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 = (σ_{f3}+ σ_{p})/ε_{0}= (σ_{f3}- P)/ε_{0}= (σ_{f3}- ε_{0}x_{e}E)/ε_{0}.

σ_{f3}= ε_{0}(1 + x_{e})E = εE.

Outside the sphere in air at r = a: E = σ_{f2}/ε_{0}, σ_{f2}= ε_{0}E.

2πa^{2}E(ε + ε_{0}) = q. E = q/(2π(ε + ε_{0})a^{2}),

C = q/(2π(ε + ε_{0})a^{2}).

__Azimuthal symmetry__

A dielectric sphere of radius a has a uniform isotropic permittivity,
kε_{0}, 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 E_{0} at infinity.

Solution:

- Concepts:

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

∇^{2}Φ = 0 inside and outside the sphere since ρ_{f}= 0 inside and outside the sphere. We are looking for a solution for ∇^{2}Φ = 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),

Assume Φ_{1}= A_{0}+ A_{1}rcosθ + B_{0}/r + (B_{1}/r^{2})cosθ 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_{1}is finite at the origin, B_{0}= B_{1}= 0. Φ_{1}= A_{0}+ A_{1}rcosθ.

**E**_{1}= -**∇**Φ_{1}, E_{1r}= -A_{1}cosθ, E_{1θ}= A_{1}sinθ,**E**_{1}= -A_{1}**k**.

**D**_{1}= ε**E**_{1}= kε_{0}**E**_{1}.

(ii) Φ is continuous at r = a. A_{0}= A_{0}' + B_{0}'/a, A_{1}a = A_{1}'a + B_{1}'/a^{2}.

(iii) The radial component of**D**is continuous at r = a. (There are no free charges.)

**E**_{2}= -**∇**Φ_{2}. E_{2r}= -A_{1}'cosθ + B_{0}'/r^{2}+ 2(B_{1}'/r^{3})cosθ, E_{2θ}= A_{1}'sinθ + (B_{1}'/r^{3})sinθ,

D_{2}= ε_{0}**E**_{2}.

-kε_{0}A_{1}cosθ = -ε_{0}A_{1}'cosθ + ε_{0}B_{0}'/a^{2}+ 2ε_{0}(B_{1}'/a^{3})cosθ.

B_{0}' = 0, kA_{1}= A_{1}' - 2(B_{1}'/a^{3}) = A_{1}' - 2(A_{1}- A_{1}')

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

-A_{1}'cosθ = E_{0}cosθ, A_{1}'sinθ = -E_{0}sinθ, A_{1}' = -E_{0}.

A_{1}= 3A_{1}'/(k + 2) = -3E_{0}/(k + 2).

B_{1}' =[(k - 1)/(k + 2)]E_{0}a^{3}.

A_{0}= A_{0}' is an arbitrary constant. We choose this constant to be zero. A_{0}= A_{0}' = 0.

Φ_{inside}= -(3E_{0}/(k + 2))rcosθ = -3E_{0}z/(k + 2),

Φ_{outside}= -E_{0}rcosθ + [(k - 1)/(k + 2)](a^{3}/r^{2})E_{0}cosθ.

(b)**E**_{in}= -**∇**Φ_{in}= 3**E**_{0}/(k + 2).

Consider a homogeneous dielectric ε, of infinite
extent, in which there is a uniform field E_{0}. 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 σ_{p} on
the walls.

(c) The field outside the cavity.

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:

Let the center of the cavity be at the origin. Then the problem has azimuthal symmetry.

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 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 Φ_{1}= A_{0}+ A_{1}rcosθ + B_{0}/r + (B_{1}/r^{2})cosθ 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)**E**_{2}=**E**_{0}at r >> a, Φ_{2}= -E_{0}z = -E_{0}rcosθ at r >> a, A_{1}' = -E_{0}, we choose A_{0}' = 0.

Then, from symmetry, A_{0}= 0.

(ii) Φ_{1}is finite at the origin, B_{0}= B_{1}= 0. Φ_{1}= A_{1}rcosθ,

**E**_{1}= -**∇**Φ_{1}, E_{1r}= -A_{1}cosθ, E_{1θ}= A_{1}sinθ,**E**_{1}= -A_{1}**k**.

**D**_{1}= ε_{0}**E**_{1}.

(iii) Φ is continuous at r = a. B_{0}' = 0, A_{1}a = -E_{0}a + B_{1}'/a^{2}.

(iv) The radial component of**D**is continuous at r = a. (There are no free charges.)

**E**_{2}= -**∇**Φ_{2}. E_{2r}= -A_{1}'cosθ + 2(B_{1}'/r^{3})cosθ, E_{2θ}= A_{1}'sinθ + (B_{1}'/r^{3})sinθ,**D**_{2}= kε_{0}**E**_{2}= e**E**_{2}.

-ε_{0}A_{1}cosθ = kε_{0}E_{0}cosθ + 2kε_{0}(B_{1}'/a^{3})cosθ.

A_{1}= -kE_{0}- 2k(B_{1}'/a^{3}) = -kE_{0}- 2k(A_{1}+ E_{0})

A_{1}= -3E_{0}k/(2k + 1) = -3E_{0}ε/(ε_{0 }+ 2ε)

B_{1}' = [(ε_{0 }- ε)/(ε_{0 }+ 2ε)]E_{0}a^{3}.

Φ_{1}= [-3E_{0}ε/(ε_{0 }+ 2ε)]rcosθ.

On the surface Φ(a) = [-3E_{0}ε/(ε_{0 }+ 2ε)]acosθ.

(b) σ_{p}= ε_{0}(E_{2r}(a)- E_{1r}(a) ) = ε_{0}E_{0}cosθ[1 + 2(ε_{0 }- ε)/(ε_{0 }+ 2ε)_{ }+ 3ε/(ε_{0 }+ 2ε)]

σ_{p}= ε_{0}E_{0}cosθ[1 + (2ε_{0 }+ ε))/(ε_{0 }+ 2ε)] = ε_{0}E_{0}cosθ[3(ε_{0 }+ ε))/(ε_{0 }+ 2ε)].

(c) Φ_{2}= -E_{0}rcosθ + [(ε_{0 }- ε)/(ε_{0 }+ 2ε)]E_{0}a^{3}cosθ/r^{2},

E_{2r}= -∂Φ_{2}/∂r = E_{0}cosθ[1 + 2[(ε_{0 }- ε)/(ε_{0 }+ 2ε)]a^{3}/r^{3}],

E_{2θ}= -(1/r)∂Φ_{2}/∂θ = -E_{0}sinθ[1 - [(ε_{0 }- ε)/(ε_{0 }+ 2ε)]a^{3}/r^{3}].

(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**/(3ε_{0}).

(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}^{∞}[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:

∇^{2}Φ = 0 inside and outside the sphere since ρ_{free}= 0 inside and outside the sphere. We are looking for a solution for ∇^{2}Φ = 0, subject to the boundary conditions. - Details of the calculation:

(a) Place the center of the sphere at the origin.

Then ρ_{b }= -**∇**∙**P**= 0, σ_{b }=**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 Φ_{1}= A_{0}+ A_{1}rcosθ inside,

Φ_{2}= 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. A_{0}= B_{0}'/a, A_{1}a = B_{1}'/a^{2}.

(ii) The radial component of**D**is continuous at r = a. (There are no free charges.)

**E**_{1}= -**∇**Φ_{1}, E_{1r}= -A_{1}cosθ, E_{1θ}= A_{1}sinθ,**E**_{1}= -A_{1}**k**.

**D**_{1}= ε_{0}**E**_{1}+**P**

**E**_{2}= -**∇**Φ_{2}. E_{2r}= B_{0}'/r^{2}+ 2(B_{1}'/r^{3})cosθ, E_{2θ}= (B_{1}'/r^{3})sinθ,**D**_{2}= ε_{0}**E**_{2}.

-ε_{0}A_{1}cosθ + P cosθ = ε_{0}B_{0}'/a^{2}+ 2ε_{0}(B_{1}'/a^{3})cosθ.

-ε_{0}A_{1}+ P = 2ε_{0}(B_{1}'/a^{3}), B_{0}' = A_{0}= 0.

-ε_{0}A_{1}+ P = 2ε_{0}A_{1}. A_{1}= P/(3ε_{0}).

**E**_{1}= -A_{1}**k**=**P**/(3ε_{0}).

(b) U = ½CV^{2}, C = ε_{0}A/d = ε_{0}πR^{2}/d, U = ½(ε_{0}πR^{2}/d)V^{2}.

(c) The dielectric sphere is placed in a previously uniform electric field**E**_{0}. ( E_{0}= V/d)

The field inside the sphere will be**E**_{in}= -**∇**Φ_{in}= 3**E**_{0}/(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 = Q^{2}/2C) is proportional to 1/C for a capacitor with constant Q.

The energy stored in the capacitor decreases.

Two very large metal plates are held a distance d apart,
one at potential zero and the other at potential V_{0}. 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}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(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 = V_{0}on the upper plate.

Φ = A_{1}'rcosθ + (B_{1}'/r^{2})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 = A_{1}'a + B_{1}'/a^{2}. B_{1}' = -A_{1}'a^{3}.

(ii)**E**= -**∇**Φ = -E_{0}**k**at r >> a, i.e. far away from the hemispherical boss.

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

Φ(z = d) = V. E_{0}d = V, E_{0}= V/d

Φ = E_{0}rcosθ - (E_{0}a^{3}/r^{2})cosθ.**E**= -**∇**Φ. E_{r}= -E_{0}cosθ - 2(E_{0}a^{3}/r^{3})cosθ, E_{θ}= E_{0}sinθ - (E_{0}a^{3}/r^{3})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.

E_{r}(a) = -3E_{0}cosθ.

**j**= σ**E**is the current density between the plate.

The current flowing to the hemisphere is

∫_{0}^{π /2}j(θ)2πa^{2}sinθ dθ = 6πσE_{0}a^{2}∫_{0}^{π /2}cosθ sinθ dθ = 3πσ E_{0}a^{2}= 3πσ Va^{2}/d = I.

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

A long cylindrical rod of radius a and dielectric constant
k is placed in a uniform electric field **E**_{0}
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}^{∞}(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

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**= E**i**. Assume that all B_{n}and B_{n}' and all A_{n}and A_{n}' 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) = A_{1i}cos(φ)ρ.

ρ > a: -**∇**Φ = E**i**as ρ --> infinity. Φ(ρ,φ) = A_{1o}cos(φ)ρ + A_{1o}'cos(φ)/ρ, A_{1o}= -E.

ρ = a: Φ is continuous,**D**∙**n**is continuous.

A_{1i}cos(φ)a = -Ecos(φ)a + A_{1o}'cos(φ)/a, A_{1i}+ E = A_{1o}'/a^{2}.

εA_{1i}cos(φ) = -ε_{0}Ecos(φ) - ε_{0}A_{1o}'cos(φ)/a^{2}, (ε/ε_{0})A_{1i}+ E = -A_{1o}'/a^{2}.

Algebra: A_{1i}= -E(2ε_{0}/(ε + ε_{0})), A_{1o}' = a^{2}E[(ε - ε_{0})/(ε + ε_{0})]

Φ_{inside}= -E(2ε_{0}/(ε + ε_{0}))cos(φ)ρ, Φ_{outside}= -Ecos(φ)ρ + a^{2}E[(ε - ε_{0})/(ε + ε_{0})]cos(φ)/ρ.

(b)**E**= -**∇**Φ. E_{ρ}= ∂Φ/∂ρ, E_{φ}= (1/ρ)∂Φ/∂φ.**E**_{in}= E(2ε_{0}/(ε + ε_{0}))(cos(φ)[**ρ**/ρ] - sin(φ)[**φ**/φ]) = (2ε_{0}/(ε + ε_{0}))E**i**.**E**_{out}= E**i**+ E(a^{2}/ρ^{2})[(ε - ε_{0})/(ε + ε_{0})](cos(φ)[**ρ**/ρ] + sin(φ)[**φ**/φ]).

(c)**∇∙P**= -ρ_{bound}= -ε_{0}x_{e}**∇∙E**= 0 since**E**= constant inside the rod.**P∙n**= σ_{bound}= ε_{0}x_{e}**E**_{in}**∙**[**ρ**/ρ] = (ε + ε_{0})**E**_{in}**∙**[**ρ**/ρ] = (2ε_{0}(ε - ε_{0})/(ε + ε_{0}))Ecos(φ).