__Uniqueness theorem__

Prove, carefully explaining your reasoning, that the solution of
**∇**∙**E **= ρ/ε_{0},** ∇**×**E** = 0, for
**E** is unique. You may assume that the sources of
**E** are bounded in
space and that therefore the field vanishes at sufficiently large distances from the
sources.

Solution:

- Concepts:

The uniqueness theorem, Green's first identity

∫_{V}(Φ**∇**^{2}ψ**∇**Φ**∙∇**ψ)_{S}**(**∂ψ/∂n) dS

Green's first identity is derived starting with Green's theorem:

∫_{V}**∇∙A**dV = ∮_{S}**A∙n**dS

Set**A**= Φ**∇**ψ, with Φ, ψ arbitrary scalar fields.

**∇∙A**=**∇∙**(Φ**∇**ψ) = Φ∇^{2}ψ +**∇**Φ**∙∇**ψ

**A∙n**= Φ**∇**ψ**∙n**= Φ**(**∂ψ/∂n) - Reasoning:

We are asked to prove the uniqueness theorem. - Details of the calculation:

**∇**×**E**= 0 <==>**E**= -**∇**Φ,**E**is the gradient of some scalar function.

**∇**·**E**= ρ/ε_{0}<==> ∇^{2}Φ = -ρ/ε_{0}.Consider the volume V bounded by surfaces at infinity. Assume we know ρ(

**r**) everywhere inside V and we know that Φ = 0 at the boundaries.

Assume that two solutions for Φ exist that satisfy the same boundary conditions.

∇^{2}Φ_{1}= -ρ/ε_{0}, ∇^{2}Φ_{2}= -ρ/ε_{0}.

Let U = Φ_{1}- Φ_{2}, then ∇^{2}U = 0 and U = 0 at the boundaries.

Apply Green's first identity with Φ = ψ = U.

∫_{V}(U**∇**^{2}U**∇**U**∙∇**U)_{S}**(**∂U/∂n) dS

The right hand side is zero. ∇^{2}U = 0. Therefore

∫_{V}|**∇**U|dV = 0.^{2}

Since |∇U|^{2}is a non-negative everywhere, this implies**∇**U = 0 everywhere inside V and U is constant. Φ_{1}and Φ_{2}differ at most by a constant. This constant is zero since Φ_{1}= Φ_{2}at the boundaries. Φ is unique,**E**= -**∇**Φ is unique.

__Boundary value problems, azimuthal symmetry__

A sphere of radius R carries a surface charge density
σ = σ_{0}cosθ.

(a) Derive the exact potential everywhere (both inside and
outside the sphere) assuming that it vanishes as r goes to infinity.

(b) Calculate the dipole moment of the charge distribution and
deduce the approximate form of the potential at points far from the sphere (r >> R).

(c) Compare (a) and (b). What can you conclude about the higher
multipoles?

Solution:

- Concepts:

Boundary value problems, Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry. - Reasoning:

Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere. To find the specific solution we apply boundary conditions. - Details of the calculation:

Assume Φ_{1}(r,θ) = ArP_{1}(cosθ) = Arcosθ inside the sphere and Φ_{2}(r,θ) = (B/r^{2})P_{1}(cosθ) = (B/r^{2})cosθ outside the sphere. The symmetry of the situation suggests this form of the solution. If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution. The potential vanishes as r goes to infinity.

Boundary conditions:

Φ is continuous across the boundary. Φ_{1}(R,θ) = Φ_{2}(R,θ). A = B/R^{3}.

(**E**_{2}-**E**_{1})·**n**_{2}= σ/ε_{0}, dΦ_{2}/dr - dΦ_{1}/dr|_{R}= -σ/ε_{0}. -2(B/R^{3})cosθ - Acosθ = -σ_{0}cosθ/ε_{0}.

-2A -A = -σ_{0}/ε_{0}. A = σ_{0}/3ε_{0}.

Φ_{1}(r,θ) = (σ_{0}/3ε_{0}) r cosθ inside the sphere and Φ_{2}(r,θ) = (σ_{0}R^{3}/3ε_{0}r^{2})cosθ outside the sphere.

(b) dipole moment:**p**= ∫σ(**r**)**r**dA =**k**σ_{0}∫cosθ Rcosθ 2πR^{2}sinθdθ =**k**σ_{0}(4π/3)R^{3}.

dipole potential: Φ_{dip}= (1/(4πε_{0}))**p∙r**/r^{3}= (σ_{0}R^{3}/3ε_{0}r^{2})cosθ.

The potential outside the sphere is a pure dipole potential. The higher multipoles are zero.

A hollow ball of radius R has a surface charge distribution which produces a
potential on the surface of V(R,θ) = k(cos^{2}θ + 1), where k is a
constant and θ is the usual polar angle relative to the z-axis.

(a)
Find the potential at all points in space.

(b) Show that the charge
distribution on the ball is σ(R,θ) = (kε_{0}/R)(5cos^{2}θ -
(1/3)).

Solution:

- Concepts:

General solution to Laplace's equation in situations with azimuthal symmetry, boundary value problems - Reasoning:

Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ) is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere. To find the specific solution we apply boundary conditions. The potential on the surface of the ball is given. - Details of the calculation:

(a) Assume

Φ_{1}(r,θ) = A_{0}+ A_{2}r^{2}P_{2}(cosθ) inside the sphere and

Φ_{2}(r,θ) = (B_{0}/r) + (B_{2}/r^{3})P_{2}(cosθ) outside the sphere.

The symmetry of the situation suggests this form of the solution. If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution. The potential vanishes as r goes to infinity.

Boundary conditions:

Φ_{1}(r,θ) is finite at r = 0 and Φ_{2}(r,θ) goes to zero as r goes to infinity.

Φ is continuous across the boundary at r = R.

A_{0}= B_{0}/R, A_{2}= B_{2}/R^{5}.

Φ(R,θ) = A_{0}+ A_{2}R^{2}P_{2}(cosθ) = A_{0}+ A_{2}R^{2}(3cos^{2}θ -1)/2 = k(cos^{2}θ + 1).

3A_{2}R^{2}/2 = k, A_{0}- A_{2}R^{2}/2 = k, A_{2}= 2k/(3R^{2}), A_{0}= 4k/3.

Φ_{1}(r,θ) = 4k/3 + (2k/(3R^{2}))r^{2}P_{2}(cosθ), Φ_{2}(r,θ) = 4kR/(3r) + (2kR^{3}/(3r^{3}))P_{2}(cosθ) .

(b) (**E**_{2}-**E**_{1})·**n**_{2}= σ/ε_{0}, ∂Φ_{2}/∂r - ∂Φ_{1}/∂r |_{R}= -σ/ε_{0}.

∂Φ_{2}/∂r = -4kR/(3r^{2}) - (2kR^{3}/r^{4})P_{2}(cosθ), ∂Φ_{1}/∂r = (4k/(3R^{2}))rP_{2}(cosθ).

σ(R,θ)/ε_{0}= 4k/(3R) + (2k/R)P_{2}(cosθ) + (4k/(3R))P_{2}(cosθ)

= 4k/(3R) + 10k/(3R))P_{2}(cosθ).

σ(R,θ) = (kε_{0}/R)(4/3 + (10/3)(3cos^{2}θ - 1)/2) = (kε_{0}/R)(5cos^{2}θ - (1/3)).

Consider a hollow sphere of radius R with a surface
potential Φ(R,θ) = k sin^{2}θ, where k is a constant and θ is the usual
polar angle relative to the z-axis.

Find the potential everywhere inside the sphere.

Hint:

P_{0}(x) = 1

P_{1}(x) = x

P_{2}(x) = (3x^{2} - 1)/2

P_{3}(x) = (5x^{3} - 3x)/2

Solution:

- Concepts:

General solution to Laplace's equation in situations with azimuthal symmetry, boundary value problems - Reasoning:

Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ)

is the most general solution inside and outside of the sphere, since ρ = 0 inside and outside of the sphere. To find the specific solution we apply boundary conditions. The potential on the surface of the sphere is given. - Details of the calculation:

Φ(R,θ) = k sin^{2}θ = k(1 - cos^{2}θ) = (2k/3)(P_{0}- P_{2}(cosθ)).

Assume

Φ_{1}(r,θ) = A_{0}+ A_{2}r^{2}P_{2}(cosθ) inside the sphere.

The symmetry of the situation suggests this form of the solution. If we can satisfy the boundary conditions, the uniqueness theorem guaranties that we have found the only solution.

Boundary conditions:

Φ_{1}(r,θ) is finite at r = 0 .

Φ is continuous across the boundary at r = R.

A_{0}= 2k/3, A_{2}R^{2}= -2k/3.

Φ_{1}(r,θ) = 2k/3 - (2k/3)(r^{2}/R^{2})P_{2}(cosθ) = 2k/3 - (2k/3)(r^{2}/R^{2})( 1 - (3/2)sin^{2}θ)

is the potential inside the sphere.

A conducting sphere of radius a is located in an electric field that is
uniform at infinity, i.e. **E** = E_{0}
**k** at infinity.
Put the origin of the coordinate system at the center of a sphere and set the
potential Φ(0) = 0.

Solve for the electric potential and the electric field everywhere by
boundary value methods.

Solution:

- Concepts:

Most general solution to Laplace's equation, boundary conditions - Reasoning:

Φ_{1}= 0,**E**_{1}= 0 inside the sphere since the interior of a conductor in electrostatics is field-free. ∇^{2}Φ_{2}= 0 and outside the sphere since ρ = 0 outside the sphere. We are looking for a solution for ∇^{2}Φ_{2}= 0, subject to the boundary conditions. - Details of the calculation:

We have azimuthal symmetry.

Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ) is the most general solution of ∇^{2}Φ = 0 for problems with azimuthal symmetry.

Assume Φ_{1}= 0 inside, Φ_{2}= A_{0}' + A_{1}'rcosθ + B_{0}'/r + (B_{1}'/r^{2})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 . 0 = A_{1}'a + B_{1}'/a^{2}.

(ii)**E**_{2}=**E**_{0}at r >> a.

-A_{1}'cosθ = E_{0}cosθ, A_{1}'sinθ = -E_{0}sinθ, A_{1}' = -E_{0}, B_{1}' = E_{0}a^{3}.

Φ_{2}= -E_{0}rcosθ + (E_{0}a^{3}/r^{2})cosθ.

**E**_{2}= -**∇**Φ_{2}. E_{2r}= E_{0}cosθ + 2(E_{0}a^{3}/r^{3})cosθ, E_{2θ}= -E_{0}'sinθ + (E_{0}a^{3}/r^{3})sinθ.

**E**_{2}=**E**_{0}+ 2(E_{0}a^{3}/r^{3})cosθ (**r**/r) + (E_{0}a^{3}/r^{3})sinθ (**θ**/θ).

**E**_{2}= external field + field due to the polarized sphere.

Two thin wires, each of length L, lie on the z-axis. Wire 1 has a positive
charge per unit length +λ and extends from (0, 0, L) to (0,0,0). Wire 2 has a
negative charge per unit length -λ and extends from (0,0,0) to (0,0,-L).

(a) Find the potential Φ on the z-axis at z > L.

(b) The potential Φ(r,θ) at a arbitrary points in space with r > L can be
expanded in terms of Legendre polynomials, Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}
+ B_{n}/r^{n+1}]P_{n}(cosθ).

Find the expansion coefficients.

Note: ln(1 + x) = ∑_{n=1}^{∞}(-1)^{n+1}x^{n}/n

Solution:

- Concepts:

Symmetry, Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ) is the general solution to Laplace's equation for problems with azimuthal symmetry. - Reasoning:

Φ(**r**) = [1/(4πε_{0})]∫ρ(**r**')dV'/|**r**-**r**'|. We can evaluate this integral for**r**= r**k**.

For r > L, Φ(**r**) satisfies Laplace's equation, ∇^{2}Φ(**r**) = 0. The general solution is

Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n-1}]P_{n}(cosθ).

We can find the coefficients by comparing the general solution to the solution found by integration on the z-axis. - Details of the calculation:

(a) Φ(z) = [1/(4πε_{0})]∫_{0}^{L}λdz'/(z -_{0})] ∫_{-L}^{0}λdz'/(z -

= [1/(4πε_{0})][-λ∫_{z}^{z-L}dx/x + λ∫_{z+L}^{z}dx/x] = [1/(4πε_{0})] [-λln((z - L)/z) + λln(z/(z + L))]

= -[λ/(4πε_{0})]ln[(z^{2}- L^{2})/z^{2}].

Φ(z) = -[λ/(4πε_{0})]ln(1 - u), with u = L^{2}/z^{2}.

(b) For r > L ∑_{n=0}^{∞}(B_{n}/r^{n+1})P_{n}(cosθ). Only the odd parity terms contribute and all A_{n}are zero since Φ = 0 at infinity.

We can therefore write Φ(r,θ) = ∑_{n=1}^{∞}(B_{2n-1}/r^{2n})P_{2n-1}(cos(θ)).

On the positive z-axis Φ(z) = ∑_{n=1}^{∞}(B_{2n-1}/z^{2n}), since P_{n}(1) = 1.

Expanding ln(1 - u) = ∑_{n=1}^{∞}(-1)^{n+1}(-u)^{n}/n we have, equating the two expressions for Φ(z),

∑_{n=1}^{∞}(B_{2n-1}/z^{2n}) = -[λ/(4πε_{0})] ∑_{n=1}^{∞}(-1)^{n+1}(-u)^{n}/n = [λ/(4πε_{0})] ∑_{n=1}^{∞}(-1)^{2n+2}u^{n}/n

= [λ/(4πε_{0})] ∑_{n=1}^{∞}L^{2n}/(nz^{2n}).

Equating term with equal powers of z we have B_{2n-1}/z^{2n}= [λ/(4πε_{0})] L^{2n}/(nz^{2n}).

B_{2n-1 }= [λ/(4πε_{0})] L^{2n}/n.

Φ(r,θ) = ∑_{n=1}^{∞}([λ/(4πε_{0})] L^{2n}/n)(1/r^{2n})P_{2n-1}(cos(θ)).

A charge array consists of two charges, each of magnitude +q, located on the
z-axis at (0, 0, ±a).

(a) Find the potential V(0,0,z) at an arbitrary
point z > a; then expand V(0, 0,z) in a power series in z.

(b) Using
this series as a "boundary condition", find the potential V(r,θ,φ) at an
arbitrary location (r,θ,φ) with r > a. An infinite series is acceptable.

(c) Characterize the first three terms in 1/r^{n}.

Solution:

- Concepts:

The electrostatic potential - Reasoning:

In regions with charge density ρ = 0 we have ∇^{2}V = 0. So in the region r > a we have

∇^{2}V = 0. For a problem with azimuthal symmetry, the most general solution for ∇^{2}V = 0 is

V(r,θ) = Σ_{n=0}^{∞}(A_{n}r^{n}+ B_{n}r^{-(n+1)})P_{n}(cosθ).

On the positive z-axis P_{n}(cosθ) = 1 and V(z) = Σ_{n=0}^{∞}(A_{n}z^{n}+ B_{n}z^{-(n+1)}).

If we know the potential on the z-axis, we can solve for the expansion coefficients by equating term of equal power in z. - Details of the calculation:

(a) V(z) = kq/(z - a) + kq/(z + a) = (kq/z)[(1 - a/z)^{-1}+ (1 + a/z)^{-1}].

Assume a/z < 1.

For x < 1 we have (1 ± x)^{-1}= 1 ∓ x + x^{2}∓ x^{3}+ ... .

V(z) = (2kq/z)(1 + (a/z)^{2}+ (a/z)^{ 4}+ ...) = 2kq(1/z + a^{2}/z^{3}+ a^{4}/z^{ 5}+ ...).

(b) V(z) = 2kq(1/z + a^{2}/z^{3}+ a^{4}/z^{ 5}+ ...) = Σ_{n=0}^{∞}(A_{n}z^{n}+ B_{n}z^{-(n+1)}).

A_{n}= 0 for all n. B_{n}= 0 for n = odd, B_{n}= 2kqa^{n}for n = even.

V(r,θ) = 2kqΣ_{n=even}a^{n}r^{-(n+1)}P_{n}(cosθ).

(c) First term: Monopole term ∝ 1/r = 2kq/r.

Second term: Dipole term ∝ 1/r^{2}= 0.

Third term: Quadrupole term ∝ 1/r^{3}= 2kqa^{2}P_{2}(cosθ)/r^{3}.

__Boundary value problems, other symmetry__

A rectangular box occupying the region 0 ≤ x ≤ A, 0 ≤ y ≤ B, 0 ≤ z ≤ C, has
an electrostatic potential V(x,y,z) = 0 on all faces except the face z =
C, where the potential is V_{0}. Starting from Laplace's equation,
find the potential V everywhere inside.

Solution:

- Concepts:

General solution to Laplace's equation, boundary conditions - Reasoning:

ρ = 0 inside the box,**∇**^{2}V = 0 inside the box. - Details of the calculation:

In Cartesian coordinates**∇**^{2}V = ∂^{2}V/∂x^{2}+ ∂^{2}V/∂y^{2}+ ∂^{2}V/∂z^{2}= 0.

Separation of variables: V(x,y,z) = F_{1}(x)F_{2}(y)F_{3}(z).

(1/F_{1})∂^{2}F_{1}/∂x^{2}+ (1/F_{2})∂^{2}F_{2}/∂y^{2}+ (1/F_{3})∂^{2}F_{3}/∂z^{2}= 0.

Since each term in the sum is a function of a different independent variable, each term must be equal to a constant and the constants have to sum to zero.

Let (1/F_{1})∂^{2}F_{1}/∂x^{2}= -a^{2}, ∂^{2}F_{1}/∂x^{2}+ a^{2}F_{1}= 0.

(1/F_{2})∂^{2}F_{2}/∂y^{2}= -b^{2}, ∂^{2}F_{2}/∂y^{2}+ b^{2}F_{2}= 0.

(1/F_{3})∂^{2}F_{3}/∂z^{2}= (a^{2}+ b^{2}), ∂^{2}F_{3}/∂z^{2}- (a^{2 }+ b^{2})F_{3}= 0.

Solutions that can fulfill the boundary conditions on all faces require

F_{1}(x) = sin(ax), a = nπ/A, F_{2}(y) = sin(by), b = mπ/B, n, m = 1. 2, 3, ... .

F_{3}(z) = sinh(cz) c = π(n^{2}/A^{2}+ m^{2}/B^{2})^{½}.

The solution is

V(x,y,z) = ∑_{n=1}^{∞}∑_{m=1}^{∞}D_{nm}sinh(c_{nm}z) sin(nπx/A) sin(mπy/B),

subject to the boundary condition V(x,y,C) = V_{0}.

If p, q = odd then ∫_{0}^{A}dx∫_{0}^{B}dy V_{0}sin(pπx/A) sin(qπy/B) = [4V_{0}AB/(pqπ^{2})]

[V_{0}AB/(pqπ^{2})] = D_{pq}sinh(c_{pq}C) ¼AB.

Therefore D_{nm }= 16V_{0}/(nmπ^{2}sinh(c_{nm}C)), n, m = odd.

Semi-infinite planes at φ = 0 and φ = π/6 are separated by an infinitesimal
insulating gap as shown in the figure. For V(φ = 0) = 0 and V(φ = π/6) =
100 V, calculate V and **E** in the region between the plates.

Solution:

- Concepts:

Laplace's equation in cylindrical coordinates - Reasoning:

Since the plates are semi-infinite, V can only depend on φ.

**∇**^{2}V = (1/ρ)∂/∂ρ(ρ∂V/∂ρ) + (1/ρ^{2})∂^{2}V/∂φ^{2}+ ∂^{2}V/∂z^{2}= 0 becomes

(1/ρ^{2})∂^{2}V/∂φ^{2}= 0.

Since ρ = 0 is excluded due to the insulating gap, we can write ∂^{2}V/∂φ^{2}= 0. - Details of the calculation:

Integrating twice gives V = Aφ + B.

Boundary conditions:

V(φ = 0) = 0 --> B = 0.

V(φ = π/6) = 100 V --> A = 600 V/π.

Therefore V(φ) = 600*(φ/π) V between the plates.

**E**= -**∇**V = -(1/ρ)∂V/∂φ^{ }(**φ**/φ) = 600/(ρπ) V/m with ρ measured in m.

A long cylinder of radius a and relative permittivity ε carries a free surface
charge density σ_{f} = σ_{0}cos^{2}(φ).

(a) Find the potential Φ inside and outside of the cylinder.

(b) Find the electric field **E** inside and outside the cylinder.

(c) Find the polarization surface charge density of the cylinder.

Let
Φ = Φ(ρ,φ), 0 ≤ θ ≤ 2π in
**cylindrical coordinates**.

Then the most general solution for ∇^{2}Φ =
0 is

Φ(ρ,φ) = ∑_{n=1}^{∞}(A_{n}cos(nφ)
+ B_{n}sin(nφ))ρ^{n} + ∑_{n=1}^{∞}(A'_{n}cos(nφ)
+ B'_{n}sin(nφ))ρ^{-n} + a_{0} + b_{0}lnρ.

Solution:

- Concepts:

Boundary value problems,

Φ(ρ,φ) = ∑_{n=1}^{∞}(A_{n}cos(nφ) + B_{n}sin(nφ))ρ^{n}+ ∑_{n=1}^{∞}(A'_{n}cos(nφ) + B'_{n}sin(nφ))ρ^{-n}+ a_{0}+ b_{0}lnρ - Reasoning:

Φ = Φ(ρ,φ) (0 ≤ φ ≤ 2π) independent of z. The problem involves the whole azimuthal range. - Details of the calculation:

(a) σ = σ_{0}cos^{2}(φ) = σ_{0}(½ + ½cos(2φ)).

Assume Φ_{1}(ρ,φ) = A_{2}cos(2φ)ρ^{2}inside,

Φ_{2}(ρ,φ) = A'_{2}cos(2φ)ρ^{-2}+ a_{0}+ b_{0}lnρ outside.

We assume all other coefficients are zero. If we find a solution with this assumption, it is the only solution.

Boundary conditions:

ρ < a: Φ is finite at the origin. We choose Φ(0) = 0.

ρ = a: Φ is continuous, A_{2}cos(2φ)a^{2}= A'_{2}cos(2φ)a^{-2}+ a_{0}+ b_{0}lna**.**A

_{2}a^{4}= A'_{2}, a_{0}= -b_{0}lna**.**(

**D**_{2}-**D**_{1})·**n**_{2}_{f}at ρ = a.

(**D**_{2}-**D**_{1})∙(**ρ**/ρ) = σ_{0}(½ + ½cos(2φ)) at ρ = a.

E_{ρ1}= -∂Φ_{1}/∂ρ = -2A_{2}cos(2φ)ρ,^{ }E_{ρ2}= -∂Φ_{2}/∂ρ = 2A_{2}cos(2φ)ρ^{-3}- b_{0}/ρ.

D_{ρ1}= -ε2A_{2}cos(2φ)ρ,^{ }D_{ρ2}= ε_{0}2A_{2}a^{4}cos(2φ)ρ^{-3}- ε_{0}b_{0}/ρ.

(ε_{0}+ ε)2A_{2}cos(2φ)a - ε_{0}b_{0}/a = σ_{0}(½ + ½cos(2φ)).

A_{2}= σ_{0}/(4a(ε_{0}+ ε)), b_{0}= -aσ_{0}/2ε_{0. }Φ_{1}(ρ,φ) = σ_{0}/cos(2φ)ρ^{2}/(4a(ε_{0}+ ε))._{ }Φ_{2}(ρ,φ) = a^{4}σ_{0}cos(2φ)ρ^{-2}/(4a(ε_{0}+ ε)) - (½aσ_{0}/ε_{0})ln(ρ/a**).**_{ }(b)**E**= -**∇**Φ. E_{ρ}= ∂Φ/∂ρ, E_{φ}= (1/ρ)∂Φ/∂φ.

**E**_{in}= -2A_{2}cos(2φ)ρ [**ρ**/ρ] + 2A_{2}sin(2φ)ρ^{2}[**φ**/φ]

= -σ_{0}cos(2φ)ρ/(2a(ε_{0}+ ε)) [**ρ**/ρ] + σ_{0}sin(2φ)ρ^{2}/(2a(ε_{0}+ ε)) [**φ**/φ].**E**_{out}= (2A_{2}a^{4}cos(2φ)ρ^{-3}- b_{0}/ρ)[**ρ**/ρ] + 2A_{2}a^{4}sin(2φ)ρ^{-2}[**φ**/φ]

= (σ_{0}a^{4}cos(2φ)ρ^{-3}/(2a(ε_{0}+ ε)) + aσ_{0}/(2ε_{0}ρ) [**ρ**/ρ] + σ_{0}a^{4}sin(2φ)ρ^{-2}/(2a(ε_{0}+ ε)) [**φ**/φ].

(c) (**E**_{2}-**E**_{1})·[**ρ**/ρ]_{f}+ σ_{p})/ε_{0}at ρ = a.

σ_{0}cos(2φ)/(2(ε_{0}+ ε)) + σ_{0}/(2ε_{0}) + σ_{0}cos(2φ)/(2(ε_{0}+ ε))

= σ_{0}cos(2φ)/(ε_{0}+ ε) + σ_{0}/(2ε_{0}) = (σ_{f}+ σ_{p})/ε_{0}.

σ_{p}= ε_{0}σ_{0}cos(2φ)/(ε_{0}+ ε) - σ_{0}½cos(2φ) = ½(ε_{0}- ε)σ_{0}cos(2φ)/(ε_{0}+ ε).

Or = σ_{p}=**P**·[**ρ**/ρ] evaluated at ρ = a.= ε

P_{0}*χ*_{e}**E**_{in}= (ε_{0}- ε_{0})**E**_{in},

σ_{p}= ½(ε_{0}- ε)σ_{0}cos(2φ)/(ε_{0}+ ε).