1. 4. Multipoles and Dielectrics 4A. Multipole Expansion Revisited
Switching to Cartesian Coordinates
Before we defined the multipole moments of a charge distribution
Let’s work out these explicitly for l = 1, 2, 3
Write it out in Cartesian coordinates

2. Dipole and Quadrupole Moment
Define total charge
Dipole moment
Quadrupole moment
Then the multipole moments can be written
Contains same information as the qlm’s

3. The Potential in Terms of p’s and Q’s
The potential is then
Substitute in the qlm’s and write out the potential explicitly
We find:
Comments:
The multipole moments depend on choice of origin
If you shift the charge distribution by an amount a, these components shift by
So the leading order non-vanishing term is independent of choice of origin
Note that qlm’s have 2l + 1 different components for each value of l
q and p have 1 and 3 components, and Q seems to have six components
It has only 5 independent components because

4. Electric Field from a Dipole
Consider a dipole oriented in an arbitrary direction
The electric field will be:
We can put it at an arbitrary point x0 instead of the origin
n is unit vector from x0 to x

5. Energy in Terms of Multipole Moments
A charge distribution in a background potential has energy
If the potential changes slowly in a region with no other charge, then
Substituting in, we find
Since no charge,
We therefore have

6. Energy, Dipoles, and Torque
Recall that E = –
A dipole’s energy depends only on electric field
For example, the energy between two dipoles will be given by
An object feels torque when energy depends on angle
The torque is related to energy by
If you write it as a vector, this works out to
Torque tries to make dipole align with E-field

7. Sample Problem 3.1
Two dipoles have identical dipole moments p oriented in the z-direction. If they are separated by a distance r at an angle  compared to the z-axis, find the force on one.
Energy of two dipoles:
Force on upper dipole:  r

8. Sample Problem 3.2 (1)
A molecule in a dilute gas has dipole moment of magnitude p and is at temperature T. What is the expectation value of the dipole moment in the presence of an electric field E in the z-direction? If the electric field is weak, show that it is proportional to E
Need to know some statistical mechanics
Because it is a gas, presumably the molecules don’t interact with each other
Label states by i, with energy Wi and some quantity of interest fi in state i, then
Where  = 1/kBT
The states in this case are all possible angles, so
Polarization will be
The energies are
So we have

9. Sample Problem 3.2 (2)
A molecule in a dilute gas has dipole moment of magnitude p and is at temperature T. What is the expectation value of the dipole moment in the presence of an electric field E in the z-direction? If the electric field is weak, show that it is proportional to E
Do the  integral
x and y directions vanish

10. Sample Problem 3.2 (3)
A molecule in a dilute gas has dipole moment of magnitude p and is at temperature T. What is the expectation value of the dipole moment in the presence of an electric field E in the z-direction? If the electric field is weak, show that it is proportional to E
Strong field limit: Ep >> 1
Weak field limit: Ep << kBT:

11. 4B. Dielectrics Space Averaging
The formulas at right can be use as descriptions of the microscopic fields
In the presence of materials, there can be very large electric fields inside
For example, 0.1 nm from a proton, the electric field is
Much larger than external fields that we would normally encounter
They also vary spatially very quickly
These don’t really interest us very much
To get rid of them, do some space averaging
On a scale large compared to atoms
Then easy to show

12. Dealing With Bound Charges
Now, there are two contribution to electric fields
Net charge that is in conductors, insulators, or free charges 
Net neutral charges in insulators, or bound charges b
The bound charges would normally have no net charge, so think of it as overlapping regions of cancelling charge:
Because the charges cancel, normally any volume would have Qb(V) = 0
But if the charge density of type i shifts by an amount di(x), then a small part of the charge may leave the volume
At a point on the surface x, if there is displacement di(x), then dot this into the normal to find the amount of charge leaving the volume
Add it up over the surface

13. Electric Polarization and Electric Displacement
Define the electric polarization
Then the bound charge in a volume V is
Use divergence theorem
The bound charge in a volume is also
Therefore,
We note that
Define the electric displacement as
Then in macroscopic media, we have

14. Constitutive Equations
These two equations alone are not generally sufficient to find E and D
We need to find a relationship between E and D
Normally, P will be a function of E
Not always!
This is called a constitutive equation
For not too large a field, you can generally Taylor expand P:
Almost never does the first term contribute
Give these coefficients names:
For most materials you can use parity to argue second term vanishes
Third term is negligible unless electric field is really big

15. Susceptibility and Dielectric Constant
The constant ij are the electric susceptibility tensor
Can be shown it is symmetric
If it is not proportional to the identity matrix, then P will not be parallel to E
This does happen in lots of interesting crystals
However, it will be parallel in liquids, gasses, amorphous solids, and many crystals
Such as those with cubic symmetry
We then define the dielectric constant
Then we have:
When this is true, then E P

16. Sample Problems 3.3
Redo all previous problems if we replace vacuum with a material with dielectric constant .
We previously were effectively solving:
Now we are solving:
Solution: copy all previous answers and replace 0 by 

17. Comments on Susceptibility
Air and other gases have susceptibility that are small, so
Liquids and solids can have susceptibilities of order 1, or even much higher
Under ordinary circumstances (at zero frequency),
Later we will often be working with oscillating fields
Then the susceptibility can depend on the frequency
It can even be complex!
We’ll find out what that means later

18. Gauss’s Law, Potential, etc.
Since   E = 0, we still can write
But we must redo Gauss’s Law:
Consider the charge in any volume V
Use the divergence theorem:

19. Discontinuity at a Boundary
Consider a surface (locally flat) with a free surface charge 
How do E and D change across the boundary?
Consider a small thin box of area A crossing the boundary
Since it is small, assume D is constant over top surface and bottom surface
Use Gauss’s Law on this small box
Charge inside the box is A
Since box is thin, ignore lateral surface
Normally, no charge on such a surface
Consider a small loop of length L penetrating the surface
Use the identity
Ends are short, so only include the lateral part A L
– L

20. Dielectric Next to Conductor
Conductors have E = D = 0 inside
Since E|| is continuous, E|| = 0 in the dielectric
Since D = , we would have D =  in the dielectric
If D  E, then D

21. Bound Charges
When we are in a homogenous, isotropic medium, in a region with no free charge,
The bound charge is given by
No bound charges in interior of homogenous isotropic regions if no free charge
On the surface, there will be bound charges
Consider a small, thin box of area A penetrating the surface
The charge inside will be
Use the divergence theorem
Lateral surface is small
No polarization on the exterior
Normal vector n on box is opposite to normal vector to surface A

22. Sample Problems 3.4 (1)
A conducting sphere of radius a with charge Q on it is surrounded by a dielectric with dielectric constant  of radius b. Find the E, D, and the potential everywhere, and the surface bound charge at b . Q
Inside the conductor, E = D = 0
By symmetry, D and E are radially outwards and depend only on distance r from the center
Gauss’s law applied to any sphere outside the conductor:
In both vacuum and the dielectric, D and E are proportional, so
While we’re at it, let’s get the bound surface charge density:

23. Sample Problems 3.4 (2)
A conducting sphere of radius a with charge Q on it is surrounded by a dielectric with dielectric constant  of radius b. Find the E, D, and the potential everywhere, and the surface bound charge at b. Q
Find the potential from
Clearly, potential will depend only on r, so
Integrate
We want () = 0, so C1 = 0
We want it continuous at r = b, so

24. 4C. Solving Problems with Dielectrics
Guess and Check
It is generally hard to find exact solutions to problems with dielectrics
“Guess and check” will often work, if you can make a good guess
How does one make intelligent guesses?
In the limit  = 0, a dielectric acts like vacuum
In the limit  = , a dielectric acts like a conductor
Often, can use conductors as inspiration
What do you have to check in problems with dielectrics?
Always must have   E = 0 and D =  within any region
Usually automatic for the right guesses
On dielectric boundaries, must have: E|| and D continuous
E|| continuous can be enforced by having a continuous potential
On conducting boundaries, must have E|| = 0 (equivalent to  constant) D
E|| E

25. Sample Problems 3.5 (1)
Two long coaxial cylinders of radii a and b are half-filled with a dielectric with dielectric constant  between them. If a charge per unit length  is put on the inner one and – on the outer one, find the voltage difference a b
Guess: fields (E or D) will be radial outwards
Guess: within each region they probably depend on just r
Q: Which one of the following makes sense?
The fields are parallel at the boundary
Therefore E will be continuous across it
So E is the one that works
So the displacement field is:
Now use Gauss’s Law on a cylindrical surface of radius r and length L

26. Sample Problems 3.5 (2)
Two long coaxial cylinders of radii a and b are half-filled with a dielectric with dielectric constant  between them. If a charge per unit length  is put on the inner one and – on the outer one, find the voltage difference a b
Is our answer correct?
It’s proportional to the answer without the dielectric, so clearly within each region
And therefore
E|| is continuous across the boundary
D = 0 is continuous at the boundary
E|| = 0 at the conductors
So it is correct
The potential difference is then

27. Point Charge With Two Dielectricsq
Suppose you have a point charge q a distance h above a dielectric region, so all of z < 0 has constant 
If it were a conductor instead, we would have:
Charge q at z = h plus image charge –q at z = – h for z > 0
No field at all for z < 0
If it were vacuum instead, we would have
Charge q at z = h for z > 0
Charge q at z = h for z < 0
We guess that the correct solution is somewhere between these extremes
For z > 0, conjecture original charge at z = h and image charge q' at z = –h
For z < 0, it looks like original charge diminished to q'' at z = +h h q' h

28. Guess and Check q h
We have a conjectured form for the fields, but is it right?
In the region z > 0, this is field from two charges, so we would automatically have:
In the region z < 0, we have
Still have to check boundary conditions! q' h

29. Matching at the Boundaryq h
Electric fields at the boundary are:
Match E|| at boundary:
Match D at boundary:
Two equation in two unknowns q' h

30. Sample Problems 3.6 (1)
A dielectric sphere of radius a and dielectric constant  is placed in a uniform electric field of magnitude E. Find the potential  everywhere.
Let’s do this with potentials
Let external field be in the z-direction, then
The external potential is
We know the solution for a conducting sphere:
And the solution for  = 0:
We therefore conjecture a solution of the form:

31. Sample Problems 3.6 (2)
A dielectric sphere of radius a and dielectric constant  is placed in a uniform electric field of magnitude E. Find the potential  everywhere.
We have no free charge, so in regions of constant , we need
The general solution to this in spherical coordinates is
By inspection,  will satisfy Laplace’s equation
It clearly has correct asymptotic value as r  
The potential must be continuous at r = a, so
Since  continuous, it will automatically satisfy E|| continuous
We still have to match D

32. Sample Problems 3.6 (3)
A dielectric sphere of radius a and dielectric constant  is placed in a uniform electric field of magnitude E. Find the electric field everywhere.
We calculate D in each region:
Solve the equations:

33. 4C. Energy in the Presence of Dielectrics
One Step at a Time
Suppose we have a collection of dielectrics, initially with no charges
Now imagine adding some charge free charges (x) a little at a time
The work required is
The small change in charge causes a change in D:
Use product rule
Use divergence theorem and fact that
Potential at infinity is zero, so

34. Necessity of Linearity
Now, assume the response is linear:
Then we have
And therefore:
Add up all the little charges
Assume that W = 0 if there is no fields
If desired, you can integrate by parts again to get an equivalent formula
Both of these formulas, however, assume linearity.