Download Dielectric sphere in an external field

Dielectric sphere in an external field
A sphere of radius a is made of a dielectric material with permittivity ǫr . The sphere
is placed in an uniform, external electric field
E0 .
E0
ε
Ed
a
a
ε
a) Find the electric field in the whole space.
Now consider a spherical cavity of radius a inside an infinite dielectric with permittivity ǫr . The system is immersed into an electric field which
far away from the cavity is uniform and equal to Ed .
b) Find the electric field in the whole space.
1
Solution
a) We know that a sphere with uniform polarization P generates a field which is uniform inside the
sphere itself and is given by
Eind = −
P
.
3ǫ0
(1)
(To recover this result, one can model the polarized sphere as two charged spheres of opposite charge
densities ±ρ which are mutually displaced by a small distance δ such that P = ρδ.) Outside the
sphere we have the field of a dipole p = (4πa3 /3)P placed in the center of the sphere.
Let us make the ansatz that the dielectric sphere becomes uniformly polarized in the external
field. Then, the “total” field E inside the sphere is the sum of the external field E0 and the field Eind
induced by the polarization:
E = E0 + Eind = E0 −
P
.
3ǫ0
(2)
Now, in a dielectric P = ǫ0 χE where χ = ǫr − 1. Thus
E = E0 −
χ
E.
3
(3)
Solving for E we obtain
E=
3
1
E0 =
E0
1 + χ/3
ǫr + 2
(r < a) .
(4)
Notice that |E| < |E0 |, i.e. the field inside the sphere is smaller than the field outside due to the
screening by the dielectric. The polarization can now be written explicitly as
P = ǫ0 χE =
3ǫ0 (ǫr − 1)
E0 .
ǫr + 2
(5)
The field outside the sphere is the sum of E0 and the dipole field:
E = E0 + k0
3(p · r̂)r̂ − p
r3
(r > a) ,
(6)
where r̂ = r/r and k0 = (4πǫ0 )−1 . Substituting for p we find
E = E0 +
ǫr − 1
1 a 3
[3(E0 · r̂)r̂ − E0 ]
[3(P · r̂)r̂ − P] = E0 +
3ǫ0 r
ǫr + 2
(r > a) .
(7)
It can be verified that the expression found for the electric field satisfies the boundary conditions
on the surface of the sphere
Ek (r = a− ) = Ek (r = a+ ) ,
ǫE⊥ (r = a− ) = ǫ0 E⊥ (r = a+ ) .
(8)
Of course one may find the solution “backwards”, i.e. as a boundary value problem. This corresponds
to assuming that the field is given by (6) for r > a and is uniform and equal to some uniform value
2
Es for r < a, with p and Es as parameters to be determined by imposing (8) plus the requirement
that the field becomes E0 at infinity.
b) Let us make the ansatz that the field has the same functional form as for the sphere, with
parameters Ec and pc :
Ec
(r < a) ,
E=
(9)
−3
E0 + k0 [3(pc · n)n − pc ] r
(r > a) .
The boundary conditions at the surface of the spherical cavity are
Ek (r = a− ) = Ek (r = a+ ) ,
ǫ0 E⊥ (r = a− ) = ǫE⊥ (r = a+ ) .
(10)
The boundary value problem is thus formally identical to that of the case a), except for the exchange
ǫ ↔ ǫ0 (and, trivially, for the slightly different symbols used for the parameters to be found). Since
the same equations (with the same boundary conditions) have the same solutions, we immediately
obtain the solution for the cavity using the solution for the sphere obtained in a) with the replacement
ǫr = ǫ/ǫ0 → ǫ0 /ǫ = 1/ǫr .
(11)
In particular, the field inside the cavity is
Ec =
3
1/ǫr + 2
Ed ,
|Ec | > |Ed | ,
i.e. it is larger than the faraway field.
3
(12)