Polarization Density

6.1.1

The layer of polarized material shown in cross-section in Fig. P6.1.1, having thickness d and surfaces in the planes y = d and y = 0, has the polarization density P = P_o cos x(i_x + i_y).

Figure P6.1.1

(a)	Determine the polarization charge density throughout the slab.
(b)	What is the surface polarization charge density on the layer surfaces?

Laws and Continuity Conditions with Polarization

6.2.1

For the polarization density given in Prob. 6.1.1, with P_o(t) = P_o cos t:

(a)	Determine the polarization current density and polarization charge density.
(b)	Using J_p and _p, show that the differential charge conservation law, (10), is indeed satisfied.

Permanent Polarization

6.3.1^*

A layer of permanently polarized material is sandwiched between plane parallel perfectly conducting electrodes in the planes x = 0 and x = a, respectively, having potentials = 0 and = -V. The system extends to infinity in the y and z directions.

(a) Given that P = P_o cos x i_x, show that the potential between the electrodes is equation GIF #6.116

(b) Given that P = P_o cos y i_y, show that the potential between the electrodes is equation GIF #6.117

6.3.2

The cross-section of a configuration that extends to infinity in the z directions is shown in Fig. P6.3.2. What is the potential distribution inside the cylinder of rectangular cross-section? floating figure GIF #27

Figure P6.3.2

6.3.3^*

A polarization density is given in the semi-infinite half-space y < 0 to be P = P_o cos [(2 / )x]i_y. There are no other field sources in the system and P_o and are given constants.

(a)	Show that _p = 0 and _sp = P_o cos (2 x/ ).
(b)	Show that

6.3.4

A layer in the region -a < y < 0 has the polarization density P = P_o i_y sin (x - x_o). In the planes y = a, the potential is constrained to be = V cos x, where P_o, and V are given constants. The region 0 < y < a is free space and the system extends to infinity in the x and z directions. Find the potential in regions (a) and (b) in the free space and polarized regions, respectively. (If you have already solved Prob. 5.6.12, you can solve this problem by inspection.) floating figure GIF #28

Figure P6.3.5

6.3.5^*

Figure P6.3.5 shows a material having the uniform polarization density P = P_o i_z, with a spherical cavity having radius R. On the surface of the cavity is a uniform distribution of unpaired charge having density _su = _o. The interior of the cavity is free space, and P_o and _o are given constants. The potential far from the cavity is zero. Show that the electric potential is

6.3.6

The cross-section of a groove (shaped like a half-cylinder having radius R) cut from a uniformly polarized material is shown in Fig. P6.3.6. The material rests on a grounded perfectly conducting electrode at y = 0, and P_o is a given constant. Assume that the configuration extends to infinity in the y direction and find in regions (a) and (b), respectively, outside and inside the groove.

Figure P6.3.6

6.3.7

The system shown in cross-section in Fig. P6.3.7 extends to infinity in the x and z directions. The electrodes at y = 0 and y = a + b are shorted. Given P_o and the dimensions, what is E in regions (a) and (b)? floating figure GIF #30

Figure P6.3.7 floating figure GIF #31

Figure P6.3.8

6.3.8^*

In the two-dimensional configuration shown in Fig. P6.3.8, a perfectly conducting circular cylindrical electrode at r = a is grounded. It is coaxial with a rotor of radius b which supports the polarization density P = [P_o r cos ( - )].

(a) Show that the polarization charge density is zero inside the rotor.

(b) Show that the potential functions ^I and ^II respectively in the regions outside and inside the rotor are equation GIF #6.120 equation GIF #6.121

(c) Show that if = t, where is an angular velocity, the field rotates in the direction with this angular velocity.

6.3.9

A circular cylindrical material having radius b has the polarization density P = [P_o (r^m+1/b^m) cos m ], where m is a given positive integer. The region b < r < a, shown in Fig. P6.3.9, is free space. floating figure GIF #32

Figure P6.3.9

(a)	Determine the volume and surface polarization charge densities for the circular cylinder.
(b)	Find the potential in regions (a) and (b).
(c)	Now the cylinder rotates with the constant angular velocity . Argue that the resulting potential is obtained by replacing ( - t).
(d)	A section of the outer cylinder is electrically isolated and connected to ground through a resistance R. This resistance is low enough so that, as far as the potential in the gap is concerned, the potential of the segment can still be taken as zero. However, as the rotor rotates, the charge induced on the segment is time varying. As a result, there is a current through the resistor and hence an output signal v_o. Assume that the segment subtends an angle /m and has length l in the z direction, and find v_o.

6.3.10^*

Plane parallel electrodes having zero potential extend to infinity in the x - z planes at y = 0 and y = d.

(a)	In a first configuration, the region between the electrodes is free space, except for a segmented electrode in the plane x = 0 which constrains the potential there to be V(y). Given V(y), what is the potential distribution in the regions 0 < x and x < 0, regions (a) and (b), respectively?
(b)	Now the segmented electrode is removed and the region x < 0 is filled with a permanently polarized material having P = P_o i_x, where P_o is a given constant. What continuity conditions must the potential satisfy in the x = 0 plane?
(c)	Show that the potential is given by (The method used here to represent is used in Example 6.6.3.)

6.3.11

In Prob. 6.1.1, there is a perfect conductor in the plane y = 0 and the region d < y is free space. What are the potentials in regions (a) and (b), the regions where d < y and 0 < y < d, respectively?

Polarization Constitutive Laws

6.4.1

Suppose that a solid or liquid has a mass density of = 10³ kg/m³ and a molecular weight of M_o = 18 (typical of water). [The number of molecules per unit mass is Avogadro's number (A_o = 6.023 x 10²⁶ molecules/kg-mole) divided by M_o.] This material has a permittivity = 2_o and is subject to an electric field intensity E = 10⁷ v/m (approaching the highest field strength that can be sustained without breakdown on scales of a centimeters in liquids and solids). Assume that each molecule has a polarization qd where q = e = 1.6 x 10^-19 C, the charge of an electron). What is |d|?

Fields in the Presence of Electrically Linear Dielectrics

6.5.1^*

The plane parallel electrode configurations of Fig. P6.5.1 have in common the fact that the linear dielectrics have dielectric "constants" that are functions of x, = (x). The systems have depth c in the z direction.

(a)	Show that regardless of the specific functional dependence on x, E is uniform and simply i_y v/d.
(b)	For the system of Fig. P6.5.1a, where the dielectric is composed of uniform regions having permittivities _a and _b, show that the capacitance is
(c)	For the smoothly inhomogeneous capacitor of Fig. P6.5.1b, = _o (1 + x/l). Show that

Figure P6.5.1

6.5.2

In the configuration shown in Fig. P6.5.1b, what is the capacitance C if = _a (1 + cos x), where 0 < < 1 and are given constants?

6.5.3^*

The region of Fig. P6.5.3 between plane parallel perfectly conducting electrodes in the planes y = 0 and y = l is filled by a uniformly inhomogeneous dielectric having permittivity = _o [1 + _a(1 + y/l)]. The electrode at y = 0 has potential v relative to that at y = l. The electrode separation l is much smaller than the dimensions of the system in the x and z directions, so the fields can be regarded as not depending on x or z.

(a)	Show that D_y is independent of y.
(b)	With the electrodes having area A, show that the capacitance is

Figure P6.5.3

6.5.4

The dielectric in the system of Prob. 6.5.3 is replaced by one having permittivity = _p exp (- y/d), where _p is constant. What is the capacitance C?

6.5.5

In the two configurations shown in cross-section in Fig. P6.5.5, circular cylindrical conductors are used to make coaxial capacitors. In Fig. P6.5.5a, the linear dielectric has a wedge shape with interfaces with the free space region that are surfaces of constant . In Fig. P6.5.5b, the interface is at r = R.

(a) Determine E(r) in regions (1) and (2) in each configuration, showing that simple fields satisfy all boundary conditions on the electrode surfaces and at the interfaces between dielectric and free space.

(b) For lengths l in the z direction, what are the capacitances?

Figure P6.5.5

6.5.6^*

For the configuration of Fig. P6.5.5a, the wedge-shaped dielectric is replaced by one that fills the gap (over all as well as over the radius b < r < a) with material having the permittivity = _a + _b cos² , where _a and _b are constants. Show that the capacitance is equation GIF #6.126

Dielectrics

6.6.1^*

An insulating sphere having radius R and uniform permittivity _s is surrounded by free space, as shown in Fig. P6.6.1. It is immersed in an electric field E_o(t)i_z that, in the absence of the sphere, is uniform.

(a)	Show that the potential is where A = (_s - _o )/(_s + 2_o ) and B = -3_o /(_s + 2_o ).
(b)	Show that, in the limit where _s , the electric field intensity tangential to the surface of the sphere goes to zero. Thus, the surface becomes an equipotential.
(c)	Show that the same solution is obtained for the potential outside the sphere as in the limit _s if this boundary condition is used at the outset.

Figure P6.6.1

6.6.2

An electric dipole having a z-directed moment p is situated at the origin, as shown in Fig. P6.6.2. Surrounding it is a spherical cavity of free space having radius a. Outside of the radius a is a linearly polarizable dielectric having permittivity . floating figure GIF #37

Figure P6.6.2

(a)	Determine and E in regions (a) and (b) outside and inside the cavity.
(b)	Show that in the limit where , the electric field intensity tangential to the interface of the dielectric goes to zero. That is, in this limit, the effect of the dielectric on the interior fields is the same as if the dielectric were a perfect conductor.
(c)	Show that the same interior potential is obtained as in the limit if this boundary condition is used at the outset.

6.6.3^*

In Example 6.6.1, an artificial dielectric is made from an array of perfectly conducting spheres. Here, an artificial dielectric is constructed using an array of rods, each having a circular cross-section with radius R. The rods run parallel to the capacitor plates and hence perpendicular to the imposed electric field intensity. The spacing between rod centers is s, and they are in a square array. Show that, for s large enough so that the fields induced by the rods do not interact, the equivalent electric susceptibility is _c = 2 (R/s)².

6.6.4

Each of the conducting spheres in the artificial dielectric of Example 6.6.1 is replaced by the dielectric sphere of Prob. 6.6.1. Again, with the understanding that the spacing between spheres is large enough to justify ignoring their interaction, what is the equivalent susceptibility of the array?

6.6.5^*

A point charge finds itself at a height h above an infinite half-space of dielectric material. The charge has magnitude q, the dielectric has a uniform permittivity , and there are no unpaired charges in the volume of the dielectric or on its surface. The Cartesian coordinates x and z are in the plane of the dielectric interface, while y is directed perpendicular to the interface and into the free space region. Thus, the charge is at y = h. The field in the free space region can be taken as the superposition of a particular solution due to the point charge and a homogeneous solution due to a charge q_b at y = -h below the interface. The field in the dielectric can be taken as that of a charge q_a at y = h.

(a)	Show that the potential is given by where r = x² + (y h)² + z² and the magnitudes of the charges turn out to be
(b)	Show that the charge is attracted to the dielectric with the force

6.6.6

The half-space y > 0 is filled by a dielectric having uniform permittivity _a, while the remaining region 0 > y is filled by a dielectric having the uniform permittivity _b. Running parallel to the interface between these dielectrics along the line where x = 0 and y = h is a uniform line charge of density . Determine the potentials in regions (a) and (b), respectively.

6.6.7^*

If the permittivities are nearly the same, so that (1 - _a /_b) is small, the qualitative approach to determining the field distribution given in connection with Fig. 6.6.7 can be made quantitative. That is, if is small, the polarization charge induced by the imposed field can be determined to a good approximation and that charge, in turn, used to find the change in the applied field. Consider the following approximate approach to finding the fields in and around the dielectric cylinder of Example 6.6.2.

(a) In the limit where is zero, the field is equal to the applied field, both inside and outside the cylinder. Write this field in polar coordinates.

(b) Show that this field gives rise to _sp = _b E_o cos at the surface of the cylinder.

(c) Find the field due to this induced polarization surface charge and add it to the imposed field to show that, with the first-order contribution of the induced polarization surface charge, the field is equation GIF #6.131

(d) Expand the exact fields given by (21) and (22) to first order in and show that they are in agreement with this result.

6.6.8

As an illustration of how identification of the induced polarization charge can be used in a qualitative determination of the fields, consider the fields between the plane parallel electrodes of Fig. P6.6.8. In Fig. P6.6.8a, there are two layers of dielectric.

(a)	In the limit where = (1 - _a /_b) is zero, what is the imposed E?
(b)	What is the _sp induced by this field at the interface between the dielectrics.
(c)	For _a > _b, sketch the field lines in the two regions. (You should be able to see, from the superposition of the fields induced by this _sp and that imposed, which of the fields is the greater.)
(d)	Now consider the more complicated geometry of Fig. 6.6.8b and carry out the same steps. Based on your deductions, draw a sketch of _sp and E for the case where _b > _a.

Figure P6.6.8

6.6.9

The configuration of perfectly conducting electrodes and perfectly insulating dielectrics shown in Fig. P6.6.9 is similar to that shown in Fig. 6.6.8 except that at the left and right, the electrodes are "shorted" together and the top electrode is also divided at the middle. Thus, the shaped electrode is grounded while the shaped one is at potential V. floating figure GIF #39

Figure P6.6.9

(a)	Determine in regions (a) and (b).
(b)	With the permittivities equal, sketch and E. (Use physical reasoning rather than the mathematical result.)
(c)	Assuming that the permittivities are nearly equal, use the result of (b) to deduce _sp on the interface between dielectrics in the case where _a /_b is somewhat greater than and then somewhat less than 1. Sketch E deduced as the sum of the fields induced by these surface charges and the imposed field.
(d)	With _a much greater that _b, draw a sketch of and E in region (b).
(e)	With _a much less than _b, sketch and E in both regions.

Smoothly Inhomogeneous Electrically Linear Dielectrics

6.7.1^*

For the two-dimensional system shown in Fig. P6.7.1, show that the potential in the smoothly inhomogeneous dielectric is equation GIF #6.132

Figure P6.7.1

6.7.2

In Example 6.6.3, the dielectrics to right and left, respectively, have the permittivities _a = _p exp (- x) and _b = _p exp ( x). Determine the potential throughout the dielectric regions.

6.7.3

A linear dielectric has the permittivity equation GIF #6.133

An electric field that is uniform far from the origin (where it is equal to E_o i_y) is imposed.

(a) Assume that /_o is not much different from unity and find _p.

(b) With this induced polarization charge as a guide, sketch E.