) then we do not know E ( r < a ) . Note also that (here) neither ρ Bound ( r < a ) nor
σ Bound ( r = a ) have been specified.
CAUTIONARY STATEMENTS ABOUT THE ELECTRIC DISPLACEMENT D ( r )
AND THE ELECTRIC POLARIZATION Ρ ( r )
Inside Class-A dielectric materials, the so-called constitutive (a.k.a. auxiliary) relation between the
three fields D ( r ) ≡ ε 0 E ( r ) + Ρ ( r ) holds/is true/valid.
Coulomb’s Law is true for ETot ( r ) , because E ( r ) is a conservative field, i.e. it is derivable from a
(
)
scalar potential E ( r ) = −∇V ( r ) , and the ∇ × E ( r ) = 0 (always) in electrostatics problems:
E (r ) =
1
4πε 0
or:
E (r ) =
or:
E (r ) =
ρTot ( r ′ )
∫
1
4πε 0
1
4πε 0
v′
∫
S′
∫
C′
r2
rˆ dτ ′, with ρTot ( r ) = ρ Bound ( r ) + ρ free ( r )
σ Tot ( r ′ )
r
2
λ ( r′)
r2
encl
encl
rˆ dA′, with QTot
= QBound
+ Q encl
free
rˆ d ′
The same/analogous thing is not true for the electric displacement, D ( r ) nor is it true for the
electric polarization, Ρ ( r ) , because neither D ( r ) nor Ρ ( r ) are conservative, and neither is
derivable from (the negative gradient of) a scalar potential. As consequences of these facts:
ρ free ( r )
ρ Bound ( r )
1
1
ˆ dτ ′ and Ρ ( r ) ≠
D (r ) ≠
r
rˆ dτ ′
2
∫
∫
r2
4π v′ r
4π v′
σ free ( r )
σ Bound ( r )
1
1
′
ˆ
r
D (r ) ≠
dA
r
and
Ρ
≠
rˆ dA′
(
)
r2
4π ∫S ′ r 2
4π ∫S ′
λ free ( r ′ )
λBound ( r ′ )
1
1
ˆ d and Ρ ( r ) ≠
D (r ) ≠
r
rˆ d
2
∫
∫
4π C ′
r2
4π C ′ r
16
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
E ( r ) is a fundamental field. E ( r ) is a conservative field.
D ( r ) and Ρ ( r ) are not fundamental fields. D ( r ) and Ρ ( r ) are not conservative fields.
D ( r ) and Ρ ( r ) are auxiliary fields.
While D ( r ) = ε 0 E ( r ) + Ρ ( r ) ⇒ ∇i D ( r ) = ε 0∇i E ( r ) + ∇iΡ ( r ) holds/is true/valid for Class-A
dielectrics, the divergence of a vector field on its own is insufficient to uniquely determine/fullyspecify the nature of a vector field.
Both ∇i A ( r ) and ∇ × A ( r ) must be specified in order to uniquely determine the A ( r ) -field.
(
(
)
Now ∇ × E ( r ) = 0 always E ( r ) and FE ( r ) are conservative
)
But ∃ many situations where
⎧⎪∇ × D ( r ) ≠ 0 ⎫⎪
⎛ has permanent electric polarization ⎞
⎨
⎬ e.g . a bar electret ⎜
⎟
⎝ - analogous to bar magnet!!!
⎠
⎪⎩∇ × Ρ ( r ) ≠ 0 ⎪⎭
D ( r ) and Ρ ( r ) are auxiliary fields associated with matter – dielectric materials in particular.
BOUNDARY CONDITIONS ON D, E and Ρ for DIELECTRIC MATERIALS
BOUNDARY CONDITIONS ON THE ELECTRIC DISPLACEMENT, D AT AN INTERFACE
Suppose we concern ourselves with what happens at the boundary/interface of two dielectric
materials, e.g. (air and water) or (glass and plastic)
Gaussian pillbox centered
D1
n̂1
on dielectric interface.
SIDE VIEW:
Shrink height h of pillbox
S1′
to zero/infinitesimally small.
Dielectric
Material # 1:
ε1 , K1e , χ1e
Free charge surface density,
Boundary/
σ free exists on interface
Interface
h
ΔS
σ free
Dielectric
Material # 2:
S3′
ε 2 , K 2e , χ 2e
n̂3
S2′
n̂2
D2
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
17
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Gaussian Surface S ′ = S1′ + S 2′ + S3′
∫
D ( r )idA = Q
S′
enclosed
free
Lecture Notes 10
Prof. Steven Errede
ΔS = area of disk at interface boundary
= ∫ σ free ( r ) dS ′ = σ free ΔS
S′
= ∫ D1 ( r )inˆ1dS1′ + ∫ D2 ( r )inˆ2 dS 2′ +
S1
S1
∫
S1
D3 ( r )inˆ3dS3′ = σ free ΔS
=0
Now nˆ1 = − nˆ2
Shrink Gaussian Pillbox
to zero height @ interface/
boundary of the two dielectrics
n̂1
D1
θ1
Dielectric
Medium #1
Interface
θ2
Dielectric
Medium #2
D2
n̂2
D1 inˆ1
interface
= − D1 ( r ) cos θ1 interface ≡ − D1n ( r ) interface =
⇒ ∫ D1 ( r )inˆ1dS ′ = − D1n ΔS imterface
S1
D2 inˆ2
interface
= + D2 ( r ) cos θ 2 interface ≡ + D2 n ( r ) interface =
⇒ ∫ D2 ( r )inˆ2 dS ′ = − D2 n ΔS interface
Normal component of D1 ( r )
evaluated at/on the interface
Normal component of D2 ( r )
evaluated at/on the interface
S2
But:
∫
S1
dS1′ = ∫ dS 2′ = ΔS
S2
∴ ⎡⎣ − D1n ( r ) + D2 n ( r ) ⎤⎦ ΔS
= σ free ΔS
interface
= σ free
or: ⎡⎣ D2 n ( r ) − D1n ( r ) ⎤⎦
interface
If σ free = 0 at interface, then D1n ( r ) interface = D2 n ( r ) interface
The normal component of D ( r ) is discontinuous across a dielectric interface when σ free is
present, by an amount σ free
18
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
BOUNDARY CONDITIONS ON THE ELECTRIC FIELD, E AT AN INTERFACE
We have already shown (see pages 12-13 of these lecture notes) that taking the contour integral
∫ E ( r )id = 0 across an interface between two dielectrics told us that the tangential components
C
of E are continuous across a dielectric interface: E1t
E1 ( r )
Medium 1
= E2t
interface
interface
1
4
2
Medium 2
Contour C
Just above &
below interface.
θ2
E2 ( r )
3
∫ E ( r )i d
C
with: E1 i
Thus: E1t
= E1 i 1 + E2 i
interface
interface
= E1t
= E2t
2
interface
interface
Shrink height h of
contour C to 0,
θ1
+ E3 i
3
+ E4 i
and − E3 i
4
= 0 where
interface
= − E2t
1
=−
3
interface
or: E1 sin θ1 interface = E2 sin θ 2 interface
=
The tangential components
of E are continuous across
a dielectric interface
If e.g. medium #1 is a conductor, then E1 = 0 inside the conductor.
If E1 = 0 inside the conductor, then D1 = ε E1 = ε 0 E1 + P1 = 0 ⇒ D1 = 0 and P1 = 0 inside conductor
∴ For conductor-dielectric interface:
Material #1 is conductor and material #2 dielectric medium, then:
D1 = E1 = P1 = 0 and D2 n = σ free and E2t = 0
Note that the potential V ( r ) interface physically must be continuous at an interface between two
materials, whether they are dielectrics or otherwise!
Also: From Gauss’ Law for E :
∫
S′
E ( r )idA′ =
enclosed
QTot
ε0
At a dielectric interface, as drawn on page 17 above, we see that:
σ
+σ
σ
The normal components
[ E2 n − E1n ] interface = Tot = bound free
ε0
ε0
of E are discontinuous
across a dielectric interface
by the amount σ Tot ε 0
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
19
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
BOUNDARY CONDITIONS ON THE ELECTRIC POLARIZATION Ρ AT AN INTERFACE
From Gauss’ Law for E :
∫
S′
E ( r )idA′ =
enclosed
QTot
Now: D ( r ) ≡ ε 0 E ( r ) + Ρ ( r ) so: E ( r ) =
∴
∫ E ( r )idA′ = ∫
or:
∫
S′
S′
ε0
1
ε0
=
enclosed
Q enclosed
+ QBound
free
D (r ) −
ε0
1
ε0
Ρ (r )
⎛1
⎞
1
1
enclosed
+ QBound
D ( r ) − Ρ ( r ) ⎟idA′ = ( Q enclosed
)
⎜
free
S′ ε
ε
ε
0
0
⎝ 0
⎠
enclosed
+ QBound
D ( r )idA′ − ∫ Ρ ( r )idA′ = Q enclosed
free
S′
But we already know that:
∫
S′
D ( r )idA′ = Q enclosed
and
free
∫ Ρ ( r )idA′ = −Q
S′
enclosed
Bound
Take a (shrunken) Gaussian pillbox centered on the interface as shown in figure below:
=Ρ1 n
So:
∫ Ρ ( r )idA′ = −Q
S′
enclosed
Bound
=Ρ 2 n
Get: [Ρ1 inˆ1 + Ρ 2 inˆ2 ]ΔS = −σ Bound ΔS But: nˆ2 = −nˆ1
= −σ Bound
Thus: ⎡⎣ Ρ 2 n ( r ) − Ρ1n ( r ) ⎤⎦
interface
The normal components of Ρ ( r ) are discontinuous
at an interface by the amount −σ bound
Since: D ( r ) = ε 0 E ( r ) + Ρ ( r ) we can also write this out for normal and tangential components as:
Dni ( r ) = ε 0 Eni ( r ) + Ρ ni ( r ) and Dti ( r ) = ε 0 Eti ( r ) + Ρ ti ( r )
Both of these component relations are valid on each side of interface, i.e. for the ith media, i = 1, 2.
⎧ D2 n − D1n = σ free and P2 n − P1n = −σ bound ⎫
⎪
⎪
Then: ⎨
1
1
⎬ at the interface of two dielectrics
⎪ E2 n − E1n = ε σ ToT = ε (σ free + σ bound ) ⎪
0
0
⎩
⎭
The tangential relations for fields at the interface are: D2t − D1t = P2t − P1t ⇐ Not necessarily = 0!
and: E2t − E1t = 0 ALWAYS (for electrostatics)!!!
20
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
RELATIONSHIPS BETWEEN ρ free ( r ) AND ρ Bound ( r )
FREE & BOUND VOLUME CHARGE DENSITIES
Since: D ( r ) = ε 0 E ( r ) + Ρ ( r ) then: Ρ ( r ) = D ( r ) − ε 0 E ( r )
However, for Class-A dielectrics: D ( r ) = ε E ( r ) or: E ( r ) =
Thus: Ρ ( r ) = D ( r ) − ε 0 E ( r ) = D ( r ) −
1
ε
D (r )
ε0
⎛ ε − ε0 ⎞
D (r ) = ⎜
⎟ D ( r ) but: ε = K eε 0
ε
⎝ ε ⎠
⎛ K −1 ⎞
∴ Ρ (r ) = ⎜ e ⎟ D (r )
⎝ Ke ⎠
⎛ K −1 ⎞
⎛
1 ⎞
Now: ∇iΡ ( r ) = ⎜ e ⎟ ∇i D ( r ) = ⎜1 −
⎟ ∇i D ( r )
⎝ Ke ⎠
⎝ Ke ⎠
But:
∇iΡ ( r ) = − ρ Bound ( r ) and ∇i D ( r ) = + ρ free ( r ) , also ε o∇i E ( r ) = ρTot ( r ) = ρ free ( r ) + ρ Bound ( r )
1⎞
⎛
∴ − ρ Bound ( r ) = ⎜ 1 − ⎟ ρ free ( r )
⎝ K⎠
⇐ NOTE: ρbound ( r ) is opposite charge sign to ρ free ( r ) !!!
⎛
1 ⎞
1
ρ free ( r )
Then: ρTot ( r ) = ρ free ( r ) + ρ Bound ( r ) = ρ free ( r ) − ⎜1 −
⎟ ρ free ( r ) = +
Ke
⎝ Ke ⎠
1
ρ free ( r )
Thus: ρTot ( r ) = +
Ke
The total volume charge density is reduced by the amount 1 K e inside a dielectric.
⇒ Dielectric material screens out charge!!!
NOTE: if K e → ∞ then ρ Tot ( r ) → 0 (perfect screening!!!)
K e → ∞ also implies χ e → ∞ (infinite electric susceptibility) because K e = 1 + χ e
K e → ∞ also implies ε → ∞ (infinite electric permittivity) because K e = ε ε o
Thus, we see (again) that ρ Bound ( r ) {partially} cancels out ρ free ( r )
Since −∇iΡ ( r ) = ρbound ( r ) , can only get ∇iΡ ( r ) ≠ 0 if ρ free ( r ) ≠ 0 !!!
IMPORTANT NOTE:
There is NO universal relationship between σ free & σ bound .
Sometimes, but not always, a relationship does exist between σ bound & σ free , but it is not universal
(i.e. valid for any/all situations).
It is NOT necessary to have σ free ≠ 0 in order to have a non-zero σ bound present on a dielectric.
Example: Bound surface charge density, σ bound on a bar electret (permanently polarized material).
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
21
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
We have previously discussed (above, p. 10 of these lecture notes) the example of free and bound
surface charge densities σ free and σ bound at a dielectric-conductor interface, e.g. with the parallelplate capacitor:
On LHS Plate: σ Bound = Ρ ( r )inˆ LHS
interface
where Ρ ( r ) = Ρ o xˆ , and nˆLHS = − xˆ
⎛ Coulombs ⎞
∴ σ Bound = −Ρ o ⎜
⎟ since xˆ inˆ LHS = −1
2
⎝ meter ⎠
Now: Ρ ( r ) = D ( r ) − ε 0 E ( r )
σ Bound = Ρ ( r )inˆLHS
interface
= D ( r )inˆ LHS
interface
− ε 0 E ( r )inˆLHS
interface
⎛ Ke − 1 ⎞
⎛
1 ⎞
⎟ D ( r ) interface = − ⎜1 −
⎟ D ( r ) interface but: D ( r ) interface = σ free (here)
⎝ Ke ⎠
⎝ Ke ⎠
σ Bound = −Ρ o = − ⎜
⎛
1 ⎞
∴ σ bound = − ⎜1 −
⎟ σ free (here) Note also that σ Bound has opposite charge sign to σ free !!!
⎝ Ke ⎠
Again we remind the reader that:
There is NO universal relationship between σ free & σ bound .
Sometimes, but not always, a relationship does exist between σ bound & σ free (as we just showed),
but it is not universal (i.e. valid for any/all situations).
22
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Griffiths Example 4.5
A conducting metal sphere of radius a carries a free charge Q and is surrounded by a Class-A
dielectric sphere of radius b > a as shown in the figure below:
b
a
Q •
Find the potential V ( r ) at the center of sphere.
From Gauss’ Law for D :
enclosed
∫ D ( r ′)idA′ = Q free gives: D ( r > a ) =
S′
Now: D ( r ) = ε E ( r ) so: E ( r ) =
⇒ for a < r < b : E ( a < r < b ) =
and for r < a :
1
ε
2
rˆ =
Q
4π
rˆ for r > a
Q free
1 Q free
rˆ and for r > b : E ( r > b ) =
rˆ
2
K e 4πε o r
4πε 0 r 2
E ( r < a ) = D ( r < a ) = Ρ ( r < a ) = 0!!!
The potential at the center of sphere is therefore:
0
b ⎛ Q free ⎞
a ⎛ Q free ⎞
V ( r = 0 ) = − ∫ E ( r )i d = − ∫ ⎜
dr
−
⎟
∫b ⎜⎝ 4πε r 2 ⎟⎠ dr −
∞
∞ 4πε r 2
0
⎝
⎠
V ( r = 0) =
4π r 2
D (r )
Q free
4πε r
Q free
0
∫ ( 0 ) dr
a
=
Q
4π
⎛ 1
1
1 ⎞
+
−
⎜
⎟
ε a εb ⎠
⎝ ε 0b
⎛ 1
1
1 ⎞
Q ⎛1
1 ⎛1
1 ⎞⎞
+
−
⎜
⎟=
⎜ +
⎜ − ⎟⎟
ε a ε b ⎠ 4πε 0 ⎝ b
Ke ⎝ a
b ⎠⎠
⎝ ε 0b
If E ( r ) is known, then Ρ ( r ) is also known, because Ρ ( r ) = ε 0 χ e E ( r )
Thus, for a ≤ r ≤ b : Ρ ( r ) = ε 0 χ e E ( r ) =
∴
ρ Bound ( r ) = −∇iΡ ( r ) = −
and: σ Bound = Ρ ( r )inˆ interface
1
4π
ε 0 χ eQ free
1
rˆ =
2
4πε r
4π
⎛ χ e ⎞ Q free
⎜
⎟ 2 rˆ (i.e. inside the dielectric)
⎝ 1 + χe ⎠ r
⎛ χ e ⎞ 1 ∂ ⎛ 2 Q free ⎞
⎟ = 0 !!!
⎜
⎟ 2 ⎜r
r2 ⎠
⎝ 1 + χ e ⎠ r ∂r ⎝
⎧ 1
⎪+
⎪ 4π
=⎨
⎪− 1
⎪ 4π
⎩
⎛ χ e ⎞ Q free
⎜
⎟ 2 at r = b
⎝ 1 + χe ⎠ b
⎛ χ e ⎞ Q free
⎜
⎟ 2 at r = a
⎝ 1 + χe ⎠ a
( n.b.
nˆr =b = + rˆ )
( n.b.
nˆr = a = − rˆ )
n.b. By convention, n̂ is the outward pointing unit vector from the surface(s) of the dielectric.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
23
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
When all space is filled with a Class-A dielectric material, the E -field inside the dielectric is
reduced by factor of 1 K e from its free-space value.
For example: A point (free) electric charge q is embedded at the center of a solid Class-A dielectric
sphere of radius R as shown in the figure below:
The E -field inside the dielectric sphere, due to the point free charge at the center of the sphere is
(using Gauss’ Law for D and then using the relation E ( r ) = D ( r ) ε ):
E (r < R) =
⎛ 1 ⎞ 1 q
q
rˆ = ⎜
rˆ
⎟
2
2
4πε r
⎝ K e ⎠ 4πε o r
1
However!! Note that for r > R : E ( r > R ) =
1
q
rˆ
4πε o r 2
Outside the dielectric ( r > R ) the E -field is the same as if the dielectric sphere wasn’t there at all!
This is a consequence of Gauss’ Law for E :
enclosed
QTot
enclosed
enclosed
′
=
where QTot
= QBound
+ Q enclosed
E
r
dA
i
free
∫ ( )
S′
ε0
i.e. we get E -field contributions from all enclosed charges:
1) +q at the center of sphere
2) −σ Bound at the inner cavity surface, radius δ << R
3) +σ bound at r = R
Note that 2) and 3) cancel each other for r > R!!!
(They don’t cancel for r < R!! obviously)
Can you show that QBound ( r = δ ) = −q and QBound ( r = R ) = + q ??
24
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
THE ELECTRIC SUSCEPTIBILITY χ e OF NON-CLASS A DIELECTRICS
A crystalline dielectric material (e.g. salt, diamond, etc.) has preferred internal axes so that the
electric polarization, Ρ ( r ) is different along the different internal axes in such materials.
For crystalline materials, Ρ ( r ) is related to E ( r ) by the tensor relation: Ρ ( r ) = ε 0 χ e E ( r )
⎛ χe
⎜ xx
Where χ e = Susceptibility Tensor: χ e ≡ ⎜ χ eyx
⎜
⎜ χe
⎝ zx
(
= ε (χ
= ε (χ
)
E )
E )
Ρ x = ε 0 χ exy Ex + χ exy E y + χ exz Ez
i.e. Ρ y
Ρz
0
0
Ex + χ eyy E y + χ eyz
e yx
ezx
Ex + χ ezy E y + χ ezz
z
z
χe
xy
χe ⎞
yy
χe ⎟
χe
χe
zy
xz
yz
⎟
⎟
χ e ⎟⎠
zz
⎛ χe
⎛ Ρx ⎞
⎜ xx
⎜ ⎟
or: ⎜ Ρ y ⎟ = ε 0 ⎜ χ eyx
⎜
⎜ ⎟
⎜χ
⎝ Ρz ⎠
⎝ ezx
χe
xy
χe
yy
χe
zy
χ e ⎞ ⎛ Ex ⎞
⎟⎜ ⎟
χe ⎟ ⎜ Ey ⎟
xz
yz
⎟
χ e ⎠⎟ ⎝⎜ Ez ⎠⎟
zz
3
or: Ρi ( r ) = ε 0 ∑ χ eij E j ( r ) and noting that χ eij = χ e ji ⇒ χ e has only six independent components.
j =1
Again, if we carefully choose the coordinate axes to coincide with the internal symmetry axes of
the crystalline dielectric material, then in reality there are only 3 independent components of χ e
- the off-diagonal elements vanish; only the diagonal elements χ exx , χ eyy and χ ezz are non-zero.
For crystalline dielectric materials:
∇i D ( r ) = ρ free ( r ) is valid, D ( r ) = ε 0 E ( r ) + Ρ ( r ) is also valid.
(
)
∴ ∇i D ( r ) = ∇i ε 0 E ( r ) + Ρ ( r ) = ε 0∇i E ( r ) + ∇iΡ ( r ) = ρ free = ρTot − ρ Bound is valid
However, in a crystalline dielectric material, generally speaking D ( r ) , E ( r ) & Ρ ( r ) are NOT all
pointing in the same direction!!!
i.e. a tensor relation also exists between D ( r ) & E ( r ) : D ( r ) = ε E ( r ) = ε o K e E ( r ) where
(
)
K e ≡ ε ε o = 1 − χ e for a linear, anisotropic dielectric material and where ε = electric permittivity
tensor, thus K e = relative electric permittivity tensor (a.k.a. dielectric “constant” tensor).
⎛ Dx ⎞ ⎛ ε xx
⎜ ⎟ ⎜
⎜ Dy ⎟ = ⎜ ε yx
⎜ D ⎟ ⎜ ε zx
⎝ z⎠ ⎝
⎛ Ke
ε xy ε xz ⎞
⎜
⎟
ε yy ε yz ⎟ = ε 0 ⎜ K e
⎜
ε yz ε zz ⎠⎟
⎜ Ke
⎝
xx
K exy
yx
K eyy
zx
K ezy
K exz ⎞ ⎛ Ex ⎞
⎟⎜ ⎟
K eyz ⎟ ⎜ E y ⎟
⎟
K ezz ⎟ ⎜⎝ Ez ⎟⎠
⎠
3
3
j =1
j =1
Di ( r ) = ∑ ε ij E j ( r ) = ε 0 ∑ K eij E j ( r )
ε ij = ε ji and K ij = K ji , etc.
Again, if choose the symmetry axes of the crystal for coordinate axes, then the off-diagonal
elements of ε and K e vanish.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
25
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
For extremely high externally applied Eext -fields, the electric polarization Ρ ( r ) becomes
increasingly non-linearly related to Eext :
3
Ρi ( r ) = ∑ aij E j ( r ) +
j =1
linear
response
Ρi
3
∑ bijk E j ( r ) Ek ( r ) +
j , k =1
quadratic
response
Linear
Regime
j =i
3
∑
j , k ,l =1
cijkl E j ( r ) Ek ( r ) El ( r ) + …
cubic
response
Non-Linear
Regime
j≠i
Ej
26
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Example: Gauss’ Law for Parallel Plate Capacitor with “Class-A” Dielectric Between Plates:
Take Gaussian pillbox centered on LHS conducting plate:
ˆ ′ = Q free = ∫ σ free dS ′ = σ free ΔS
∫ D ( r )indS
S′
S′
Because of the parallel-plate geometry: E , D, Ρ are constant fields between the plates of the
capacitor (we neglect the fringe-field/edge region(s) of the parallel plate capacitor, since
d
A ( = × w) )
∴ D1 inˆ1ΔS + D2 inˆ2 ΔS = +σ free ΔS or: D1 inˆ1 + D2 inˆ2 = σ free
But: nˆ2 = −nˆ1 , thus: D1 inˆ1 − D2 inˆ1 = σ free or: D1n − D2 n = σ free
⇒ Normal components of D discontinuous across dielectric interface, by amount σ free .
Now D1 = ε 0 E1 = 0 and P1 = 0 since LHS of Gaussian pillbox ends inside the LHS plate of ||-plate
capacitor, also the conducting metal is not a dielectric, so it has no electric polarization.
∴
D2 n = σ free or D2 n = D = σ free
( D2 n ⊥
to surface of plates )
But: D = ε E for Class-A dielectric. ∴ D = ε E = σ free or: E = σ free ε = σ free K eε o
And: E = ΔV d for ||-plate capacitor, σ free = Q free A and: Q free = C ΔV
σ free
Q
= free
K eε 0 K eε 0 A
K eε 0 A
ε0 A
Thus: E = ΔV d =
So: CDiel =
∴
Q
=
ΔV
d
= Ke
d
But now recall that for no dielectric, C0 =
ε0 A
d
, K0 = 1
Cdiel
= Ke
C0
⇒ Capacitance of parallel-plate capacitor with dielectric increased by factor of K e over vacuum.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
27
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
EXAMPLE: Gauss’ Law for “Class-A” Dielectric Sphere with Point Charge Q at its Center:
Use Gauss’ Law to obtain D -field inside the dielectric and thus obtain E , Ρ inside the dielectric:
∫
S′
ˆ ′ = Q enclosed
D ( r )indS
= Q and (from above): nˆout
but: Q enclosed
free
free
r=R
= + rˆ, and: nˆin
r =δ
= − rˆ.
Note that D ( r ) is radial (i.e. no θ , ϕ dependence) due to rotational symmetry/invariance of problem.
⇒
Dinside ( r ) =
Q
rˆ for δ ≤ r ≤ R .
4π r 2
For “Class-A” dielectrics
But: Dinside ( r ) = ε 0 Einside ( r ) + Ρinside ( r ) = ε Einside ( r ) = K eε 0 Einside ( r ) with: ε = K eε 0
⇒ Einside ( r ) =
Q
4πε r
2
rˆ =
Q
1
1
rˆ = Dinside ( r ) =
Dinside ( r )
2
K eε 0
ε
4π K eε o r
Ρinside ( r ) = Dinside ( r ) − ε 0 Einside ( r ) = Dinside ( r ) −
∴
ε0
D
(r )
ε inside
⎛ K −1 ⎞
⎛ K −1 ⎞ Q
⎛ ε − ε0 ⎞
=⎜
Dinside ( r ) = ⎜ e ⎟ Dinside ( r ) = ⎜ e ⎟
rˆ
⎟
2
⎝ ε ⎠
⎝ Ke ⎠
⎝ K e ⎠ 4π r
⎛ K −1 ⎞ Q
rˆ
⇒ Ρinside ( r ) = ⎜ e ⎟
2
⎝ K e ⎠ 4π r
At the outer surface of the dielectric sphere the bound surface charge density is:
28
σ Bound
r =R
σ Bound
r =R
≡ Ρinside ( r )inˆout
r =R
= Ρinside ( r )irˆ
r =R
⎛ K −1 ⎞ Q
=⎜ e ⎟
and thus: QBound
2
⎝ K e ⎠ 4π R
⎛ K −1 ⎞ Q
= Ρinside ( r ) r = R = ⎜ e ⎟
2
⎝ K e ⎠ 4π R
r =R
= σ Bound
r=R
⎛ K −1 ⎞
4π R 2 = ⎜ e ⎟ Q on the outer surface.
⎝ Ke ⎠
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
At the inner surface of the dielectric sphere the bound surface charge density is:
σ Bound
σ Bound
r =R
r =δ
≡ Ρinside ( r )inˆin
r =δ
= Ρinside ( r )i( − rˆ )
r =δ
⎛ K −1 ⎞ Q
= −⎜ e ⎟
and thus: QBound
2
⎝ K e ⎠ 4πδ
Thus, we explicitly see that: QBound
r =δ
⎛ K −1 ⎞ Q
= − Ρinside ( r ) r =δ = − ⎜ e ⎟
2
⎝ K e ⎠ 4πδ
r =δ
= − QBound
= σ Bound
r =R
r =δ
⎛ K −1 ⎞
4πδ 2 = − ⎜ e ⎟ Q on the inner surface.
⎝ Ke ⎠
⎛ K −1 ⎞
⎛ K −1 ⎞
= − ⎜ e ⎟ Q = − ⎜ e ⎟ Q enclosed
free
⎝ Ke ⎠
⎝ Ke ⎠
⎛ K −1 ⎞ Q
rˆ
Now from above, we found that: Ρinside ( r ) = ⎜ e ⎟
2
⎝ K e ⎠ 4π r
Then inside the dielectric sphere: ρ Bound ( r ) = −∇iΡ inside ( r ) for δ < r < R .
Now ∇ in spherical coordinates: ∇ =
1 ∂ 2
r rˆ + …θˆ + …ϕˆ
2
r ∂r
these terms not important
here, since Ρ=Ρrˆ only
Then: ρ Bound ( r ) = −
1 ∂ 2
1 ∂ ⎡ 2 ⎛ Ke −1 ⎞ Q ⎤
1 ∂ ⎡⎛ K e − 1 ⎞ Q ⎤
r Ρinside ( r ) = − 2
⎢r ⎜
⎥=− 2
⎢⎜
⎟
⎟ ⎥=0
2
2
r ∂r ⎣⎝ K e ⎠ 4π ⎦
r ∂r
r ∂r ⎣ ⎝ K e ⎠ 4π r ⎦
=constant, ≠ fcn of r
⇒ ρ Bound ( r ) = 0 for δ < r < R . ⇐ Also is true since ρ free ( r ) = 0 here in this problem, for δ < r < R .
Using Gauss’ Law for E we also see that the total charge as seen by an observer inside the dielectric
enclosed
enclosed
ˆ ′ = QTot
= Q enclosed
+ QBound
.
(i.e. for δ < r < R ) is: ∫ ′ E ( r )indS
free
S
⎛ Ke − 1 ⎞
2
=
−
πδ
4
⎜
⎟Q
r =δ
r =δ
K
e
⎝
⎠
⎛ K −1 ⎞
⎛ K − Ke + 1 ⎞
1
= Q − ⎜ e ⎟Q = ⎜ e
Q for δ < r < R .
⎟Q =
Ke
Ke
⎝ Ke ⎠
⎝
⎠
enclosed
= Q for δ < r < R and QBound
= QBound
Since Q enclosed
free
enclosed
enclosed
We see that: QTot
= Q enclosed
+ QBound
free
= σ Bound
⇒ For δ < r < R the total/net charge “seen” by the E -field (e.g. using a test charge QT in the region
δ < r < R ) is reduced by a factor of 1 K e , e.g. compared to the E -field associated with a “bare”
point charge, Q located at the origin. In the region δ < r < R , the bound surface charge density
σ Bound r =δ located at the inner radius r = δ of the dielectric thus “screens” the bare charge Q located at
the origin, reducing its charge strength!
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
29
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Outside the dielectric sphere (i.e. for r > R ) again using Gauss’ Law for E we also see that the total
charge as seen by an observer outside the dielectric is:
enclosed
enclosed
Q enclosed
+ QBound
QTot
free
ˆ ′=
=
.
∫ ′ E ( r )indS
εo
S
εo
⎛ K −1 ⎞
⎛ K −1 ⎞
= −⎜ e ⎟Q + ⎜ e ⎟Q = 0
⎝ Ke ⎠
⎝ Ke ⎠
= Q + 0 = Q for r > R !!!
enclosed
= Q for r > R and QBound
= QBound
Since Q enclosed
free
enclosed
enclosed
= Q enclosed
+ QBound
Thus we see that: QTot
free
r =δ
+ QBound
r =R
⇒ For r > R the total/net charge “seen” by the E -field (e.g. using a test charge QT in the region
r > R ) is not “screened” by the dielectric – the E -field outside the dielectric is the same as the E field associated with a “bare” point charge, Q located at the origin! The bound surface charge density
σ Bound r = R (located at the outer radius r = R of the dielectric) precisely cancels the effect(s) associated
with σ Bound
r =δ
(located at the inner radius r = δ of the dielectric)!!!
Outside the dielectric sphere (i.e. for r > R):
Gauss’ Law for D :
ˆ =Q
∫ D ( r )indS
Gauss’ Law for E :
ˆ =Q
∫ E ( r )indS
S
S
enclosed
Tot
Q 1
rˆ
for r > R.
4π r 2
Q 1
rˆ for r > R.
⇒ Eoutside ( r ) =
4πε o r 2
⇒ Doutside ( r ) =
enclosed
free
εo
Then: Doutside ( r ) = ε 0 Eoutside ( r ) for r > R. Obviously, Ρ outside ( r ) = 0 for r > R !!!
30
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
THE BAR ELECTRET: The Electrostatic Analog of A Bar Magnet
An electret is a polar dielectric which has permanent polarization, Ρ . Electrets can be made
e.g. by heating a polar dielectric material (i.e. a dielectric material which has permanent molecular
dipole moments), heating it in the presence of a (very) strong uniform external electric field. The
electret is then cooled e.g. to ambient/room temperature in the presence of the external E -field.
It is then removed from the external E -field, still retaining a net, permanent polarization!
Eext
+
+
−
+
−
+
−
+
−
+
−
+ −σ b
−
+ −
+ −
+ −
+ −
+ −
+σ b −
−σ f
P
+σ f
Heat
Source
Heat polar dielectric material, then cool to room temperature.
Afterwards:
Electret retains uniform, permanent electric polarization:
Ρ = Ρ o zˆ = p Volume = Electric dipole moment per unit volume.
Ρ = Ρ o zˆ
−σ b
ẑ
+σ b
Note that since ρ Bound ( r ) = −∇iΡ ( r ) = 0 ⇒ ρ free = 0 too!!
∴ ∃ no free charge, σ free on surface (or ρ free within volume) of electret.
Outside the electret: Dout ( r ) = ε 0 Eout ( r )
Inside the electret: Din ( r ) = ε 0 Ein ( r ) + Pin
( Ρ ( r ) ≡ 0)
( r ) ( Ρ ( r ) = Ρ zˆ )
out
in
0
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
31
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Boundary Conditions on the Surfaces of the Electret:
No free charge anywhere on/within electret.
If Ρin ( r ) = Ρ o zˆ , there is no bound surface charge density on the cylindrical portion of the electret.
Ρin ( r ) = Ρ o zˆ
At Endcaps of Electret:
( nˆ = ± zˆ )
Take Gaussian Pillbox on, e.g. LHS end cap:
∫
S′
s′
ẑ
−σ b
+σ b
⊥
⊥
ˆ = Q encl
D ( r )indS
free = 0 ⇒ Dout − Din = 0
⊥
= Din⊥ Normal component of D ( ⊥ to endcaps ) is continuous across endcaps.
⇒ Dout
Stoke’s Law:
Take line integral on LHS Endcap:
∫ E ( r )id = 0 ⇒ Eout = Ein Tangential components of E continuous across endcaps.
C
Also, Gauss’ Law for Ρ :
ˆ = −Q
∫ Ρ ( r )indS
enclosed
bound
S′
⊥
⇒ Ρ out
− Ρin⊥ = −σ b or: Ρin⊥ = σ b
On Cylindrical Surface of Electret:
Gauss’ Law for D :
Stoke’s Law:
∫ E ( r )i d
C
Gauss’ Law for Ρ :
( nˆ = ρˆ radial direction )
⊥
⇒ Dout
= Din⊥
= 0 ⇒ Eout = Ein
⇒ Ρin⊥ = 0 since uniform polarization Ρin ( r ) = Ρ 0 zˆ
Outside Electret: Ρ out ( r ) = 0 , Dout ( r ) = ε o Eout ( r )
Very Important Note:
For the bar electret (or anything else with permanent electric polarization), cannot get conditions on
E )⊥ E (in or out) from D ⊥ , D except via explicit use of D ( r ) = ε 0 E ( r ) + Ρ ( r ) .
The reason for this is that the relations Din ( r ) = ε Ein ( r ) and Ρ ( r ) = ε 0 χ e E ( r ) are not valid here for
permanently polarized / electret materials!!!
The relations D = ε E and Ρ = ε 0 χ e E are valid only for “Class-A”/linear dielectrics!!!
32
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Lines of E for the Bar Electret: (n.b. E terminates on (any) charges (free or bound)!!!)
Important note:
Lines of E terminate on any charges – free or bound.
Lines of D terminate only on free charges (n.b. none here in the electret!!!).
Lines of P terminate only on bound charges.
In situations where the electret is a thin polarized sheet:
Din ( r ) = ε 0 Ein ( r ) + Ρ in ( r ) = 0
i.e. Din ( r ) = 0 ⇒ Ein ( r ) = − Ρin ( r ) ε 0
+σ b
−σ b
Ein
Ρin
Inside e.g. a long bar electret: Ρin ( r ) ≥ ε 0 Ein ( r )
Din points in direction of Ρin but Ein points in opposite direction of Ρin
Inside e.g. a thin polarized sheet: Ρin ( r ) = ε 0 Ein ( r )
Din = 0 and Ein points in opposite direction of Ρin
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
33
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
More Pix of the Bar Electret: Note direction of electric polarization here is opposite to above!
34
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
VACUUM POLARIZATION, CHARGE RENORMALIZATION, MASS RENORMALIZATION
Suppose we use the previous example of the spherical dielectric with a (free) at its center in an attempt
to understand what happens to the (physical) vacuum at very small distances from fundamental
(i.e. point-like) electrically charged particles, such as the electron.
The physical vacuum is by no means “empty” – it is actually a (strange) form of dielectric medium
(because microscopically, fundamentally it is entirely quantum mechanical in nature – seething with
(very briefly appearing & disappearing) virtual particle-antiparticle pairs of all possible kinds/types).
Nevertheless, the vacuum has two macroscopic (i.e. microscopically-averaged over) parameters
associated with it:
– The electric permittivity of free space (the vacuum): ε o = 8.85 ×10−12 Farads/meter
– The magnetic permeability of free space (the vacuum): μo = 4π × 10−7 Henries/meter
These two macroscopic properties of the vacuum are not independent of one another, because they are
linked via a third macroscopic parameter associated with the vacuum, namely the “speed of light”,
c = 3 ×108 m/s (which is a misnomer, since c is the (maximum) speed for which any fundamental force
(E&M, strong, weak, gravity) can propagate!)
1
1
or: c =
These three quantities are related to each other by: c 2 =
ε o μo
ε o μo
“Empty” space also has a fourth macroscopic property associated with it – again not independent:
– The impedance of free space (the vacuum):
o
=
μo
εo
376.8 Ω
For E&M, at the microscopic/quantum level only electrically charged particle-antiparticle pairs
contribute to the macroscopic parameters ε o and μo – e.g. all of the charged fermion-antifermion pairs
(in the context of the Standard Model of Electroweak Interactions – these would be the 3 generations
of charged leptons: e ± , μ ± , τ ± and the 3 generations of charged up and down quarks (u, d, s, c, b, t))
and also the charged W ± bosons – the electrically charged carrier/mediator of the weak force.
Now the classical E&M, macroscopic E -field for a point charged particle is E ( r ) =
The corresponding potential is: V ( r ) =
Near r
1 −e
rˆ
4πε 0 r 2
1 −e
since E ( r ) = −∇V ( r )
4πε 0 r
0 , or r ≤ re = classical electron radius = 2.8 ×10−13 cm = 2.8 fm the electric field of the
electron becomes extremely high, E ( r = re ) ~ 1.84 × 1020 Volts/m !!! Note that this is a significantly
larger field strength than those e.g. typical of atomic-scale fields, E ( r = 1Å ) ~ 1011 Volts/m.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
35
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
If we had a true macroscopic spherical dielectric, with an infinitesimally small spherical cavity of
radial size δ re = 2.8 × 10−13 cm with an electron at the center of this small spherical cavity, it might
look something like that shown in the figure below:
By Gauss’ Law the effective electric charge is reduced from that of the bare charge (e) by an amount
Qeff = Qbare K e Where K e = dielectric constant (of the physical vacuum, here).
In the region where the electric field of the charged particle is extremely high, the local field energy
density there is sufficient e.g. to produce (virtual) e + e − pairs in this region of space. These virtual
e+ e− pairs “live” only for an extremely short period of time – as allowed by the Heisenberg uncertainty
principle ( ΔE Δt ≤ ) . Because opposite (like) charges attract (repel), the e+ (e−) from the e + e − pair
that is “pulled” out of the vacuum, tends to be (on average) closer to (further from) the “bare” e−,
respectively. Thus “vacuum polarization” results. The net effect, to an observer (who also lives in the
same physical “dielectric” medium – the vacuum!!) is that the observed electric charge is reduced
from the “bare” charge of the electron!!! The observed/physical value of e is in fact the one we know
and love! eobs = 1.6021892 + 0.0000046 × 10−19 Coulombs.
Note the following “amusing” thought: suppose we were able to get “outside” our universe (i.e.
outside of the dielectric of our physical vacuum (“empty” space)!!). Then Gauss’ Law for outside
says that we should “see” the full bare charge, ebare and not the reduced charge, eobserved!!!
However, since we cannot get “outside” of the dielectric medium we live in we can never hope to
directly observe the pure/bare charge, ebare, except perhaps via extremely high energy collisions
between charged particles.
36
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 10
Prof. Steven Errede
Using the theoretical formalism of Quantum Electrodynamics (QED), one finds (to first order in
⎛
⎞
e2
1
the so-called fine structure “constant”, α ) ⎜ α =
=
⎟ that:
c
πε
4
137.036...
o
⎝
⎠
n.b. formally divergent! (but only logarithmically…)
⎛
⎛ Λ ⎞⎞
α
2
2
ebare
log ⎜
⎜⎜ 1 −
2 ⎟⎟
⎟ Λ = cutoff parameter, e.g. 10me c
π
3
m
c
⎝ e ⎠⎠
⎝
2
2
For: Λ 10me c : eobserved = 0.9992e 2bare
2
eobserved
= K e = dielectric const.
Or: e observed = 0.9996 ebare
⇒ ebare =
1.00039
eobserved small effect!!!
Another effect of vacuum polarization in the region of space immediately surrounding a bare
electron is that the apparent mass of the electron is increased, due to the observer mis-interpreting the
“cloud” of e+ e− , etc. pairs surrounding the (bare) electron as being part of the electron itself.
Experiments on a free electron measure mobs, e.g. the mass used in the Lorentz Force Law,
Fe− = −eEext − eve × Bext = me ae . However, the observed mass, mobs is the sum of the bare mass, mbare,
plus the inertia of the electron’s “self-field”. Even from classical electrodynamics, we obtain the
following mass relation:
⎛
Λ ⎞
4
where Λ = “cutoff parameter”
mobs ≈ mbare ⎜1 + α
2 ⎟
3
m
c
π
e
⎝
⎠
Again, this relation is formally divergent, but if Λ 10me c 2 ( me c 2 = 0.511 MeV )
then mobs 1.029 mbare Or: mbare 0.97mobserved
Thus, we see that (to first order in α ) that mass renormalization is a larger effect than charge
renormalization is.
©Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005 - 2008. All rights reserved.
37