Description
Problem 1) Dielectric Sphere and a charge:
This is a workshop problem the only one group got very far
in solving. It’s worthwhile to do this, as it’s a good example
of how to deal with the boundary conditions for a dielectric.
Consider an insulating sphere of radius R and dielectric
constant 𝜖 and a point charge 𝑞 on the z‐axis with a
position 𝑟⃗ᇱ ൌ 𝑑 𝑧̂.
Due to the spherical boundary conditions and the azimuthal
symmetry of the problem, we know we can use an
expansion in Legendre Polynomials.
For any azimuthally
symmetric function satisfying Laplace’s Equation:
𝐹ሺ𝑟, 𝜃ሻ ൌ ൬𝑎 𝑟
𝑏
𝑟ାଵ ൰ 𝑃ሺcos 𝜃ሻ
And for the point charge:
1
|𝑟⃗െ𝑟⃗ᇱ|
ൌ 𝑟ழ
𝑟வ
ାଵ 𝑃ሺcos 𝜃ሻ
Where 𝑟ழ ൌ 𝑟, for 𝑟൏𝑟ᇱ
, 𝑟ழ ൌ 𝑟ᇱ
, for 𝑟𝑟ᇱ
, 𝑟வ ൌ 𝑟, for 𝑟𝑟ᇱ
, and 𝑟வ ൌ 𝑟ᇱ
, for 𝑟൏𝑟ᇱ
.
A) Define the potentials:
𝜙ௌூሺ𝑟, 𝜃ሻ = The potential due to the sphere for 𝑟൏𝑅 (inside the sphere).
𝜙ௌைሺ𝑟, 𝜃ሻ = The potential due to the sphere for 𝑟𝑅 (outside the sphere).
𝜙ூሺ𝑟, 𝜃ሻ = The potential due to the charge for 𝑟൏𝑑.
𝜙ூூሺ𝑟, 𝜃ሻ = The potential due to the charge for 𝑟𝑑.
Write down Legendre polynomial expansions for the total potential 𝜙் ൌ 𝜙ௌ 𝜙 (sum of the
sphere potential and potential of q) both inside and outside the sphere, for
𝑟൏𝑑. Do not include terms that must be zero for 𝜙் to remain finite everywhere.
B) Using the fact that 𝜙் is continuous at the surface of the sphere, find a relation between the
coefficients 𝑎 and 𝑏 in the Legendre expansions 𝜙ௌூ and 𝜙ௌை.
C) Using the boundary condition that 𝑟̂⋅ 𝐷ሬሬ⃗ is continuous at the surface of the sphere (and that
𝐷ሬሬ⃗ ൌ 𝜖 𝐸ሬ⃗ ൌ െ𝜖 ∇ሬሬ⃗𝜙), solve for the coefficients 𝑎 and 𝑏 in 𝜙ௌூ and 𝜙ௌை.
𝑥
𝑧
𝜖
𝑞, 𝑟⃗ᇱ ൌ 𝑑 𝑧̂
D) The force on the point charge due to the dipole is:
𝐹 ൌ െ𝑞 𝜕 𝜙ௌைሺ𝑟, 0ሻ|ୀௗ
Solve for the force due to the dipole, 𝑙 ൌ 1, term in the expansion for 𝜙ௌை.
2) Electric Dipole Sphere
A “ferroelectric” sphere of radius 𝑅 has a constant polarization field
𝑃ሬ⃗ሺ𝑟⃗ሻ ൌ 𝑃 𝑧̂. In this problem, we’ll find the fields inside and outside of the
sphere.
The properties of electric fields and electric materials are given by:
∇ሬሬ⃗ ൈ 𝐸ሬ⃗ ൌ 0, ∇ሬሬ⃗ ⋅ 𝐷ሬሬ⃗ ൌ 𝜌, ∇ሬሬ⃗ ⋅ 𝐸ሬ⃗ ൌ 1
𝜖
൫𝜌 𝜌௨ௗ൯, 𝐷ሬሬ⃗ ൌ 𝜖𝐸ሬ⃗ 𝑃ሬ⃗,
𝐷ሬሬ⃗ ൌ 𝜖 𝐸ሬ⃗, 𝜌 ൌ െ∇ሬሬ⃗ ⋅ 𝑃ሬ⃗, 𝜎 ൌ 𝑛ො⋅𝑃ሬ⃗ , 𝐷ሬሬ⃗
ୄ, െ 𝐷ሬሬ⃗
ୄ,௨௧ ൌ 𝜎, 𝐸ሬ⃗
∥, ൌ 𝐸ሬ⃗
∥,௨௧
𝐷ሬሬ⃗ is created by free charges, 𝐸ሬ⃗ by free and “bound” charges from macroscopic dipole
moments, and the fields must also satisfy the boundary conditions. We’ll solve for 𝐸ሬ⃗ using the
bound charges, as there are no free charges in the problem.
A) Using the definitions above, determine the bound charges:
𝜌 ൌ ∇ሬሬ⃗ ⋅ 𝑃ሬ⃗, 𝜎 ൌ 𝑛ො⋅𝑃ሬ⃗
Show that 𝜌 ൌ 0 while the bound surface charge has a distribution very similar to what we found on a
conducting sphere in a uniform electric field (Workshop 9 & Hw 4).
B) Using the bound surface charge, write an integral that can be solved for the potential 𝜙ሺ𝑟⃗ሻ. This
should be an application of Coulomb’s law.
C) Solve your integral from (B). A couple of hints that will do most of the work:
cosሺ𝜃ᇱ
ሻ
|𝑟⃗െ𝑟⃗ᇱ|
ൌ 𝑟ழ
𝑟வ
ାଵ
𝑃ሺ𝑟̂
ᇱ ⋅ 𝑟̂ሻ cosሺ𝑟̂
ᇱ ⋅ 𝑟̂ሻ
ඵ𝑑Ω′ 𝑃ሺ𝑟̂
ᇱ ⋅ 𝑟ଵ̂ ሻ 𝑃ሺ𝑟̂
ᇱ ⋅ 𝑟ଶ̂ ሻ ൌ 4𝜋
2 𝑙 1 𝑃ሺ𝑟ଵ̂ ⋅ 𝑟ଶ̂ ሻ 𝛿
Where: 𝑟ழ is the smaller of 𝑟 and 𝑟′, 𝑟வ is the smaller of 𝑟 and 𝑟′, and the 2D integral over 𝑑Ω′ is over the
solid angle of 𝑟⃗′, or 𝑑Ωᇱ ൌ sin 𝜃′ 𝑑𝜃ᇱ 𝑑𝜙′.
D) Solve for the fields 𝐸ሬ⃗ and 𝐷ሬሬ⃗ both outside and inside the sphere. You’ll get something very much like
the results from Problem 1. Show that, again, the boundary conditions are met.
E) Draw pictures of the 𝐸ሬ⃗ and 𝐷ሬሬ⃗ fields. How do these compare with the fields from Problem 1?
𝑥
𝑧
𝑷ሬሬ⃗
3) Magnetic Sphere:
Interestingly, the methods used in Problem 2 can also solve for the fields due to a
magnetized sphere.
The properties of magnetic fields and magnetic materials are given by:
∇ሬሬ⃗ ൈ 𝐻ሬ⃗ ൌ 𝐽⃗
, ∇ሬሬ⃗ ൈ 𝐵ሬ⃗ ൌ 𝜇൫𝐽⃗
𝐽⃗
൯, ∇ሬሬ⃗ ⋅ 𝐵ሬ⃗ ൌ 0, 𝐵ሬ⃗ ൌ 𝜇 ൫𝐻ሬ⃗ 𝑀ሬሬ⃗൯,
𝐵ሬ⃗ ൌ 𝜇 𝐻ሬ⃗, 𝐽⃗
ൌ ∇ሬሬ⃗ ൈ 𝑀ሬሬ⃗, 𝐾ሬ⃗
ൌ 𝑛ොൈ𝑀ሬሬ⃗
𝐵ሬ⃗
ୄ, ൌ 𝐵ሬ⃗
ୄ,௨௧, 𝐻ሬ⃗
∥, െ 𝐻ሬ⃗
∥,௨௧ ൌ 𝐼
𝐻ሬ⃗ is created by free currents, 𝐵ሬ⃗ by free and “bound” currents from macroscopic magnetic
moments, and the fields must also satisfy the boundary conditions at the surface of the
materials.
To do this, we’ll use the concepts of the “Magnetic Scalar Potential” and “Magnetic Bound Charges”:
If there are no free currents, the curl of 𝐻ሬ⃗ is zero, so we can define a potential 𝜓:
∇ൈ𝐻ሬ⃗ ൌ 0 ⇒ 𝐻ሬ⃗ ൌ െ∇ሬሬ⃗ 𝜓
Considering the divergence equation:
∇ሬሬ⃗ ⋅ 𝐵ሬ⃗ ൌ ∇ሬሬ⃗ ⋅ ൫𝐻ሬ⃗ 𝑀ሬሬ⃗൯ ൌ 0 ⇒ ∇ሬሬ⃗ ⋅ 𝐻ሬ⃗ ൌ െ∇ሬሬ⃗ ⋅ 𝑀ሬሬ⃗
And the boundary condition ൫𝐵ሬ⃗
ଶ െ 𝐵ሬ⃗
ଵ൯⋅𝑛ො ൌ 0 related the field 𝐵ሬ⃗ outside and inside the magnetic
material requires:
൫𝐻ሬ⃗
ଶ 𝑀ሬሬ⃗
ଶ൯⋅𝑛ො ൌ ൫𝐻ሬ⃗
ଵ 𝑀ሬሬ⃗
ଵ൯⋅𝑛ො ⇒ ൫𝐻ሬ⃗
ଶ െ 𝐻ሬ⃗
ଵ൯⋅𝑛ොൌ𝑛ො⋅𝑀ሬሬ⃗
This means the solution for 𝐻ሬ⃗ is the same as for and electrostatics problem with the analogy of a scalar
potential and “magnetic charges” (a problem‐solving device, not real magnetic charges):
∇ሬሬ⃗ ൈ 𝐻ሬ⃗ ൌ 0, ∇ሬሬ⃗ ⋅ 𝐻ሬ⃗ ൌ 𝜌, ∇ଶ𝜓 ൌ 𝜌, 𝜌 ൌ െ∇ሬሬ⃗ ⋅ 𝑀ሬሬ⃗, 𝜎 ൌ 𝑛ො⋅𝑀ሬሬ⃗
Considering this analogy and your solution to Problem 2 above, solve for the field 𝐻ሬ⃗ and 𝐵ሬ⃗ for the
magnetized sphere.
Note: The solution to Workshop 12, Problem 1 will be (is) posted for comparison.
Problem 4: Write and solve a problem that you think could be on the E&M Qualifier.
𝑥
𝑧
𝑴ሬሬሬ⃗
