4. Multipoles and Dielectrics

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

Dipole and Quadrupole Moment • Define total charge • Dipole moment • Quadrupole moment • Then the multipolemoments can be written • Contains same information as the qlm’s

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 Qseems to have six components • It has only 5 independent components because

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

Energy in Terms of Multipole Moments • A charge distribution in abackground potential has energy • If the potential changes slowly ina region with no other charge, then • Substitutingin, we find • Since no charge, • We therefore have

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

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

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 andsome quantity of interest fi in state i, then • Where  = 1/kBT • The states in this case areall possible angles, so • Polarization will be • The energies are • So we have

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

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:

4B. Dielectrics Space Averaging • The formulas at right can be use asdescriptions 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

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 ishifts 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

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

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 thesecoefficients names: • For most materials you can use parity to argue second term vanishes • Third term is negligible unless electric field is really big

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 P E

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 

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

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:

Discontinuity at a Boundary Consider a surface (locally flat) with a free surface charge  • How do E and D changeacross the boundary? • Consider a small thin box of area Acrossing the boundary • Since it is small, assume D is constant over top surface and bottom surface • Use Gauss’s Lawon 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, soonly include the lateral part A L – L

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

Bound Charges A • 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

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 outwardsand depend only on distance r from the center • Gauss’s law applied to any sphere outside the conductor: • In both vacuum and the dielectric, Dand E are proportional, so • While we’re at it, let’s get thebound surface charge density:

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 willdepend only on r, so • Integrate • We want () = 0, so C1 = 0 • We want it continuous at r = b, so

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) E|| D D E|| E

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 • 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 b a

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 b Is our answer correct? • It’s proportional to the answer without the dielectric, so clearlywithin 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 a

Point Charge With Two Dielectrics q • Suppose you have a point charge q a distance h abovea 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 h q'

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 wewould automatically have: • In the region z < 0, we have • Still have to check boundary conditions! h q'

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

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 solutionfor a conducting sphere: • And the solution for  = 0: • We therefore conjectureasolution of the form:

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 inspherical 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

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:

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

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.

4. Multipoles and Dielectrics