Chapter 4. Electric Fields in Matter

Transcription

1 Chapter 4. Electric Fields in Matter

2 4.1.2 Induced Dipoles What happens to a neutral atom when it is placed in an electric field E? The atom now has a tiny dipole moment p, in the same direction as E. Typically, P is approximately proportional to the field. is called atomic polarizability Example 4.1 An atom consists of a point nucleus (+q) surrounded by a uniformly charged spherical cloud ( -q) of radius a. Calculate the atomic polarizability of such an atom. It is reasonable to assume that the electron cloud retains its spherical shape. Equilibrium occurs when the nucleus is displaced a distance d from the center of the sphere and pulling by the internal field produced by the electron cloud, E e : At equilibrium, E = E e

3 Polarizability tensor Carbon dioxide (CO 2 ), for instance, when the field is at some angle to the axis, you must resolve it into parallel and perpendicular components, and multiply each by the pertinent polarizability: A most general linear relation between E and p is The set of nine constants ij polarizability tensor By choose "principal" axes, we can leave just three nonzero polarizabilities: xx, yy, and zz.

4 4.1.4 Polarization What happens to a piece of dielectric material when it is placed in an electric field? A lot of little dipoles pointing along the direction of the field. The material becomes polarized. P lim pi / C/m i 0 2 : Total dipole moment per unit volume Polarization This Polarization vector will displace the electric strength in a dielectric medium: : D is known as the electric displacement : For linear dielectrics : Permittivity : Relative permittivity

5 4.2 The Field of a Polarized Object What is the field produced by P in a polarized material? (not the field that may have caused the polarization, but the field the polarization itself causes) Bound Charges The polarization P can produces bound volume and surface charges. b P b Pn

6 4.2.1 Bound Charges For a single dipole p, the potential is, A dipole moment in each volume element d : the differentiation is with respect to the source coordinates (r') Integrating by parts, Using the divergence theorem, : Volume charge density : Surface charge density

7 4.2.2 Physical Interpretation of Bound Charges -q +q If the ends have been sliced off perpendicularly, For an oblique cut, If the polarization is nonuniform, we get accumulations of bound charge within the material. The net bound charge in a given volume is equal and opposite to the amount that has been pushed out through the surface:

8 Electric field (E) produced by a uniformly polarized (P) sphere The field inside a uniformly charged sphere is A uniformly polarized sphere can be regarded as the case when two spheres are overlapped: One is positive and the other is negative in charge. p E? The field in the region of overlap between two uniformly charged spheres is d

9 4.3 The Electric Displacement Gauss's Law in the Presence of Dielectrics The effect of polarization (P) is to produce accumulations of bound charge: : within the dielectric : on the surface If any free charge exists within a dielectric, therefore, the total charge density can be written as Gauss's law reads f : free charge density D E P 2 0 ( C/ m ) Electric displacement In a typical problem, we know f, but we do not (initially) know b Similar to Gauss's law Whenever the requisite symmetry is present, we can immediately calculate D by the Gauss's law method, because it makes reference only to free charges.

10 Gauss's Law: Problem 4.15 A thick spherical shell (inner radius a, outer radius b) is made of dielectric material with a "frozen-in" polarization P(r). (No free charge inside) Find the electric field in all three regions by two different methods: (a) Identify all the bound charge, and use Gauss's law to calculate the field it produces. (divergence theorem) (b) Use the equations of electric displacement D: Much simple to use D.

11 Gauss's Law: Problem 4.16 Suppose the field inside a large piece of dielectric is E 0, so that the electric displacement is: Now a small spherical cavity is hollowed out of the material. Find the field E at the center of the cavity in terms of E 0 and P. Also find the displacement D at the center of the cavity in terms of D 0 and P. E 0 P D 0 = 0 E 0 + P E & D =? Assume the cavities are small enough that P, E 0, and D 0 are essentially uniform. Hint Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization. P -P E = E 0 (E P ) E P = - P/3 0

12 4.3.2 A Deceptive Parallel: Misleading in comparison between E and D Let s compare it with Gauss's law Only the total charge density is replaced by the free charge density f, D is substituted for o E. D is "just like" E (apart from the factor o ), except that its source is f instead of But, D is completely different from E: There is no "Coulomb's law" for D: In the case of electrostatic fields, the curl of E is always zero. But the curl of D is not always zero. Because D cannot be expressed as the gradient of a scalar. there is no "potential" for D.

13 4.3.3 Boundary Conditions D D da D a D a S a D D S D D S f 1 n2 2 n1 n n 2n S d h S V f f D D 1n 2n f f lim h h0 f C D P Ddl D l D l l D D D D D D P P 1t 2t 1t 2t t 2t Note for E: Summary of Boundary conditions E enclosed f 1n E2n 0 0 b E 1t E2t 0 f b

14 4.4 Linear Dielectrics Susceptibility, Permittivity, Dielectric Constant P E 0 e e In linear dielectrics, P is proportional to E, provided E is not too strong. : Electric susceptibility (It would be a tensor in general cases) D EP E e D E r 1 e : Permittivity 0 1 e 0 : Relative permittivity (Dielectric constant) Note that E is the total field from free charges and the polarization itself. If, for instance, we put a piece of dielectric into an external field E 0, we cannot compute P directly from P E; 0 e E E 0 E 0 produces P, P will produce its own field, this in turn modifies P, which... Breaking where?

15 Susceptibility, Permittivity, Dielectric Constant Problem 4.41 In a linear dielectric, the polarization is proportional to the field: P = 0 e E. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = E. What is the relation between atomic polarizability and susceptibility e? Note that, the atomic polarizability was defined for an isolated atom subject to an external field coming from somewhere else, E else p = E else For N atoms in unit volume, the polarization can be set There is another electric field, E self, produced by the polarization P: Therefore, the total field is The total field E finally produce the polarization P: or Clausius-Mossotti formula

16 Susceptibility, Permittivity, Dielectric Constant Example 4.5 A metal sphere of radius a carries a charge Q. It is surrounded, out to radius b, by linear dielectric material of permittivity. Find the potential V at the center (relative to infinity). To compute V, we need to know E. Because it has spherical symmetry, calculate D first. Inside the metal sphere, Outside the metal sphere, The potential at the center Note that the polarization P in the dielectric is

17 The two vectors of E and D is parallel in linear dielectric. In linear dielectrics, P E 0 e D E P E E e E 0 D 0?? since E and D are parallel??? Unfortunately, it does not in genreal: BUT, if the space is entirely filled with a homogeneous ( e is the same in all position) linear dielectric, Compare and 0 E 0 Conclusion: When all space is filled with a homogeneous linear dielectric, the field everywhere is simply reduced by a factor of one over the dielectric constant. For example, if a free charge q is embedded in a large dielectric It becomes weaker! Because P shield the charge.

18 4.4.2 Boundary Value Problems with Linear Dielectrics b P The bound charge density is proportional to the free charge density. If f = 0; b = 0 Any net charge must resident at the surface Within a dielectric when f = 0; the potential obeys Laplace's equation. D D 1n 2n f (If embedded in vacuum) OR

19 Summary on Boundary Conditions for Electrostatics Electric field f b Field displacement Potential

20 Boundary Value Problems with Linear Dielectrics Problem 4.23 Find the electric field inside the sphere ( r ). Use the following method of successive approximations: First pretend the field inside is just E 0 Write down the resulting polarization P 0 E 0 r P 0 generates a field of its own, E 1 E 1 modifies the polarization by an amount P 1 P 1 E 2 and so on. The resulting field is 1 E E0 E0 Ee /3E /3 1 e 1 1 P 0 ee E0 E P E E0 P E P if E

21 Energy in Dielectric Systems Let s prove it. As we bring in free charge, a bit at a time, the work done on the increamental free charge f is since The divergence theorem turns the first term into a surface integral, which vanishes over all of space. If the medium is a linear dielectric, Total work done = Stored energy

22 Forces on Dielectrics f + b = f x b Q 1 0w 0w C wx w( l x) f rl ( r 1) x rl ex V V d d l The energy stored in the capacitor with a constant charge Q is (the dielectric is pulled into the capacitor)