11/4/2004
Polarization Charge Distributions.doc
1/6
Polarization Charge
Distributions
Consider a hunk of dielectric material with volume V.
Say this dielectric material is immersed in an electric field
E ( r ) , therefore creating atomic dipoles with density P ( r ) .
Q: What electric potential field V ( r ) is created by
these diploes?
A: We know that:
V (r ) =
∫∫∫
V
P ( r′ ) ⋅ ( r-r′ )
4πε 0 r-r′
3
dv ′
But, it can be shown that (p. 135):
V (r ) =
=
∫∫∫
V
P ( r′ ) ⋅ ( r-r′ )
4πε 0 r-r′
1
4πε 0
∫∫∫
V
3
dv ′
−∇ ⋅ P ( r′ )
1
dv ′ +
r-r′
4πε 0
P ( r′ ) ⋅ ˆan ( r )
w r-r′ ds ′
∫∫
S
where S is the closed surface that surrounds volume V, and
ˆan ( r ) is the unit vector normal to surface S (pointing outward).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
Polarization Charge Distributions.doc
2/6
This complicated result is only important when we compare it to
the electric potential created by volume charge density ρv ( r )
and surface charge density ρs ( r ) :
V (r ) =
V (r ) =
1
4πε 0
1
4πε 0
∫∫∫
V
∫∫S
ρv ( r′ )
r-r′
ρs ( r′ )
r-r′
dv ′
ds ′
If both volume and surface charge are present, the total
electric potential field is:
V (r ) =
1
4πε 0
∫∫∫
V
ρv ( r′ )
r-r′
dv ′ +
1
4πε 0
∫∫S
ρs ( r′ )
r-r′
ds ′
Compare this expression to the previous integral involving the
polarization vector P ( r ) . It is evident that the two expressions
are equal if the following relations are true:
ρvp ( r ) = −∇ ⋅ P ( r )
and
ρsp ( r ) = P ( r ) ⋅ ˆan
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
Polarization Charge Distributions.doc
3/6
The subscript p (e.g., ρvp , ρsp ) indicates that these functions
represent equivalent charge densities due to the due to the
dipoles created in the dielectric.
In other words, the electric potential field V ( r ) (and thus
electric field E ( r ) ) created by the dipoles in the dielectric (i.e.,
P ( r ) ) is indistinguishable from the electric potential field
created by the equivalent charge densities ρvp ( r ) and ρsp ( r ) !
For example, consider a dielectric material immersed in an
electric field, such that its polarization vector P ( r ) is:
P ( r ) = 3 ˆaz
⎡C ⎤
⎢⎣ m2 ⎥⎦
E (r )
ˆaz
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
Polarization Charge Distributions.doc
4/6
Note since the polarization vector is a constant, the equivalent
volume charge density is zero:
ρvp ( r ) = −∇ ⋅ P ( r )
= −∇ ⋅ 3 ˆaz
=0
On the top surface of the dielectric (ˆan = ˆaz ), the equivalent
surface charge is:
ρsp ( r ) = P ( r ) ⋅ ˆan
= 3 ˆaz ⋅ ˆaz
=3
⎡C
⎤
⎢⎣ m 2 ⎥⎦
On the bottom of the dielectric (ˆan = −ˆaz ), the equivalent
surface charge is:
ρsp ( r ) = P ( r ) ⋅ ˆan
= −3 ˆaz ⋅ ˆaz
= −3
⎡C
⎤
⎢⎣ m 2 ⎥⎦
On the sides of the dielectric material, the surface charge is
zero, since ˆaz ⋅ ˆan = 0 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
Polarization Charge Distributions.doc
5/6
This result actually makes physical sense! Note at the top of
dielectric, there is a layer of positive charge, and at the
bottom, there is a layer of negative charge.
ρsp = 3 C/m2
ˆaz
ρvp = 0
ρsp = −3 C/m2
In the middle of the dielectric, there are positive charge
layers on top of negative charge layers. The two add together
and cancel each other, so that equivalent volume charge density
is zero.
Finally, recall that there is no perfect dielectric, all materials
will have some non-zero conductivity σ ( r ) .
As a result, we find that the total charge density within some
material is the sum of the polarization charge density and the
free charge (i.e., conducting charge) density:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
11/4/2004
Polarization Charge Distributions.doc
6/6
ρvT ( r ) = ρv ( r ) + ρvp ( r )
Where:
ρvT ( r ) total charge density
ρv ( r ) free charge density
ρvp ( r ) polarization charge density
This is likewise (as well as more frequently!) true for surface
charge density:
ρsT ( r ) = ρs ( r ) + ρsp ( r )
Jim Stiles
The Univ. of Kansas
Dept. of EECS