Ch4. (Electric Fields in Matter) 4.1(Polarization) 4.2 (The Field of a Polarized Object) 4.3 (The Electric Displacement) 4.4 (Linear Dielectrics)

4.1.1.1(Polarization) —(Dielectrics) Electric fields in matter: conductors and insulators (dielectrics) In conductors charges will be pushed to the boundary by external field. In dielectrics charges are attached to atoms or molecules. The electric field can distort the charge distribution of a dielectric atom or molecule by two principal mechanisms: stretching and rotating.

4.1.1.2 (Polarization) —(Dielectrics) Many non-conducting solids have permanent dipole moments, or become polarized when immersed in an external electric field. Materials such as these are known as dielectrics. Normally, the dipole moment is zero on large scales since atomic dipoles are oriented in random directions. Immersion of a dielectric in an electric field polarizes atoms and tends to align the atomic dipoles. The induced dipole moment or the total polarization of an atom is approximately proportional to the field:

4.1.2.1 (Induced Dipoles) e : electric susceptibility of the medium

4.1.2.2 (Induced Dipoles) Example : A primitive model for 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. In the presence of an external field E, the nucleus will be shifted slightly to the right and the electron cloud to the left (extremely small). The electric field at a distance d from the center of a uniformly charged sphere is The atomic polarizability

4.1.2.3(Induced Dipoles) Polarizability : Carbon dioxide 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 : The induced dipole moment may not be in the same directions as . For a completely asymmetrical molecule, the most general linear relation between and is : The set of nine constants constitute the polarizability tensor for the molecule.

4.1.2.4 (Induced Dipoles) Problem 4.4 A point charge q is situated a large distance r from a neutral atom of polarizability . Find the force of attraction between them. r The electric field produced by the charge q : The induced dipole moment of the atom : θ~0 The electric field produced by the dipole, at location of q : The force on q due to this field : attractive

4.1.3.1 (Alignment of Polar Molecules) +q (Permanent dipole moment) H+ H+ Force : d 1050 Polar molecule O- -q The electric field is uniform Torque : If the field is non-uniform, Assuming the dipole is very short

4.2.1.1(The Field of a Polarized Object)– (Bound Charges) What can it be if the dipole is not placed at the origin? dipole Assume that the dipole is placed at and the observed is sitting at . The dipole moment in each volume element So the potential is where

4.2.1.2(The Field of a Polarized Object)– (Bound Charges) Note that the differentiation is with respect to the source coordinates Integrating by parts Using the divergence theorem

4.2.1.3(The Field of a Polarized Object)– (Bound Charges) The first term looks like the potential of a surface charge surface charge density : Where is the normal unit vector While the second term looks like the potential of a volume charge volume charge density :

4.2.1.4(The Field of a Polarized Object)– (Bound Charges) Example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R Since is uniform  azimuthally symmetry and spherically symmetrical problem

4.2.1.5(The Field of a Polarized Object)– (Bound Charges) Boundary conditions

4.2.2.1(Physical Interpretation of Bound Charges) The dipole moment qd = PAd q = PA + + + + + + d + +  + If the polarization is non-uniform we get accumulation of bound charge within the material as well as on the surface. + + - - - + - - + - - - + + +

4.2.2.2(Physical Interpretation of Bound Charges) There is another way of analyzing the uniformly polarized sphere: (radius : R) There are two spheres of charge : a positive sphere and a negative sphere Without polarization the two are superimposed and cancel completely. But when the material is uniformly polarized, + + + + + + + + + + d - Bound surface charge b - - - - - - - - - The electric field in the region of overlap between two uniformly charged spheres

4.2.2.3(Physical Interpretation of Bound Charges) Proof Problem 2.18 The electric field inside the positive sphere + The electric field inside the negative sphere - So the total field

4.2.2.4(Physical Interpretation of Bound Charges) Meanwhile, for points outside, When we calculate the electric fields and potential via bound charges, we obtain the “macroscopic” field. the “macroscopic” field : the average field over regions large enough to contain many thousands of atoms. Ordinarily, the macroscopic field is what people mean when they speak of “the” field inside matter.

4.3.1.1 (Gauss’s Law in the Presence of Dielectrics) When we consider the “free” charges alone, the Gauss’s law is expressed as When we include the “bound” charges for the presence of dielectrics, Here we define the electric displacement : Hence the Gauss’s law is now expressed as :

4.3.1.2 (Gauss’s Law in the Presence of Dielectrics) Example 4.4 A long straight wire, carrying uniform line charge , is surrounded by rubber insulation out to a radius a. Find the electric displacement. L Drawing a cylindrical Gaussian surface, of radius s and length L. a s  therefore The formula above for D holds both within the insulation and outside it. Outside region : Inside region : since we do not know , the electric field cannot be determined!

4.3.1.3(Gauss’s Law in the Presence of Dielectrics) Problem 4.15 A thick spherical shell (inner radius a, outer radius b) is made of dielectric material with a “frozen-in” polarization Where k is a constant and r is the distance from the center. Find the electric field in all three regions by two different methods : (a). Locate all the bound charge, and use Gauss’s law to calculate the field it produces. (b). Use to find , and then get from b a

4.3.1.4 (Gauss’s Law in the Presence of Dielectrics) Solution : b (a). a For r < a : Qenc = 0, so For r > b : Qenc = 0, so For a < r < b :

4.3.1.5 (Gauss’s Law in the Presence of Dielectrics) Solution : b (b). a everywhere For r < a and r > b : For a < r < b :

4.3.1.6 (Gauss’s Law in the Presence of Dielectrics) Problem 4.16 Suppose the field inside a large piece of dielectric is , so that the electric displacement is (a). Now a small spherical cavity (Fig. a) is hollowed out of the material. Find the field at the center of the cavity in terms of and . Also find the displacement at the center of the cavity in terms of and . (b). Do the same for a long needle-shaped cavity running parallel to (Fig. b) (c). Do the same for a thin wafer-shaped cavity perpendicular to (Fig. c) a b c

4.3.1.7 (Gauss’s Law in the Presence of Dielectrics) Solution : (a). The electric field at the center of a sphere with uniformly polarization : (see Example 4.2) So the electric field at the center of a spherical cavity

4.3.1.8 (Gauss’s Law in the Presence of Dielectrics) Solution : (b). The electric field of opposite charges at the two far away ends of the needle is very small : So the electric field at the center of a long needle-shaped cavity

4.3.1.9 (Gauss’s Law in the Presence of Dielectrics) Solution : (c). The electric field of a parallel-plate capacitor with upper plate at  = P : So the electric field at the center of a thin wafer-shaped cavity

4.3.2 (A Deceptive Parallel) In vacuum In dielectrics Why? Note that while the curl of E is zero, the curl of D is in general non-zero: In general Therefore D cannot in general be written as the gradient of some scalar potential (unlike E)

4.3.3 (Boundary Conditions) In vacuum In dielectrics

4.4.1.1 (Susceptibility),(Permittivity),(Dielectric Constant) K =

4.4.1.2 (Susceptibility),(Permittivity),(Dielectric Constant) Linear dielectrics :

4.4.1.3 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 at the center. (relative to infinity) For all points r > a : b Q For a < r < b : For r > b : a Inside the metal sphere : The potential at the center :

4.4.1.4 (Susceptibility),(Permittivity),(Dielectric Constant) vacuum dielectric

4.4.1.5 (Susceptibility),(Permittivity),(Dielectric Constant) Example 4.6 A parallel-plate capacitor is filled with insulating material of dielectric constant r. What effect does this have on its capacitance? +++++++++ For upper plate ---------------- For bottom plate

4.4.1.6 (Susceptibility),(Permittivity),(Dielectric Constant) For upper plate +++++++++ ---------------- For bottom plate In dielectric

4.4.1.7 (Susceptibility),(Permittivity),(Dielectric Constant) +++++++++ +++++++++ s s ----------------

4.4.1.8 (Susceptibility),(Permittivity),(Dielectric Constant) +++++++++ +++++++++ +++++++++ s Inside the upper plate : This is true for both slabs (b). In slab 1 In slab 2 (c). In slab 1 In slab 2

4.4.1.9 (Susceptibility),(Permittivity),(Dielectric Constant) In slab 1 constant In slab 2 For both slabs In slab 1 top bottom In slab 2 top bottom

4.4.1.10 (Susceptibility),(Permittivity),(Dielectric Constant) (f). In slab 1, the total surface charge above : the total surface charge below : In slab 2, the total surface charge above : the total surface charge below :

4.4.2.1 (Boundary Value Problems with Linear Dielectrics) In a homogeneous linear dielectric the bound charge density (b) is proportional to the free charge density (f). In particular, unless free charge is embedded in the material,  =0 , and any net charge must reside at the surface. Remember the boundary conditions for dielectric :

4.4.2.2 (Boundary Value Problems with Linear Dielectrics) Example 4.7 A sphere of homogeneous linear dielectric material is placed in an otherwise uniform electric field . Find the electric field inside the sphere. There is no free charge and only bound charges exist on the surface. So the boundary conditions : Inside the sphere : Outside the sphere :

4.4.2.3 (Boundary Value Problems with Linear Dielectrics) for for

4.4.2.4 (Boundary Value Problems with Linear Dielectrics) for for for for

4.4.2.5 (Boundary Value Problems with Linear Dielectrics) Example 4.8 Suppose the entire region below the plane z = 0 is filled with uniform linear dielectric material of susceptibility . Calculate the force on a point charge q situated a distance above the origin. z q  d y r x

4.4.2.6 (Boundary Value Problems with Linear Dielectrics) The surface bound charge on the xy plane is of opposite sign to q, so the force will be attractive. Let us first calculate b : Where Ez is the z-component of the total field just outside the dielectric, at z = 0. The contribution of charge q to the z-component of the total field : The contribution of the bound charge to the z-component of the total field : Therefore

4.4.2.7 (Boundary Value Problems with Linear Dielectrics) Therefore Apart from the factor , this is exactly the same as the induced charge on an infinite conducting plane under similar circumstances. For z > 0 Using the method of images : qb at the image position z = -d For z < 0 Using the method of images : q+qb at the image position z = d

4.4.2.8 (Boundary Value Problems with Linear Dielectrics) We check the boundary condition for the image solution : Right! So the force on q is :

4.4.3.1 (Energy in Dielectric Systems) The energy stored in any electrostatic system : The energy stored in dielectric systems :

4.4.3.2 (Energy in Dielectric Systems) Suppose the dielectric material is fixed in position, and we bring in the free charge. The work done on the incremental free charge : Since If the medium is a linear dielectric, then

4.4.4.1 (Forces on Dielectrics) Fringing field d dielectric The bound surface charge on the dielectric experiences the forces F as shown. The dielectric is pulled into the region of the plates. Let W be the energy of the system. If I pull the dielectric out an infinitesimal distance dx, the energy is changed by an amount equal to the work done : So the electric force on the slab is

4.4.4.2 (Forces on Dielectrics) Let’s assume that the total charge Q on the plates is held constant, as the dielectric moves. Now, the energy stored in the capacitor is So If the potential on the plates is fixed by connecting it up to a battery, the battery also does work as the dielectric moves. So

4.4.4.3 (Forces on Dielectrics) If the potential on the plates is fixed by connecting it up to a battery, the battery also does work as the dielectric moves. So