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Transcript
11/4/2004
The Polarization Vector.doc
1/3
The Polarization Vector
Recall that in dielectric materials (i.e., insulators), the charges
are bound.
-
p
+
E (r )
As a result, atoms/molecules form electric dipoles when an
electric field is present!
Note that even for some small volume ∆v , there are many
atoms/molecules present; therefore there will be many electric
dipoles.
E (r )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
The Polarization Vector.doc
2/3
We will therefore define an average dipole moment, per unit
volume, called the Polarization Vector P ( r ) .
P (r ) ∑ pn
∆v
⎡ dipole moment C ⎤
⎢ unit volume = m 2 ⎥
⎣
⎦
where pn is one of N dipole moments in volume ∆v , centered at
position r . Note the polarization vector is a vector field. As a
result, the direction and magnitude of the Polarization vector
can change as function of position (i.e., a function of r ).
Q: How are vector fields P ( r
) and E ( r ) related??
A: Recall that the direction of each dipole moment is the same
as the polarizing electric field. Thus P ( r ) and E ( r ) have the
same direction. There magnitudes are related by a unitless
scalar value χe ( r ) , called electric susceptibility:
P ( r ) = ε 0 χe (r ) E (r
)
Electric susceptibility is a material parameter indicating the
“stretchability” of the dipoles.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
The Polarization Vector.doc
3/3
Q: Can we determine the fields created by a polarized
material?
A: Recall the electric potential field created by one dipole is:
V (r ) =
p ⋅ ( r-r′ )
4πε 0 r-r′
3
Therefore, using d p = P ( r ) dv , the electric potential field
created by a distribution of dipoles (i.e., P ( r ) ) across some
volume V is (see fig. 5.9):
V (r ) =
∫∫∫
V
P ( r′ ) ⋅ ( r-r′ )
4πε 0 r-r′
3
dv ′
Q: But I thought scalar
charge distributions ρv ( r )
and ρs ( r ) created the
electric potential field
V ( r ) . Now you are saying
that electric fields are
created by the vector field
P ( r ) !?!
A:
As we will soon see, the polarization vector P ( r )
creates equivalent charge distributions—we will get the
correct answer for V ( r ) from either source!
Jim Stiles
The Univ. of Kansas
Dept. of EECS