### Boundary value problems with dielectrics

#### Problem:

A conducting sphere of radius a carrying a charge q is submerged halfway into a non-conducting dielectric liquid of dielectric constant ε.  The other half is in air.
Will the electric field be purely radial?  Explain.

Solution:

• Concepts:
Boundary value problems, the uniqueness theorem
• Reasoning:
There are no free charges except on the boundaries and the problem has azimuthal symmetry.
• Details of the calculation:
If we place the origin of our coordinate system at the center of the sphere and let the z-axis point upward, then the problem has azimuthal symmetry.  We have 3 regions.
Region 1: the sphere
Region 2: the air
Region 3: the dielectric
In region 1 E = 0 and Φ = constant (properties of conductors).
Assume that in regions 2 and 3 Φ(r) = C/r, with C a constant.
This potential fulfills all the boundary conditions.
Φ = constant at r = a, Φ --> 0 as r --> infinity, Φ = continuous across the air-dielectric interface, and the normal component of D is continuous across the air-dielectric interface.  The uniqueness theorem guaranties that this solution is the only solution.  E = -Φ, the electric field will be purely radial.  E will only depend on r.
Think about:  What is the value of the constant C?
Outside the sphere in the dielectric at r = a:  E = (σf3 + σp)/ε0 = (σf3 - P)/ε0 = (σf3 - ε0xeE)/ε0.
σf3 = ε0(1 + xe)E = εE.
Outside the sphere in air at r = a:  E = σf20,  σf2 = ε0E.
2πa2E(ε + ε0) = q.  E = q/(2π(ε + ε0)a2),
C = q/(2π(ε + ε0)a2).

Azimuthal symmetry

#### Problem:

A dielectric sphere of radius a has a uniform isotropic permittivity, kε0, and is located in an electric field that is uniform at infinity.
(a)  Solve for the electric potential everywhere by boundary value methods.
(b)  Show that the electric field inside the sphere is uniform and find its value relative to the field E0 at infinity.

Solution:

• Concepts:
Most general solution to Laplace's equation, boundary conditions
• Reasoning:
2Φ = 0 inside and outside the sphere since ρf = 0 inside and outside the sphere.  We are looking for a solution for ∇2Φ = 0, subject to the boundary conditions.
• Details of the calculation:
We have azimuthal symmetry.
Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ).
Assume Φ1 = A0 + A1rcosθ + B0/r + (B1/r2)cosθ inside,
Φ2 = A0' + A1'rcosθ + B0'/r + (B1'/r2)cosθ outside.
We assume all higher-order coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  Φ1 is finite at the origin, B0 = B1 = 0.  Φ1 = A0 + A1rcosθ.
E1 = -Φ1,  E1r = -A1cosθ,  E = A1sinθ,  E1 = -A1k.
D1 = εE1 = kε0E1.
(ii)  Φ is continuous at r = a.  A0 = A0' + B0'/a,  A1a = A1'a + B1'/a2.
(iii)  The radial component of D is continuous at r = a.  (There are no free charges.)
E2 = -Φ2.  E2r = -A1'cosθ + B0'/r2 + 2(B1'/r3)cosθ,  E = A1'sinθ + (B1'/r3)sinθ,
D
2 = ε0E2.
-kε0A1cosθ = -ε0A1'cosθ + ε0B0'/a2 + 2ε0(B1'/a3)cosθ.
B0' = 0, kA1 = A1' - 2(B1'/a3) = A1' - 2(A1 - A1').
(iv)  E = E0 at r >> a.
-A1'cosθ = E0cosθ,  A1'sinθ = -E0sinθ,  A1' = -E0.
A1 = 3A1'/(k + 2) = -3E0/(k + 2).
B1' =[(k - 1)/(k + 2)]E0a3.
A0 = A0' is an arbitrary constant.  We choose this constant to be zero.  A0 = A0' = 0.
Φinside = -(3E0/(k + 2))rcosθ = -3E0z/(k + 2),
Φoutside = -E0rcosθ + [(k - 1)/(k + 2)](a3/r2)E0cosθ.

(b)  Ein = -Φin = 3E0/(k + 2).

#### Problem:

Consider a homogeneous dielectric ε, of infinite extent, in which there is a uniform field E0.  A spherical cavity of radius a is cut out of this dielectric.  Find:
(a)  Φ in the cavity and on its surface.
(b)  The polarization charge density σp on the walls.
(c)  The field outside the cavity.

Solution:

• Concepts:
Boundary value problems,  Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry.
• Reasoning:
Let the center of the cavity be at the origin.  Then the problem has azimuthal symmetry.
Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ) is the most general solution inside and outside of the cavity, since r = 0 inside and outside of the cavity.  To find the specific solution we apply boundary conditions.
• Details of the calculation:
(a)  Assume Φ1 = A0 + A1rcosθ + B0/r + (B1/r2)cosθ inside,
Φ2 = A0' + A1'rcosθ + B0'/r + (B1'/r2)cosθ outside.
We assume all higher-order coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  E2 = E0 at r >> a,  Φ2 = -E0z = -E0rcosθ  at r >> a,  A1' = -E0,  we choose A0' = 0.
Then, from symmetry, A0 = 0.
(ii)  Φ1 is finite at the origin, B0 = B1 = 0.  Φ1 = A1rcosθ,
E1 = -Φ1,  E1r = -A1cosθ,  E = A1sinθ,  E1 = -A1k.
D1 = ε0E1.
(iii)  Φ is continuous at r = a.  B0' = 0,  A1a = -E0a + B1'/a2.
(iv)  The radial component of D is continuous at r = a.  (There are no free charges.)
E2 = -Φ2.  E2r = -A1'cosθ + 2(B1'/r3)cosθ,  E = A1'sinθ + (B1'/r3)sinθ,  D2 = kε0E2 = eE2.
0A1cosθ = kε0E0cosθ + 2kε0(B1'/a3)cosθ.
A1 = -kE0 - 2k(B1'/a3) = -kE0 - 2k(A1 + E0).
A1 = -3E0k/(2k + 1) = -3E0ε/(ε0 + 2ε).
B1' = [(ε0 - ε)/(ε0 + 2ε)]E0a3.
Φ1 = [-3E0ε/(ε0 + 2ε)]rcosθ.
On the surface Φ(a) = [-3E0ε/(ε0 + 2ε)]acosθ.

(b)  σp = ε0(E2r(a)- E1r(a) ) = ε0E0cosθ[1 + 2(ε0 - ε)/(ε0 + 2ε) + 3ε/(ε0 + 2ε)]
σp = ε0E0cosθ[1 + (2ε0 + ε))/(ε0 + 2ε)] = ε0E0cosθ[3(ε0 + ε))/(ε0 + 2ε)].

(c)  Φ2 = -E0rcosθ + [(ε0 - ε)/(ε0 + 2ε)]E0a3cosθ/r2,
E2r = -∂Φ2/∂r = E0cosθ[1 + 2[(ε0 - ε)/(ε0 + 2ε)]a3/r3],
E = -(1/r)∂Φ2/∂θ = -E0sinθ[1 - [(ε0 - ε)/(ε0 + 2ε)]a3/r3].

#### Problem:

A dielectric sphere radius "a" and relative permittivity εr is placed in a uniform electric field.  The field region is large so the presence of the sphere does not perturb the field sources.
(a)  Assuming the field inside the sphere to be uniform and the field outside to be described by the uniform field plus a dipole field centered on the sphere, find the fields and the dipole moment induced in the sphere by applying boundary conditions at one point.
(b)  What happens if the sphere is conducting?

Solution:

• Concepts:
The dipole field, boundary conditions
• Reasoning:
Place the sphere at the origin.  Assume external field points in the z-direction,
Eext = E0k.
The external electric field will polarize the sphere.  The electric field of a uniformly polarized sphere is uniform inside the sphere and a pure dipole field outside the sphere.  Assume that the total field inside the sphere is the superposition of the field of a uniformly polarized sphere and the constant external field E0k.  The superposition of these two fields fulfills all the boundary conditions for D on the surface of the sphere.
We have a unique solution as long as inside D = εrE inside the sphere.
• Details of the calculation:
(a)  Let us first concentrate on the field produced by a polarized sphere with dipole moment p.
The field produced by the polarized sphere outside the sphere is
E(r) = (4πε0)-1(3(p∙r)r/r5 - p/r3),
with p = pk = dipole moment of the sphere.
Then at r = ak (on the z-axis) we have
E
out(sphere)= pk/(2πε0a3) = 2Pk/(3ε0), with P = p/(4πa3/3).
Eout(sphere) - Ein(sphere) = (σp0)k = (P/ε0)k.
Therefore Ein(sphere) = -Pk/(3ε0), for the field produced by the polarized sphere.
Now let us look at the total field inside and outside of the sphere.
The total field at r = ak (on the z-axis) inside the sphere is Ein = (E0 - P/(3ε0))k,  Din = εr(E0 - P/(3ε0))k.
The total field at r = ak (on the z-axis) outside the sphere is
E
out = (E0 + 2P/(3ε 0))kDout = ε0(E0 + 2P/(3ε0))k.
The normal component of D is continuous.
εr(E0 - P/(3ε0)) = ε0(E0 + 2P/(3ε0)),  (2ε0 + εr)P/(3ε0) = (εr - ε0)E0.
P = 3ε0r - ε0)E0/(2ε0 + εr), p = P4πa3/3 = 4πε0a3r - ε0)E0/(2ε0 + εr).
Ein = E00/(2ε0 + εr), Din = εrEin, p = 4πε0a3r - ε0)E0/(2ε0 + εr).
(b)  When the sphere is conducting the total field at r = ak (on the z-axis) inside the sphere is
E
in = (E0 - P/(3ε0))k = 0.
P = 3ε0E0, p = 4πε0a3E0.

#### Problem:

(a)  A spherical dielectric of radius a has a uniform polarization P in the z-direction.
Show that the electric field inside the dielectric due to the polarization is given by E = -P/(3ε0).
(b)  A large capacitor in vacuum has parallel circular plates of radius R separated by a distance d (d << R).
The capacitor is charged to a potential difference V and disconnected from the source.  Find the energy stored in the capacitor.
(c)  Subsequently, a small sphere of radius a (a << d) and dielectric constant K is placed in the center of the capacitor between the plates.
Find the electric field inside the dielectric.
(d)  Will the capacitor have less, more, or the same energy than before the dielectric was inserted?  Explain!

Solution:

• Concepts:
Boundary value problems, Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry.
• Reasoning:
2Φ = 0 inside and outside the sphere since ρfree = 0 inside and outside the sphere.  We are looking for a solution for ∇2Φ = 0, subject to the boundary conditions.
• Details of the calculation:
(a)  Place the center of the sphere at the origin.
Then ρb = -P = 0,  σb = P n = P cosθ.
We have a sphere with a surface charge density proportional to cosθ.  We solve for the potential and the field using boundary value methods.
Assume Φ1 = A0 + A1rcosθ  inside,
Φ2 = B0'/r + (B1'/r2)cosθ  outside.
We assume all higher-order coefficients are zero.  If we find a solution with this assumption, it is the only solution.
Boundary conditions:
(i)  Φ is continuous at r = a.  A0 = B0'/a,  A1a = B1'/a2.
(ii)  The radial component of D is continuous at r = a.  (There are no free charges.)
E1 = -Φ1,  E1r = -A1cosθ,  E = A1sinθ,  E1 = -A1k.
D1 = ε0E1 + P.
E2 = -Φ2.  E2r = B0'/r2 + 2(B1'/r3)cosθ,  E = (B1'/r3)sinθ,  D2 = ε0E2.
0A1cosθ + P cosθ = ε0B0'/a2 + 2ε0(B1'/a3)cosθ.
0A1 + P = 2ε0(B1'/a3),  B0' = A0 = 0.
0A1 + P = 2ε0A1.  A1 = P/(3ε0).
E1 = -A1k = P/(3ε0).

(b)  U = ½CV2,  C = ε0A/d = ε0πR2/d,  U = ½(ε0πR2/d)V2.

(c)  The dielectric sphere is placed in a previously uniform electric field E0. ( E0 = V/d)
The field inside the sphere will be Ein = -Φin = 3E0/(K + 2).

(d)  The capacitor is disconnected from the battery.  Q is constant.  The capacitance increases when a dielectric is inserted.
The energy stored in the capacitor (U = Q2/2C) is proportional to 1/C for a capacitor with constant Q.
The energy stored in the capacitor decreases.

#### Problem:

Two very large metal plates are held a distance d apart, one at potential zero and the other at potential V0.  A metal sphere of radius a is sliced in two, and one hemisphere is placed on the grounded plate, so that its potential is likewise zero.  The radius of the sphere a is very small compare to the distance d between the plates, (a << d), so that you may assume that the electric field near the upper plate is constant.  If the region between the plates is filled with a weakly conducting material of conductivity σ, what current flows to the hemisphere?

Solution:

• Concepts:
Boundary value problems, azimuthal symmetry.
• Reasoning:
Put the origin of the coordinate system at the center of the sphere. Then
Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ)
is the most general solution in the region between the plates, since the charge density ρ = 0 there.  To find the specific solution we apply boundary conditions.
• Details of the calculation:
The boundary conditions are Φ = 0 on the lower plate and the sphere and Φ = V0 on the upper plate.
Φ = A1'rcosθ  + (B1'/r2)cosθ  between the plates.
We assume all other coefficients are zero.  If we find a solution with this assumption, then the uniqueness theorem guaranties that it is the only solution.
Boundary conditions:
(i)  Φ is continuous at r = a.  0 = A1'a + B1'/a2.  B1' = -A1'a3.
(ii)  E = -Φ = -E0k at r >> a,  i.e. far away from the hemispherical boss.
-A1'cosθ = -E0cosθ,  A1'sinθ = E0sinθ,  A1' = E0,  B1' = -E0a3.
Φ(z = d) = V.  E0d = V,  E0 = V/d
Φ = E0rcosθ  - (E0a3/r2)cosθ.

E = -Φ.  Er = -E0cosθ  - 2(E0a3/r3)cosθ,  Eθ = E0sinθ - (E0a3/r3)sinθ.
E = constant field + field due to polarized sphere.
At the surface of the sphere at r = a, E is perpendicular to the surface and therefore radial.
Er(a) = -3E0cosθ.

j = σE is the current density between the plate.
The current flowing to the hemisphere is
0π /2j(θ)2πa2 sinθ dθ = 6πσE0a20π /2cosθ sinθ dθ = 3πσ E0a2 = 3πσ Va2/d = I.

Cylindical coordinates:  Φ = Φ(ρ,φ)  (0 ≤ φ  ≤ 2π) independent of z

#### Problem:

A long cylindrical rod of radius a and dielectric constant k is placed in a uniform electric field E0 with its axis perpendicular to the field direction.
(a)  Find the potential Φ inside and outside the rod.
(b)  Find the electric field E inside and outside the rod.
(c)  Find the volume and surface polarization charge density.

Solution:

• Concepts:
Boundary value problems,
Φ(ρ,φ) = ∑n=1(Ancos(nφ) +  Bnsin(nφ))ρn + ∑n=1(A'ncos(nφ) +  B'nsin(nφ))ρ-n +a0 + b0lnρ
is the general solution to Laplace's equation if Φ = Φ(ρ,φ)  (0 ≤ φ  ≤ 2π) in cylindrical coordinates.
• Reasoning:
Φ = Φ(ρ,φ)  (0 ≤ φ  ≤ 2π) independent of z.  The problem involves the whole azimuthal range.
• Details of the calculation:
(a)  Let E = Ei.  Assume that all Bn and Bn' and all An and An' with n > 1 are zero in the general solution.  If we find such a solution that satisfies all boundary conditions, then it is a unique solution.
Boundary conditions:
r < a:  Φ is finite at the origin.  Choose Φ(0) = 0.  Then Φ(ρ,φ) = A1icos(φ)ρ.
ρ > a:  -Φ = Ei as ρ --> infinity.  Φ(ρ,φ) = A1ocos(φ)ρ + A1o'cos(φ)/ρ,  A1o = -E.
ρ = a:  Φ is continuous,   Dn is continuous.
A1icos(φ)a = -Ecos(φ)a + A1o'cos(φ)/a,  A1i + E =  A1o'/a2.
εA1icos(φ) = -ε0Ecos(φ) - ε0A1o'cos(φ)/a2,  (ε/ε0)A1i + E =  -A1o'/a2.
Algebra:  A1i = -E(2ε0/(ε + ε0)), A1o' = a2E[(ε - ε0)/(ε + ε0)]
Φinside = -E(2ε0/(ε + ε0))cos(φ)ρ,  Φoutside = -Ecos(φ)ρ + a2E[(ε - ε0)/(ε + ε0)]cos(φ)/ρ.
(b)  E = -Φ.  Eρ = ∂Φ/∂ρ,  Eφ = (1/ρ)∂Φ/∂φ.
Ein = E(2ε0/(ε + ε0))(cos(φ)[ρ/ρ] - sin(φ)[φ/φ]) = (2ε0/(ε + ε0))Ei.
Eout = Ei + E(a22)[(ε - ε0)/(ε + ε0)](cos(φ)[ρ/ρ] + sin(φ)[φ/φ]).
(c)  ∇∙P = -ρbound = -ε0xe∇∙E = 0 since E = constant inside the rod.
P∙n = σbound = ε0xeEin[ρ/ρ] = (ε + ε0)Ein[ρ/ρ] = (2ε0(ε - ε0)/(ε + ε0))Ecos(φ).