1. A hydrogen atom (with the Bohr radius of half an angstrom) is
situated between two metal plates 1 mm apart, which are connected to
opposite terminals of a 500 V battery. What fraction of the atomic
radius does the separation distance d amount to, roughly? Estimate the
voltage you would need with this apparatus to ionize the atom. [Use the
value of α in Table 4.1. Moral: The displacements we’re talking about
are minute, even on an atomic scale.] Get solution
2. According to quantum mechanics, the electron cloud for a hydrogen atom in the ground state has a charge density ... where q is the charge of the electron and a is the Bohr radius. Find the atomic polarizability of such an atom. [Hint: First calculate the electric field of the electron cloud, Ee(r ); then expand the exponential, assuming r ≫a.1 Get solution
3. According to Eq. 4.1, the induced dipole moment of an atom is proportional to the external field. This is a “rule of thumb,” not a fundamental law, and it is easy to concoct exceptions—in theory. Suppose, for example, the charge density of the electron cloud were proportional to the distance from the center, out to a radius R. To what power of E would p be proportional in that case? Find the condition on ρ(r ) such that Eq. 4.1 will hold in the weak-field limit. Reference equation 4.1 p = αE. Get solution
4. A point charge q is situated a large distance r from a neutral atom of polarizability α. Find the force of attraction between them. Get solution
5. In Fig. 4.6, p1 and p2 are (perfect) dipoles a distance r apart. What is the torque on p1 due to p2?What is the torque on p2 due to p1? [In each case, I want the torque on the dipole about its own center. If it bothers you that the answers are not equal and opposite, see Prob. 4.29.] ... Reference prob 4.29 (a) For the configuration in Prob. 4.5, calculate the force on p2 due to p1, and the force on p1 due to p2. Are the answers consistent with Newton’s third law? (b) Find the total torque on p2 with respect to the center of p1, and compare it with the torque on p1 about that same point. [Hint: combine your answer to (a) with the result of Prob. 4.5.] Get solution
6. A (perfect) dipole p is situated a distance z above an infinite grounded conducting plane (Fig. 4.7). The dipole makes an angle θ with the perpendicular to the plane. Find the torque on p. If the dipole is free to rotate, in what orientation will it come to rest? Reference figure 4.7 ... Get solution
7. Show that the energy of an ideal dipole p in an electric field E is given by ... Get solution
8. Show that the interaction energy of two dipoles separated by a displacement r is ... [Hint: Use Prob. 4.7 and Eq. 3.104.] Reference prob 4.7Show that the energy of an ideal dipole p in an electric field E is given by ... Reference equation 3.104 ... Get solution
9. dipole p is a distance r from a point charge q, and oriented so that p makes an angle θ with the vector r from q to p. (a) What is the force on p? (b) What is the force on q? Get solution
10. A sphere of radius R carries a polarization P ( r ) = kr, where k is a constant and r is the vector from the center. (a) Calculate the bound charges σb and ρb. (b) Find the field inside and outside the sphere. Get solution
11. A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Get solution
12. Calculate the potential of a uniformly polarized sphere (Ex. 4.2) directly from Eq. 4.9. Reference equation 4.9 ... Reference example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R. Get solution
13. A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder. Show that the field outside the cylinder can be expressed in the form ... [Careful: I said “uniform,” not “radial”!] Get solution
14. When you polarize a neutral dielectric, the charge moves a bit, but the total remains zero. This fact should be reflected in the bound charges σb and ρb. Prove from Eqs. 4.11 and 4.12 that the total bound charge vanishes. Reference equation 4.11 ... Reference equation 4.12 ... Get solution
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 (Fig. 4.18). (There is no free charge in the problem.) Find the electric field in all three regions by two different methods: ... 6The polarization drops abruptly to zero outside the material, so its derivative is a delta function (see Prob. 1.46). The surface bound charge is precisely this term—in this sense it is actually included in ρb, but we ordinarily prefer to handle it separately as σb. (a) Locate all the bound charge, and use Gauss’s law (Eq. 2.13) to calculate the field it produces. (b) Use Eq. 4.23 to find D, and then get E from Eq. 4.21. [Notice that the second method is much faster, and it avoids any explicit reference to the bound charges.] Reference equation 4.23 ... Reference equation 4.21 ... Reference equation 2.13 ... Reference Prob. 1.46 (a) Show that ... [Hint: Use integration by parts.] (b) Let θ(x) be the step function: ... Show that dθ/dx = δ(x). Get solution
16. Suppose the field inside a large piece of dielectric is E0, so that the electric displacement is ... (a) Now a small spherical cavity (Fig. 4.19a) is hollowed out of the material. Find the field at the center of the cavity in terms of E0 and P. Also find the displacement at the center of the cavity in terms of D0 and P. Assume the polarization is “frozen in,” so it doesn’t change when the cavity is excavated. (b) Do the same for a long needle-shaped cavity running parallel to P (Fig. 4.19b). (c) Do the same for a thin wafer-shaped cavity perpendicular to P (Fig. 4.19c). Assume the cavities are small enough that P, E0, and D0 are essentially uniform. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization.] Figure 4.19 ... Get solution
17. For the bar electret of Prob. 4.11, make three careful sketches: one of P, one of E, and one of D. Assume L is about 2a. [Hint: E lines terminate on charges; D lines terminate on free charges.] Reference prob 4.11 A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Get solution
18. The space between the plates of a parallel-plate capacitor (Fig. 4.24) is filled with two slabs of linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5. The free charge density on the top plate is σ and on the bottom plate −σ. Reference fig 4.24 ... (a) Find the electric displacement D in each slab. (b) Find the electric field E in each slab. (c) Find the polarization P in each slab. (d) Find the potential difference between the plates. (e) Find the location and amount of all bound charge. (f) Now that you know all the charge (free and bound), recalculate the field in each slab, and confirm your answer to (b). Get solution
19. Suppose you have enough linear dielectric material, of dielectric constant ?r, to half-fill a parallel-plate capacitor (Fig. 4.25). By what fraction is the capacitance increased when you distribute the material as in Fig. 4.25(a)? How about Fig. 4.25(b)? For a given potential difference V between the plates, find E, D, and P, in each region, and the free and bound charge on all surfaces, for both cases. ... Get solution
20. A sphere of linear dielectric material has embedded in it a uniform free charge density ρ. Find the potential at the center of the sphere (relative to infinity), if its radius is R and the dielectric constant is ϵr . Get solution
21. A certain coaxial cable consists of a copper wire, radius a, surrounded by a concentric copper tube of inner radius c (Fig. 4.26). The space between is partially filled (from b out to c) with material of dielectric constant ϵr, as shown. Find the capacitance per unit length of this cable. ... Get solution
22. A very long cylinder of linear dielectric material is placed in an otherwise uniform electric field E0. Find the resulting field within the cylinder. (The radius is a, the susceptibility χe, and the axis is perpendicular to E0.) Get solution
23. Find the field inside a sphere of linear dielectric material in an otherwise uniform electric field E0 (Ex. 4.7) by the following method of successive approximations: First pretend the field inside is just E0, and use Eq. 4.30 to write down the resulting polarization P0. This polarization generates a field of its own, E1 (Ex. 4.2), which in turn modifies the polarization by an amount P1, which further changes the field by an amount E2, and so on. The resulting field is E0 + E1+ E2 +· · · . Sum the series, and compare your answer with Eq. 4.49. A sphere of homogeneous linear dielectric material is placed in an otherwise uniform electric field E0 (Fig.4.27). Find the electric field inside the sphere. ... Reference example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R. Reference equation 4.30 ... Get solution
24. An uncharged conducting sphere of radius a is coated with a thick insulating shell (dielectric constant ϵr ) out to radius b. This object is now placed in an otherwise uniform electric field E0. Find the electric field in the insulator. Get solution
25. Suppose the region above the x y plane in Ex. 4.8 is also filled with linear dielectric but of a different susceptibility χ ′ e. Find the potential everywhere. Reference Ex. 4.8 Suppose the entire region below the plane z = 0 in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility χe. Calculate the force on a point charge q situated a distance d above the origin. Fig. 4.28 ... Get solution
26. A spherical conductor, of radius a, carries a charge Q (Fig. 4.29). It is surrounded by linear dielectric material of susceptibility χe, out to radius b. Find the energy of this configuration (Eq. 4.58). ... Reference equation 4.58 ... Get solution
27. Calculate W, using both Eq. 4.55 and Eq. 4.58, for a sphere of radius R with frozen-in uniform polarization P (Ex. 4.2). Comment on the discrepancy. Which (if either) is the “true” energy of the system? Reference equation 4.55 ... Reference equation 4.58 ... Reference example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R. Get solution
28. Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (susceptibility χe, mass density ρ). The inner one is maintained at potential V, and the outer one is grounded (Fig. 4.32). To what height (h) does the oil rise, in the space between the tubes? ... Get solution
29. (a) For the configuration in Prob. 4.5, calculate the force on p2 due to p1, and the force on p1 due to p2. Are the answers consistent with Newton’s third law? (b) Find the total torque on p2 with respect to the center of p1, and compare it with the torque on p1 about that same point. [Hint: combine your answer to (a) with the result of Prob. 4.5.] Reference prob 4.5 Get solution
30. An electric dipole p, pointing in the y direction, is placed midway between two large conducting plates, as shown in Fig. 4.33. Each plate makes a ... small angle θ with respect to the x axis, and they are maintained at potentials ±V. What is the direction of the net force on p? (There’s nothing to calculate, here, but do explain your answer qualitatively.) Get solution
31. A point charge Q is “nailed down” on a table. Around it, at radius R, is a frictionless circular track on which a dipole p rides, constrained always to point tangent to the circle. Use Eq. 4.5 to show that the electric force on the dipole is ... Notice that this force is always in the “forward” direction (you can easily confirm this by drawing a diagram showing the forces on the two ends of the dipole). Why isn’t this a perpetual motion machine?21 Reference equation 4.5 ... Get solution
32. Earnshaw’s theorem (Prob. 3.2) says that you cannot trap a charged particle in an electrostatic field. Question: Could you trap a neutral (but polarizable) atom in an electrostatic field? (a) Show that the force on the atom is ... (b) The question becomes, therefore: Is it possible for E2 to have a local maximum (in a charge-free region)? In that case the force would push the atom back to its equilibrium position. Show that the answer is no. [Hint: Use Prob. 3.4(a).]22 Reference prob 3.4 (a) Show that the average electric field over a spherical surface, due to charges outside the sphere, is the same as the field at the center. (b) What is the average due to charges inside the sphere? Get solution
33. A dielectric cube of side a, centered at the origin, carries a “frozenin” polarization P = kr, where k is a constant. Find all the bound charges, and check that they add up to zero. Get solution
34. The space between the plates of a parallel-plate capacitor is filled with dielectric material whose dielectric constant varies linearly from 1 at the bottom plate (x = 0) to 2 at the top plate (x = d). The capacitor is connected to a battery of voltage V. Find all the bound charge, and check that the total is zero. Get solution
35. A point charge q is imbedded at the center of a sphere of linear dielectric material (with susceptibility χe and radius R). Find the electric field, the polarization, and the bound charge densities, ρb and σb. What is the total bound charge on the surface? Where is the compensating negative bound charge located? Get solution
36. At the interface between one linear dielectric and another, the electric field lines bend (see Fig. 4.34). Show that ... assuming there is no free charge at the boundary. [Comment: Eq. 4.68 is reminiscent of Snell’s law in optics.Would a convex “lens” of dielectric material tend to “focus,” or “defocus,” the electric field?] ... Get solution
37. A point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius R and dielectric constant ϵr ). Find the electric potential inside and outside the sphere. Get solution
38. Prove the following uniqueness theorem: A volume V contains a specified free charge distribution, and various pieces of linear dielectric material, with the susceptibility of each one given. If the potential is specified on the boundaries S of V (V = 0 at infinity would be suitable) then the potential throughout V is uniquely determined. [Hint: Integrate ∇ · (V3D3) over V.] Get solution
39. A conducting sphere at potential V0 is half embedded in linear dielectric material of susceptibility χe, which occupies the region z (a) Write down the formula for the proposed potential V(r ), in terms of V0, R, and r . Use it to determine the field, the polarization, the bound charge, and the free charge distribution on the sphere. (b) Show that the resulting charge configuration would indeed produce the potential V(r ). (c) Appeal to the uniqueness theorem in Prob. 4.38 to complete the argument. (d) Could you solve the configurations in Fig. 4.36 with the same potential? If not, explain why. Reference figure 4.35 ... Reference figure 4.36 ... Reference Prob. 4.38 Prove the following uniqueness theorem: A volume V contains a specified free charge distribution, and various pieces of linear dielectric material, with the susceptibility of each one given. If the potential is specified on the boundaries S of V (V = 0 at infinity would be suitable) then the potential throughout V is uniquely determined. [Hint: Integrate ∇ · (V3D3) over V.] Get solution
40. According to Eq. 4.5, the force on a single dipole is (p · ∇)E, so the net force on a dielectric object is Equation 4.69 ... [Here Eext is the field of everything except the dielectric. You might assume that it wouldn’t matter if you used the total field; after all, the dielectric can’t exert a force on itself. However, because the field of the dielectric is discontinuous at the location of any bound surface charge, the derivative introduces a spurious delta function, and it is safest to stick with Eext.] Use Eq. 4.69 to determine the force on a tiny sphere, of radius R, composed of linear dielectric material of susceptibility χe, which is situated a distance s from a fine wire carrying a uniform line charge λ. Reference equation 4.5 ... Get solution
41. In a linear dielectric, the polarization is proportional to the field: P = ϵ0χeE. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = αE. Question: What is the relation between the atomic polarizability α and the susceptibility χe? Since P (the dipole moment per unit volume) is p (the dipole moment per atom) times N (the number of atoms per unit volume), P = Np = NαE, one’s first inclination is to say that Reference equation 4.70 ... And in fact this is not far off, if the density is low. But closer inspection reveals a subtle problem, for the field E in Eq. 4.30 is the total macroscopic field in the medium, whereas the field in Eq. 4.1 is due to everything except the particular atom under consideration (polarizability was defined for an isolated atom subject to a specified external field); call this field Eelse. Imagine that the space allotted to each atom is a sphere of radius R, and show that Equation 4.71 and 4.72 ... Equation 4.72 is known as the Clausius-Mossotti formula, or, in its application to optics, the Lorentz-Lorenz equation. Reference equation 4.1 ... Reference equation 4.30 ... Get solution
42. Check the Clausius-Mossotti relation (Eq. 4.72) for the gases listed in Table 4.1. (Dielectric constants are given in Table 4.2.) (The densities here are so small that Eqs. 4.70 and 4.72 are indistinguishable. For experimental data that confirm the Clausius-Mossotti correction term see, for instance, the first edition of Purcell’s Electricity and Magnetism, Problem 9.28.)23 Reference equation 4.72 ... Reference equation 4.70 ... Reference table 4.1 ... Reference table 4.2 ... Get solution
43. The Clausius-Mossotti equation (Prob. 4.41) tells you how to calculate the susceptibility of a nonpolar substance, in terms of the atomic polarizability α. The Langevin equation tells you how to calculate the susceptibility of a polar substance, in terms of the permanent molecular dipole moment p. Here’s how it goes: (a) The energy of a dipole in an external field E is u = −p · E = −pE cos θ (Eq. 4.6), where θ is the usual polar angle, if we orient the z axis along E. Statistical mechanics says that for a material in equilibrium at absolute temperature T , the probability of a given molecule having energy u is proportional to the Boltzmann factor, ... The average energy of the dipoles is therefore ... Where ... and the integration is over all orientations (θ : 0 → π; φ : 0 → 2π). Use this to show that the polarization of a substance containing N molecules per unit volume is Equation 4.73 ... That’s the Langevin formula. Sketch P/Np as a function of pE/kT . (b) Notice that for large fields/low temperatures, virtually all the molecules are lined up, and the material is nonlinear. Ordinarily, however, kT is much greater than pE. Show that in this régime the material is linear, and calculate its susceptibility, in terms of N, p, T , and k. Compute the susceptibility of water at 20?C, and compare the experimental value in Table 4.2. (The dipole moment of water is 6.1 × 10−30 C·m.) This is rather far off, because we have again neglected the distinction between E and Eelse. The agreement is better in low-density gases, for which the difference between E and Eelse is negligible. Try it for water vapor at 100?C and 1 atm. Reference equation 4.6 ... Reference prob 4.41 In a linear dielectric, the polarization is proportional to the field: P = ϵ0χeE. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = αE. Question: What is the relation between the atomic polarizability α and the susceptibility χe? Since P (the dipole moment per unit volume) is p (the dipole moment per atom) times N (the number of atoms per unit volume), P = Np = NαE, one’s first inclination is to say that Reference equation 4.70 ... And in fact this is not far off, if the density is low. But closer inspection reveals a subtle problem, for the field E in Eq. 4.30 is the total macroscopic field in the medium, whereas the field in Eq. 4.1 is due to everything except the particular atom under consideration (polarizability was defined for an isolated atom subject to a specified external field); call this field Eelse. Imagine that the space allotted to each atom is a sphere of radius R, and show that Equation 4.71 and 4.72 ... Equation 4.72 is known as the Clausius-Mossotti formula, or, in its application to optics, the Lorentz-Lorenz equation. Reference equation 4.1 ... Reference equation 4.30 ... Get solution
2. According to quantum mechanics, the electron cloud for a hydrogen atom in the ground state has a charge density ... where q is the charge of the electron and a is the Bohr radius. Find the atomic polarizability of such an atom. [Hint: First calculate the electric field of the electron cloud, Ee(r ); then expand the exponential, assuming r ≫a.1 Get solution
3. According to Eq. 4.1, the induced dipole moment of an atom is proportional to the external field. This is a “rule of thumb,” not a fundamental law, and it is easy to concoct exceptions—in theory. Suppose, for example, the charge density of the electron cloud were proportional to the distance from the center, out to a radius R. To what power of E would p be proportional in that case? Find the condition on ρ(r ) such that Eq. 4.1 will hold in the weak-field limit. Reference equation 4.1 p = αE. Get solution
4. A point charge q is situated a large distance r from a neutral atom of polarizability α. Find the force of attraction between them. Get solution
5. In Fig. 4.6, p1 and p2 are (perfect) dipoles a distance r apart. What is the torque on p1 due to p2?What is the torque on p2 due to p1? [In each case, I want the torque on the dipole about its own center. If it bothers you that the answers are not equal and opposite, see Prob. 4.29.] ... Reference prob 4.29 (a) For the configuration in Prob. 4.5, calculate the force on p2 due to p1, and the force on p1 due to p2. Are the answers consistent with Newton’s third law? (b) Find the total torque on p2 with respect to the center of p1, and compare it with the torque on p1 about that same point. [Hint: combine your answer to (a) with the result of Prob. 4.5.] Get solution
6. A (perfect) dipole p is situated a distance z above an infinite grounded conducting plane (Fig. 4.7). The dipole makes an angle θ with the perpendicular to the plane. Find the torque on p. If the dipole is free to rotate, in what orientation will it come to rest? Reference figure 4.7 ... Get solution
7. Show that the energy of an ideal dipole p in an electric field E is given by ... Get solution
8. Show that the interaction energy of two dipoles separated by a displacement r is ... [Hint: Use Prob. 4.7 and Eq. 3.104.] Reference prob 4.7Show that the energy of an ideal dipole p in an electric field E is given by ... Reference equation 3.104 ... Get solution
9. dipole p is a distance r from a point charge q, and oriented so that p makes an angle θ with the vector r from q to p. (a) What is the force on p? (b) What is the force on q? Get solution
10. A sphere of radius R carries a polarization P ( r ) = kr, where k is a constant and r is the vector from the center. (a) Calculate the bound charges σb and ρb. (b) Find the field inside and outside the sphere. Get solution
11. A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Get solution
12. Calculate the potential of a uniformly polarized sphere (Ex. 4.2) directly from Eq. 4.9. Reference equation 4.9 ... Reference example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R. Get solution
13. A very long cylinder, of radius a, carries a uniform polarization P perpendicular to its axis. Find the electric field inside the cylinder. Show that the field outside the cylinder can be expressed in the form ... [Careful: I said “uniform,” not “radial”!] Get solution
14. When you polarize a neutral dielectric, the charge moves a bit, but the total remains zero. This fact should be reflected in the bound charges σb and ρb. Prove from Eqs. 4.11 and 4.12 that the total bound charge vanishes. Reference equation 4.11 ... Reference equation 4.12 ... Get solution
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 (Fig. 4.18). (There is no free charge in the problem.) Find the electric field in all three regions by two different methods: ... 6The polarization drops abruptly to zero outside the material, so its derivative is a delta function (see Prob. 1.46). The surface bound charge is precisely this term—in this sense it is actually included in ρb, but we ordinarily prefer to handle it separately as σb. (a) Locate all the bound charge, and use Gauss’s law (Eq. 2.13) to calculate the field it produces. (b) Use Eq. 4.23 to find D, and then get E from Eq. 4.21. [Notice that the second method is much faster, and it avoids any explicit reference to the bound charges.] Reference equation 4.23 ... Reference equation 4.21 ... Reference equation 2.13 ... Reference Prob. 1.46 (a) Show that ... [Hint: Use integration by parts.] (b) Let θ(x) be the step function: ... Show that dθ/dx = δ(x). Get solution
16. Suppose the field inside a large piece of dielectric is E0, so that the electric displacement is ... (a) Now a small spherical cavity (Fig. 4.19a) is hollowed out of the material. Find the field at the center of the cavity in terms of E0 and P. Also find the displacement at the center of the cavity in terms of D0 and P. Assume the polarization is “frozen in,” so it doesn’t change when the cavity is excavated. (b) Do the same for a long needle-shaped cavity running parallel to P (Fig. 4.19b). (c) Do the same for a thin wafer-shaped cavity perpendicular to P (Fig. 4.19c). Assume the cavities are small enough that P, E0, and D0 are essentially uniform. [Hint: Carving out a cavity is the same as superimposing an object of the same shape but opposite polarization.] Figure 4.19 ... Get solution
17. For the bar electret of Prob. 4.11, make three careful sketches: one of P, one of E, and one of D. Assume L is about 2a. [Hint: E lines terminate on charges; D lines terminate on free charges.] Reference prob 4.11 A short cylinder, of radius a and length L, carries a “frozen-in” uniform polarization P, parallel to its axis. Find the bound charge, and sketch the electric field (i) for L ≫ a, (ii) for L ≫ a, and (iii) for L ≈ a. [This is known as a bar electret; it is the electrical analog to a bar magnet. In practice, only very special materials—barium titanate is the most “familiar” example—will hold a permanent electric polarization. That’s why you can’t buy electrets at the toy store.] Get solution
18. The space between the plates of a parallel-plate capacitor (Fig. 4.24) is filled with two slabs of linear dielectric material. Each slab has thickness a, so the total distance between the plates is 2a. Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5. The free charge density on the top plate is σ and on the bottom plate −σ. Reference fig 4.24 ... (a) Find the electric displacement D in each slab. (b) Find the electric field E in each slab. (c) Find the polarization P in each slab. (d) Find the potential difference between the plates. (e) Find the location and amount of all bound charge. (f) Now that you know all the charge (free and bound), recalculate the field in each slab, and confirm your answer to (b). Get solution
19. Suppose you have enough linear dielectric material, of dielectric constant ?r, to half-fill a parallel-plate capacitor (Fig. 4.25). By what fraction is the capacitance increased when you distribute the material as in Fig. 4.25(a)? How about Fig. 4.25(b)? For a given potential difference V between the plates, find E, D, and P, in each region, and the free and bound charge on all surfaces, for both cases. ... Get solution
20. A sphere of linear dielectric material has embedded in it a uniform free charge density ρ. Find the potential at the center of the sphere (relative to infinity), if its radius is R and the dielectric constant is ϵr . Get solution
21. A certain coaxial cable consists of a copper wire, radius a, surrounded by a concentric copper tube of inner radius c (Fig. 4.26). The space between is partially filled (from b out to c) with material of dielectric constant ϵr, as shown. Find the capacitance per unit length of this cable. ... Get solution
22. A very long cylinder of linear dielectric material is placed in an otherwise uniform electric field E0. Find the resulting field within the cylinder. (The radius is a, the susceptibility χe, and the axis is perpendicular to E0.) Get solution
23. Find the field inside a sphere of linear dielectric material in an otherwise uniform electric field E0 (Ex. 4.7) by the following method of successive approximations: First pretend the field inside is just E0, and use Eq. 4.30 to write down the resulting polarization P0. This polarization generates a field of its own, E1 (Ex. 4.2), which in turn modifies the polarization by an amount P1, which further changes the field by an amount E2, and so on. The resulting field is E0 + E1+ E2 +· · · . Sum the series, and compare your answer with Eq. 4.49. A sphere of homogeneous linear dielectric material is placed in an otherwise uniform electric field E0 (Fig.4.27). Find the electric field inside the sphere. ... Reference example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R. Reference equation 4.30 ... Get solution
24. An uncharged conducting sphere of radius a is coated with a thick insulating shell (dielectric constant ϵr ) out to radius b. This object is now placed in an otherwise uniform electric field E0. Find the electric field in the insulator. Get solution
25. Suppose the region above the x y plane in Ex. 4.8 is also filled with linear dielectric but of a different susceptibility χ ′ e. Find the potential everywhere. Reference Ex. 4.8 Suppose the entire region below the plane z = 0 in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility χe. Calculate the force on a point charge q situated a distance d above the origin. Fig. 4.28 ... Get solution
26. A spherical conductor, of radius a, carries a charge Q (Fig. 4.29). It is surrounded by linear dielectric material of susceptibility χe, out to radius b. Find the energy of this configuration (Eq. 4.58). ... Reference equation 4.58 ... Get solution
27. Calculate W, using both Eq. 4.55 and Eq. 4.58, for a sphere of radius R with frozen-in uniform polarization P (Ex. 4.2). Comment on the discrepancy. Which (if either) is the “true” energy of the system? Reference equation 4.55 ... Reference equation 4.58 ... Reference example 4.2 Find the electric field produced by a uniformly polarized sphere of radius R. Get solution
28. Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (susceptibility χe, mass density ρ). The inner one is maintained at potential V, and the outer one is grounded (Fig. 4.32). To what height (h) does the oil rise, in the space between the tubes? ... Get solution
29. (a) For the configuration in Prob. 4.5, calculate the force on p2 due to p1, and the force on p1 due to p2. Are the answers consistent with Newton’s third law? (b) Find the total torque on p2 with respect to the center of p1, and compare it with the torque on p1 about that same point. [Hint: combine your answer to (a) with the result of Prob. 4.5.] Reference prob 4.5 Get solution
30. An electric dipole p, pointing in the y direction, is placed midway between two large conducting plates, as shown in Fig. 4.33. Each plate makes a ... small angle θ with respect to the x axis, and they are maintained at potentials ±V. What is the direction of the net force on p? (There’s nothing to calculate, here, but do explain your answer qualitatively.) Get solution
31. A point charge Q is “nailed down” on a table. Around it, at radius R, is a frictionless circular track on which a dipole p rides, constrained always to point tangent to the circle. Use Eq. 4.5 to show that the electric force on the dipole is ... Notice that this force is always in the “forward” direction (you can easily confirm this by drawing a diagram showing the forces on the two ends of the dipole). Why isn’t this a perpetual motion machine?21 Reference equation 4.5 ... Get solution
32. Earnshaw’s theorem (Prob. 3.2) says that you cannot trap a charged particle in an electrostatic field. Question: Could you trap a neutral (but polarizable) atom in an electrostatic field? (a) Show that the force on the atom is ... (b) The question becomes, therefore: Is it possible for E2 to have a local maximum (in a charge-free region)? In that case the force would push the atom back to its equilibrium position. Show that the answer is no. [Hint: Use Prob. 3.4(a).]22 Reference prob 3.4 (a) Show that the average electric field over a spherical surface, due to charges outside the sphere, is the same as the field at the center. (b) What is the average due to charges inside the sphere? Get solution
33. A dielectric cube of side a, centered at the origin, carries a “frozenin” polarization P = kr, where k is a constant. Find all the bound charges, and check that they add up to zero. Get solution
34. The space between the plates of a parallel-plate capacitor is filled with dielectric material whose dielectric constant varies linearly from 1 at the bottom plate (x = 0) to 2 at the top plate (x = d). The capacitor is connected to a battery of voltage V. Find all the bound charge, and check that the total is zero. Get solution
35. A point charge q is imbedded at the center of a sphere of linear dielectric material (with susceptibility χe and radius R). Find the electric field, the polarization, and the bound charge densities, ρb and σb. What is the total bound charge on the surface? Where is the compensating negative bound charge located? Get solution
36. At the interface between one linear dielectric and another, the electric field lines bend (see Fig. 4.34). Show that ... assuming there is no free charge at the boundary. [Comment: Eq. 4.68 is reminiscent of Snell’s law in optics.Would a convex “lens” of dielectric material tend to “focus,” or “defocus,” the electric field?] ... Get solution
37. A point dipole p is imbedded at the center of a sphere of linear dielectric material (with radius R and dielectric constant ϵr ). Find the electric potential inside and outside the sphere. Get solution
38. Prove the following uniqueness theorem: A volume V contains a specified free charge distribution, and various pieces of linear dielectric material, with the susceptibility of each one given. If the potential is specified on the boundaries S of V (V = 0 at infinity would be suitable) then the potential throughout V is uniquely determined. [Hint: Integrate ∇ · (V3D3) over V.] Get solution
39. A conducting sphere at potential V0 is half embedded in linear dielectric material of susceptibility χe, which occupies the region z (a) Write down the formula for the proposed potential V(r ), in terms of V0, R, and r . Use it to determine the field, the polarization, the bound charge, and the free charge distribution on the sphere. (b) Show that the resulting charge configuration would indeed produce the potential V(r ). (c) Appeal to the uniqueness theorem in Prob. 4.38 to complete the argument. (d) Could you solve the configurations in Fig. 4.36 with the same potential? If not, explain why. Reference figure 4.35 ... Reference figure 4.36 ... Reference Prob. 4.38 Prove the following uniqueness theorem: A volume V contains a specified free charge distribution, and various pieces of linear dielectric material, with the susceptibility of each one given. If the potential is specified on the boundaries S of V (V = 0 at infinity would be suitable) then the potential throughout V is uniquely determined. [Hint: Integrate ∇ · (V3D3) over V.] Get solution
40. According to Eq. 4.5, the force on a single dipole is (p · ∇)E, so the net force on a dielectric object is Equation 4.69 ... [Here Eext is the field of everything except the dielectric. You might assume that it wouldn’t matter if you used the total field; after all, the dielectric can’t exert a force on itself. However, because the field of the dielectric is discontinuous at the location of any bound surface charge, the derivative introduces a spurious delta function, and it is safest to stick with Eext.] Use Eq. 4.69 to determine the force on a tiny sphere, of radius R, composed of linear dielectric material of susceptibility χe, which is situated a distance s from a fine wire carrying a uniform line charge λ. Reference equation 4.5 ... Get solution
41. In a linear dielectric, the polarization is proportional to the field: P = ϵ0χeE. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = αE. Question: What is the relation between the atomic polarizability α and the susceptibility χe? Since P (the dipole moment per unit volume) is p (the dipole moment per atom) times N (the number of atoms per unit volume), P = Np = NαE, one’s first inclination is to say that Reference equation 4.70 ... And in fact this is not far off, if the density is low. But closer inspection reveals a subtle problem, for the field E in Eq. 4.30 is the total macroscopic field in the medium, whereas the field in Eq. 4.1 is due to everything except the particular atom under consideration (polarizability was defined for an isolated atom subject to a specified external field); call this field Eelse. Imagine that the space allotted to each atom is a sphere of radius R, and show that Equation 4.71 and 4.72 ... Equation 4.72 is known as the Clausius-Mossotti formula, or, in its application to optics, the Lorentz-Lorenz equation. Reference equation 4.1 ... Reference equation 4.30 ... Get solution
42. Check the Clausius-Mossotti relation (Eq. 4.72) for the gases listed in Table 4.1. (Dielectric constants are given in Table 4.2.) (The densities here are so small that Eqs. 4.70 and 4.72 are indistinguishable. For experimental data that confirm the Clausius-Mossotti correction term see, for instance, the first edition of Purcell’s Electricity and Magnetism, Problem 9.28.)23 Reference equation 4.72 ... Reference equation 4.70 ... Reference table 4.1 ... Reference table 4.2 ... Get solution
43. The Clausius-Mossotti equation (Prob. 4.41) tells you how to calculate the susceptibility of a nonpolar substance, in terms of the atomic polarizability α. The Langevin equation tells you how to calculate the susceptibility of a polar substance, in terms of the permanent molecular dipole moment p. Here’s how it goes: (a) The energy of a dipole in an external field E is u = −p · E = −pE cos θ (Eq. 4.6), where θ is the usual polar angle, if we orient the z axis along E. Statistical mechanics says that for a material in equilibrium at absolute temperature T , the probability of a given molecule having energy u is proportional to the Boltzmann factor, ... The average energy of the dipoles is therefore ... Where ... and the integration is over all orientations (θ : 0 → π; φ : 0 → 2π). Use this to show that the polarization of a substance containing N molecules per unit volume is Equation 4.73 ... That’s the Langevin formula. Sketch P/Np as a function of pE/kT . (b) Notice that for large fields/low temperatures, virtually all the molecules are lined up, and the material is nonlinear. Ordinarily, however, kT is much greater than pE. Show that in this régime the material is linear, and calculate its susceptibility, in terms of N, p, T , and k. Compute the susceptibility of water at 20?C, and compare the experimental value in Table 4.2. (The dipole moment of water is 6.1 × 10−30 C·m.) This is rather far off, because we have again neglected the distinction between E and Eelse. The agreement is better in low-density gases, for which the difference between E and Eelse is negligible. Try it for water vapor at 100?C and 1 atm. Reference equation 4.6 ... Reference prob 4.41 In a linear dielectric, the polarization is proportional to the field: P = ϵ0χeE. If the material consists of atoms (or nonpolar molecules), the induced dipole moment of each one is likewise proportional to the field p = αE. Question: What is the relation between the atomic polarizability α and the susceptibility χe? Since P (the dipole moment per unit volume) is p (the dipole moment per atom) times N (the number of atoms per unit volume), P = Np = NαE, one’s first inclination is to say that Reference equation 4.70 ... And in fact this is not far off, if the density is low. But closer inspection reveals a subtle problem, for the field E in Eq. 4.30 is the total macroscopic field in the medium, whereas the field in Eq. 4.1 is due to everything except the particular atom under consideration (polarizability was defined for an isolated atom subject to a specified external field); call this field Eelse. Imagine that the space allotted to each atom is a sphere of radius R, and show that Equation 4.71 and 4.72 ... Equation 4.72 is known as the Clausius-Mossotti formula, or, in its application to optics, the Lorentz-Lorenz equation. Reference equation 4.1 ... Reference equation 4.30 ... Get solution
