X. Things We Skipped

1. X. Things We Skipped XA. Capacitance Revisited Two Component Capacitors • In general, we defined capacitance (chapter 1) as the proportionality constant between the voltage and the charge for a group of conductors • For a general combination of conductors, Cij is defined by: • A very common actual situation is two conductorswith equal and opposite charge on them • In this case we commonly define the capacitance asthe difference between the voltages on the two conductors • With some work, you can show that • I don’t think this formula is ever useful • Though we discussed them before dielectrics, they work well with dielectrics too +Q –Q

2. Energy in Capacitors • In general, the energy in a capacitor canbe most easily computed from the formula: • For a two-component capacitor, the charge is on the two componentsof the capacitor, and on each of them the potential is constant, so • This can be rewritten in terms of the capacitance in two ways:

3. Sample Problem X.1 A parallel plate capacitor has dimensions l w and separation d much smaller than the size. If a dielectric with dielectric constant  fills the space to a distance xalong the length l, what is the capacitance? l • We need to find electric field • Away from edges, it will point straight across the gap • Since E|| is continuous, it will have the same valuein the dielectric and in vacuum • We therefore have • The potential difference is therefore • The displacement is • The surface charge on either surface is • Multiplying by the area, the charge is l – x x

4. Sample Problem X.2 For the capacitor problem in problem X.1, write the total energy for (a) fixed potential difference V, and (b) fixed charge Q. In each of these cases, if the dielectric is free to move, will it be pulled into or expelled from the gap? l • The first part is trivial: • Since for most materials,  > 0,capacitance increases as x increases • If V is fixed, increasing x increases the energy • If Q is fixed, increasing x decreases the energy • Systems tend to go to lower energy • This suggests, incorrectly, that dielectric is (a) expelled and (b) pulled in • In order to maintain constant V, there must be an external source of power • Like a battery • In reality, in both cases the dielectric is drawn in l – x x

5. Sample Problem X.3 (’11 test) (1) A parallel plate capacitor has length l, width w, and separation d, and is filled with two triangular wedges of dielectric constants 1 and 2. Find the capacitance. Assume d is small compared to the other dimensions. l • This is tricky because the boundary is at a weird angle • Fortunately, if d is small, then the boundary is almost parallelto the conducting plates • Keeping in mind that D is continuous, this suggests a constantD pointing straight across the boundary, depending only on x • The electric field will depend on both x and z: • The vertical height of each of theseregions is given, as a function of x, by • The potential difference, is

6. Sample Problem X.3 (’11 test) (2) A parallel plate capacitor has length l, width w, and separation d, and is filled with two triangular wedges of dielectric constants 1 and 2. Find the capacitance. Assume d is small compared to the other dimensions l • The potential is a constant, so we now solve this for D • The surface charge density on the conductor is • We now integrate this over the surface to get the charge: • Make a “u substitution” • So capacitance is

7. XB. Conductivity Conductivity • We have mostly ignored the effects of conductivity, treating objects as perfect conductors or perfect insulators • Perfect conductors: E = 0; Perfect insulators: J = 0 • Realistic materials have finite conductivity • Most materials, at reasonable electric fields, follow Ohm’s law: •  is the (microscopic) conductivity; unit: m-1ohm-1 = m-1-1 • Consider a macroscopic material in the shape of a cylinderwith conductivity  with length L and cross-sectional area A • Apply a potential difference V across it • Electric field is • The current will be • Define the resistance as • Unit is ohm ( = V/A) • Then we have macroscopic Ohm’s law L

8. Sample Problem X.4 (’14 and ’15 exam) (1) Two very large metal plates are held at a distance d apart. One is at potential 0, and the other at potential V0. A metal sphere of radius a (a << d) is sliced in two, and one hemisphere is placed on the grounded plate so that it’s potential is also zero. If the region between the plates is filled with a material with conductivity , determine an expression for the current that flows to the hemisphere. • We need a picture • Since the sphere is small, it will only have a local effect • Background field from just the capacitor • Because the plates are large,we get a constant field • But it won’t be constant near the hemisphere • Goal – find a potential that • Vanishes on the surface z = 0 and r = a • Has asymptotic constant E-field • Because E = – , we must have asymptotically

9. Sample Problem X.4 (’14 and ’15 exam) (2) Two very large metal plates are held at a distance d apart. One is at potential 0, and the other at potential V0. A metal sphere of radius a (a << d) is sliced in two, and one hemisphere is placed on the grounded plate so that it’s potential is also zero. If the region between the plates is filled with a material with conductivity , determine an expression for the current that flows to the hemisphere. • We’ll work in spherical coordinates at the center of the hemisphere • General solutionin sphericalcoordinates is • Somehow, you have to remember that • Since z = rcos, we can therefore get right asymptotic properties just with A10 • This suggests a solution of the form • Need it to vanish at r = a • Solution is easy to find:

10. Sample Problem X.4 (’14 and ’15 exam) (3) Two very large metal plates are held at a distance d apart. One is at potential 0, and the other at potential V0. A metal sphere of radius a (a << d) is sliced in two, and one hemisphere is placed on the grounded plate so that it’s potential is also zero. If the region between the plates is filled with a material with conductivity , determine an expression for the current that flows to the hemisphere. • We can get the current if we knew E: • The electric field is: • We want it on the surface, where it has to be radial anyway • We integrate the current over the surface

11. Sample Problem X.2½ (1) For the capacitor problem in problem X.1, the dielectric is slid into the capacitor while a battery maintains the capacitor at constant voltage V. (a) How much does the energy increase in the capacitor? (b) How much work does the battery do? l • The capacitance increases • Therefore the chargein the capacitor increases • Therefore the energy increases • The battery is continuously moving the charge from one side to the other • The work required to move small charge Q across potential V: • The total work done is therefore l – x x

12. Sample Problem X.2½ (2) For the capacitor problem in problem X.1, the dielectric is slid into the capacitor while a battery maintains the capacitor at constant voltage V. (a) How much does the energy increase in the capacitor? (b) How much work does the battery do? l l – x x • Why aren’t these numbers equal? • We are doing more work than the increase in the capacitor energy • There is also a force pulling the dielectric into the capacitor • This force is doing mechanical work on the dielectric

13. Finite Frequency Conductivity • In the steady state, there is no particular reason for conductivity to be independent of frequency • We will, as always, assume all quantities are proportional to e –it • We always take the real part at the end • Ohm’s Law will now be frequency dependent: • Assume we have a linear medium as well: • Take a look at Ampere’s Law: • This becomes • One way to think of this is an effective complex susceptibility

14. Free or Bound? • What is the difference between free and bound charges? • In a steady electric field, free charges move an unlimited distance, but bound charges move only a finite distance • In the presence of an oscillating field, even a free charge will only move a finite distance • It is therefore an artificial choice to distinguishbetween free and bound charges • All that matters is the effective dielectric constant • If you treat all charges as free, then we would have • If you treat all charges as bound, then effectively the bound charges are just included in  • The presence of actual free electrons shows up as a divergence in Im() at  = 0

15. Energy Absorption • Let’s, for the moment, treat all charge asif it is explicit, rather than bound • Then we have • We recall that electric fields do work on charges: • The power per unit volume will, therefore, be • Recall that actual fields are found bymultiplying by e–it and taking the real part, so • If we do a time average, the terms with e2it go away, so • The real part of  causes energy loss/absorption • If we work with eff, absorption given you information about Im(eff)

16. Sample Problem X.5 Suppose we have a thin gas of charged particles of charge q and mass m with number density n, so thin they don’t collide. Find the conductivity as a function of . • Assume we have an electric field • This will cause the charged particles to accelerate due to • Their positions will thensatisfy the equation • Integrating, we find a velocity • The current is therefore given by • Conductivity is defined by Ohm’s Law • The conductivity is therefore:

17. Plasma Frequency • For a plasma of electrons, the effective dielectric constant is • Define the plasma frequency as • Then we have • Consider an EM wave passing through a thin plasma. We have • The plasma contributes little to , so this is

18. Plasma Frequency (2) • As written, this applies only to free electrons at low density • We ignored any binding of the electrons • We ignored any collisions • At high frequency, electrons move only a tiny distance • Any restoring force is negligible • Collisions are also negligible • At sufficiently high frequency, all electrons are effectively free • What about the nuclei? Why did we ignore them? • Electron mass is 103 or more times lighter than the nuclei

19. XC. The Kramers-Kronig Relations Are All Dielectric Constants Possible? • Let’s work in a notation where we incorporate everything into : • We can find the polarization using the formula • Are all possible complex functions () possible? • The reaction of P (the motion of the charges) must come after the stimulus E • Let G() be the polarization that results a time  after a stimulus E • Then we have • Note that G must be real • Let both P and E be periodic • Substituting, we have • We therefore have

20. Analyticity and Properties of () • Take the complex conjugateof this expression • We therefore have • If we are in an insulator, G must fall off to zero as  goes to infinity • Because the charges have to return to their starting point eventually • This leads to a smooth, differentiable function for all real  • We can also consider this formula for complex  • We note that if Im() > 0, this expression will converge, and we will get a smooth, differentiable function • There are no “poles” or other problems for Im()  0 • We can use contour integration anywhere in the upper half of the complex plane

21. Using Contour Integration • Consider thefollowing integral: • The “P” means principal value • Integral has a pole at ' =  • The “P” means to symmetrically skip the singularity • To find this integral, close contour in the uhp • Justified since  – 0 small when Im() > 0 • We now distort this contour into a tiny half-circle around  • We now write

22. Kramers-Kronig Relations • To get the Kramers-Kronig relations, take real and imaginary parts of this equation • Then substitute   –  for the negative values of , and use

23. Comments on Kramers-Kronig Relations • We note that if we want dispersion (  0), we must have an imaginary part to  • Though not necessarily at the same frequency • It is often easier to measure absorption than it is the dielectric constant • Then Kramers-Kronig can then be used to find the real part

24. Sample Problem X.6 (1) Suppose that absorption measurements indicated that the imaginary part of the permittivity is as given at right.What is the real part of the permittivity? • We use one of theKramers-Kronig relations:

25. Sample Problem X.6 (2) Suppose that absorption measurements indicated that the imaginary part of the permittivity is as given at right.What is the real part of the permittivity? • The principal part only affects the logarithmic term

26. XD. Waves with Complex Numbers Poynting Vector and Complex Notation • Recall for real waves we had • For complex waves, E and Hhave implied time-dependence • The time-averaged Poynting vector is then • We therefore have • Also recall

27. Complex Wave Numbers • Suppose we had a complex permittivity  • Frequency and wave number k are related by: • Because  is complex, k will have an imaginary part • Our electric fieldwill look like • And our magnetic field will look like • Poynting vector is • We note that intensity is falling exponentially, due to absorption

28. Square Roots of Complex Numbers • How do we take the square root of a complex number? • First, write the complex number in the form • The argument can be found from • Then to find the square root, we simply write Sample Problem X.7 What’s the square root of i?

29. Good Conductors • The effective dielectric constant is given by • Many materials have high conductivity, so that • Free electrons move much farther than bound electrons • Especially at low frequency • Conductivity tends to also be real, at low frequencies • In this case, we approximate • As frequencies increase, the current starts to lag the electric field • This implies that  becomes complex • Indeed, at high frequencies, we already talked about the plasma frequency

30. Sample Problem X.8 A wave with frequency  is attempting to propagate through a good conductor with conductivity . At what distance is the intensity down by a factor of 103? • The effective permittivity is • We therefore have • We define the skin depth • The Poynting vector will have space dependence • The intensity will have fallen by 103 when • So the answer is

31. Index of Refraction and So On • The index of refraction of a material was defined as • If  is complex, we can still useall our formulas from before • For example, we can find reflection and transmission probabilities using • The complex results can result in changes in expected behavior • The presence of complex numbers can lead to a phase shift at the boundary • Define the phase shift as the argument Perpendicular Parallel

32. Sample Problem X.9 (1) A wave with frequency is normally incident from vacuum on a good conductor with conductivity  and susceptibility 0. What fraction of the power is reflected, and what is the phase shift for that reflected light? • The conductor has effective permittivity • The index of refraction is then • For normal incidence,both formulas have • For good conductors,  >> 0

33. Sample Problem X.9 (2) A wave with frequency is normally incident from vacuum on a good conductor with conductivity  and susceptibility 0. What fraction of the power is reflected, and what is the phase shift for that reflected light? • The intensity is • Now work on the phase