Field of a polarized object: Suppose we have a piece of polarized materiali.e. an object containing lots of microscopic dipoles. To find the field produced by the polarized object kept in vacuum, we work with the polarizationP~ which is dipole moment per unit volume,
~
p = P dτ~ 0 Using the potential for a single dipole, we write,
Vd = 1 4π0
~ p·ˆr
r2 ⇒ VP = 1 4π0
Z
V
P~(r0)·rˆ r2 dτ0
We then use the relation ∇0(1/r) = ˆr/r2 (and not −ˆr/r2, why? We are taking derivative w.r.t source coordinate r0 and ~r=~rfield−~r0.) Hence,
VP = 1 4π0
Z
V0
P~ · ∇0 1
r
dτ0
Using the identity ∇ ·(f ~A) = f∇ ·A~+∇f ·A, gives,~ VP = 1
4π0
"
Z
V0
∇0· P~ r
! dτ0−
Z
V0
∇0·P~ r dτ0
#
Finally, using the divergence theorem we obtain, VP = 1
4π0
"
Z
S0
P~ ·d~s0 r +
Z
V0
−(∇0·P~) r dτ0
#
(17) VP ≡ 1
4π0 Z
S0
σb
r ds0+ 1 4π0
Z
V0
ρb
r dτ0 (18)
The boundsurface and volume charge density is defined as,
σb = P~ ·n,ˆ ρb = −∇0·P~ (19) Total polarization i.e. bound charge is
Qb= Z
S0
σbds0+ Z
V0
ρbdτ0 = 0. (20)
For interpretation of bound charges, please see Griffiths section 4.2.2 and Reitz page 79-80.
The electric field E~P is obviously−∇VP, derivative on field (unprimed) coordinate.
Example 21. A sphere of radius R carries a polarization P~ =k~r, where kis a constant and ~r is the vector from the center. Calculate σb andρb.
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Gauss’s law in presence of dielectrics
We consider presence of two kind of charges in the medium, (i) bound, that are results of polarization, and (ii) free, that are not results of polarization but got there somehow.
Within dielectric, the total charge density is,
ρ = ρb+ρf (21)
The electric field E~ is total field, not just that due to the polarization, and from Gauss’s law,
∇ ·E~ = ρ 0
⇒ 0∇ ·E~ = ρb+ρf = −∇ ·P~ +ρf
We can combine the two divergence terms as: ∇ ·(0E~ +P) =~ ρf. Defining electric displacementas
D~ ≡ 0E~ +P~ (22)
the Gauss’s law in dielectric medium reads,
∇ ·D~ = ρf =⇒ Z
S
D~ ·d~s = Qenclf (23)
E~ versus D: The Gauss’s law in vacuum and in dielectric look almost identical except for~ the use of electric displacement in place of electric field and free charge instead of total charge.
∇ ·E~ = ρ 0
and ∇ ·D~ =ρf But still they are not the same thing.
? There is no Coulomb’s law for D~ i.e.
D~ 6= 1 4π
Z ρf r2rdτ.ˆ
Because∇ ·E~ alone does not describe electric field completely, it also requires∇ ×E.~
? The curl ofD~ is not necessarily zero,
∇ ×D~ = ∇ ×(0E~ +P~) = ∇ ×P~ 6= 0
? No potential can be defined for D:~ D~ 6=−∇V
Electrostatic energy in medium: The expression for field energy is, W = 0
2 Z
E2dτ −→ W = 1 2
Z
E~ ·Ddτ~ (24) For derivation see Griffith’s section 4.4.3.
Boundary condition: The electrostatic boundary conditions give us discontinuity ofD~ across charged surface as,
E~above−E~below = σ 0
ˆ
n −→ D~above−D~below = σfˆn. (25)
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Linear dielectrics: Provided electric field E~ is not too strong, polarization P~ is linearly proportional to total field 0E, for homogeneous and non-homogeneous linear dielectric,~
P~ = χe(0E),~ χe= susceptibility (26) P~ = [χ](0E),~ [χ] = susceptibility tensor (27) TheE~ =E~0+E~p, whereE~0 is the external field that cause polarization andE~p is field due to polarization. In linear medium,
D~ = 0E~ +P~ = 0(1 +χe)E~ (28) So, electric displacement D~ is also proportional to E~ → D~ = ~E. The new constant is permitivityof the material andK isdielectric constant,
≡0(1 +χe) ⇒ K =
0 = 1 +χe Besides, we can show
ρb =−∇ ·P~ =−∇ · 0χeD~
!
=− χe
1 +χeρf
implying volume bound charge density in a homogeneous linear dielectric is proportional to the density of free charge.
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