2 A Dielectric Sphere in a Uniform Electric Field


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1 Dielectric Problems and Electric Susceptability Lecture 10 1 A Dielectric Filled Parallel Plate Capacitor Suppose an infinite, parallel plate capacitor with a dielectric of dielectric constant ǫ is inserted between the plates. The field is perpendicular to the plates and to the dielectric surfaces. Thus use Gauss Law to find the field between the plates in the dielectric. For a cylindrical Gaussian surface surrounding an area of the plate surface; D d A = qfree Note the field is only between the plates with a value obtained by superposition of the field from both plates. E = σ free /ǫ The potential between the plates is therefore; V = E d l = σd/ǫ; where d is the plate separation. The capacitance is; C = Q/V = (Area)ǫ/d For a given value of V, the dielectric reduces the total field between the plates, so the capacitor stores additional charge on the plates for the same applied voltage. This is developed further in the sections below. However for the moment, you should consider how energy is conserved when it is determined using the energy density stored in the electric field. 2 A Dielectric Sphere in a Uniform Electric Field In a previous lecture the field due to a conducting sphere in a uniform electric field was found. This field caused charge to move so that there was no E component parallel to the spherical surface and no field inside the conductor. In the case of a dielectric sphere with dielectric constant ǫ = ǫ 0 ǫ r, Figure 1, charge cannot move, but charge polarization in the material occurs. Therefore there can be no free charge in, or on, the sphere. Then apply Laplace s equation with vanishing free charge density and the appropriate boundary conditions at the spherical surface. 1
2 ε o E = Eo r cos( θ ) z^ E = Eo r cos( θ ) z^ ε E = Eo r cos( θ ) z^ E = Eo r cos( θ ) z^ Figure 1: The geometry of a dielectric sphere placed in a uniform field 2 V = ρ/ǫ The above equation is solved using separation of variables with the boundary conditions; V = E 0 z = E 0 r cos(θ) as r ; Now E is finite as r = 0; and E = (ǫ /ǫ 0 ) E, E = E at r = a In spherical coordinates the solution to Laplaces s equation with azimuthal symmetry when applying separation of variables has the form; For r < a For r > a V = κ A l r l P l (x) V = κ B l r (l+1) P l (x) Then, x = cos(θ). Apply the boundary conditions to obtain the equation; V = (κ)[a 0 + A 1 r cos(θ)] for r < a V = (κ/r)[b 0 + B 1 /r] cos(θ) V 0 rcos(θ) for r > a 2
3 These solutions match the boundary conditions when r and r 0. All other coifficients, A l,b l, must vanish. Then match the potential and field when r = a. Use E = ǫ V so that; ǫ V r in = ǫ 0 V r out By definition, ǫ r = ǫ/ǫ 0 which gives; ǫ r Al l a l 1 P l = B l (l + 1) a (l+2) P l E 0 cos(θ) The requirement that tangential E is continuous is equivalent to the continuity of the potential. Al a l P l = B l a (l+1) P l E 0 P l Equate the constants for each value of l. A 0 = B 0 /a and B 0 a 2 = 0 A 1 a = B 1 a 2 E 0 a and ǫ r A 1 = 2B 1 a 3 + E 0 A 2 a 2 = B 2 a 3 and ǫ r aa 2 = 3B 2 a 4 This means that ; B 0 = 0; A 0 = 0 A 1 = B 1 /a 3 E 0 All other values of A l and B l are zero. Finally; B 1 = (ǫ r 1) (ǫ r + 2) a3 E 0 A 1 = 3 (ǫ r + 2) E 0 The potential is then; r < a V = 3 E 0 (ǫ r + 2) r cos(θ) 3
4 σ E z θ de Figure 2: The geometry used to find the field at the center of the polarized sphere r > a V = E 0 rcos(θ) + (ǫ r 1) (ǫ r + 2) E 0 (a/r) 3 r cos(θ) 3 Polarization of the Dielectric Sphere in a Uniform Electric Field. The field inside the dielectric sphere, as obtained in the last section, is; E = V with V in = 3E 0 (ǫ r + 2) r cos(θ). Thus E = 3E 0 (ǫ r + 2) ẑ The field is uniform and in the ẑ direction. The volume polarizationcharge density is given by ρ = P = 0, since the field within the sphere is constant. The Polarization is given by; P = (ǫ ǫ 0 ) E Thus the volume charge density vanishes, but there is a surface charge density given by; σ = P ˆn = P cos(θ) where ˆn is the outward surface normal. The field inside the sphere is due to the surface charge which forms a dipole field, Figure 2, in addition to the applied field. Calculate this 4
5 z P y E E E E x Figure 3: The geometry used to find the field at the center of the polarized sphere field at the center of the sphere 2. The field due to a small element as shown in the figure is; de z = κ P cos(θ) r 2 cos(θ) r 2 dω Integrate over the solid angle dω; E z = P 3ǫ0 Although this field was found at the center of the sphere, it is the same for all points in the sphere, since the field inside the sphere is constant as obtained in the solution using separation of variables. Note this solution also shows the polarization field is directed opposite to the applied field, Figure Connection between the Electric Susceptibility and Atomic Polarizability An applied field induces a polarization in a dielectric material. To better understand this process consider the polarization at the center of a polarized spherical dielectric. Then combine this polarization with the applied field to obtain the field for a Class A dielectric; D = ǫ 0 E + P = ǫ E In this case the polarization is independent of surface effects. Assume the applied field is E 0, 5
6 and E is the field due to the polarized material. The total field inside the dielectric is the superposition of these fields. Now the total field (applied plus induced) causes the polarization, so the effect is nonlinear due to this self interaction. The field in the dielectric may be approximated by the applied field, vece 0 and the field due to polarization as calculated in the last section. E total = E 0 + E pol with E pol = P/3ǫ 0 The dipole moment of the sphere is obtained from the atomic polarizability, α. p = α E total = α(ǫ 0 E0 + P 3 ) The polarization is the dipole moment per unit volume, P = N p, where N is the number density of dipoles. Now solve for the polarization. P/N = α( E 0 + P/3ǫ 0 ) The electric susceptibility, χ e, is defined by; χ e = ǫ 0 (ǫ r 1) P = ǫ 0 χ e E Substitute for P in the above equation to obtain the following equation Nα = 3ǫ 0 ǫ r 1 ǫ r + 2 The above is the ClausisMossotti equation. The polarization is; P = ǫ 0 χ e Etotal = (ǫ ǫ 0 ) E total Now replace the total field in the material, E total, with applied field, E 0. This results in a first order polarization, P 1. P 1 = ǫ 0 χ e E 0 = (ǫ r 1)ǫ 0 E 0 However, this does not equal the above value for the final polarization, ie P P 1. Thus the polarization itself acts to create new polarization (ie a nonlinear effect). Iterate the above equations, beginning with the first order polarization. This gives an electric field, E 1, due 6
7 to the polarization, which then produces the next order in the iteration of the polarization. E 1 = P 1 3ǫ = (ǫ r 1) 0 3 E 0 Thus an incremental polarization, P 2 is obtained. P 2 = (ǫ r 1)ǫ 0 E 1 = (ǫ r 1) 2 3 and thus an additional E field; E 2 = ( ǫ r 1 3 ) 2 E 0 Continuing the iterations; ǫ 0 E o P = 3 n ( ǫ r 1 3 ) n ǫ 0E0 = 3(ǫ r 1) ǫ r + 2 ǫ 0E 0 To summarize, the following solution is found for the field in the interior of a dielectric sphere in an applied uniform electric field. E total = 3E 0 ǫ r + 2 From the definition of the electric displacement, D = ǫ 0 E + P, so that P is; P = ǫ 0 (ǫ r 1) E For the dielectric sphere; P = 3 (ǫ r 1) ǫ r + 2 ǫ 0 E 0 5 Energy in a Dielectric Return to a parallel plate capacitor filled with a dielectric constant, ǫ, and plate separation, d. IN review, the capacitance is ; C = Q/V V = Ed C = Q/Ed 7
8 Use Gauss s Law to get E; D d A = Qfree E = σ/ǫ C = ǫ(area)/d Without the dielectric the capacitance is Therefore; C 0 = ǫ 0 (Area)/d C = ǫ r C 0 = ǫ 0E t + P E t C 0 In the above, E t is the total field in the capacitor. From this obtain; D = ǫ r ǫ 0 Et = (ǫ 0 Et + P) Then P points in opposition to E ǫ r = (ǫ 0 + P/E t ) = ǫ 0 (1 + χ e ) The above equation connects the permittivity (dielectric constant) to the susceptibility, χ e. The energy of a parallel plate capacitor is obtained by; W = 1/2 CV 2 = 1/2 ǫ r C 0 V 2 W = (ǫ/2) dτ E 2 When one keeps the same voltage across the capacitor, there is an increase in energy W = ǫ r W 0 in a dielectric of the capacitor. Look at this additional energy. The differential energy to polarize a dipole p in the direction of the applied field is ; dw = F d x = q E d x = Edp dw = E dp Then use the dipole density, N, to obtain the potential energy per unit volume due to the polarization, P = N p. 8
9 ε l d h Figure 4: A parallel plate capacitor dipped into a dielectric liquid dw = dp E = de ǫ0 (ǫ r 1)E W = (1/2)ǫ 0 (ǫ r 1)E 2 This is to be added to the energy density of the vacuum field (ǫ 0 /2)E 2 and finally gives the expected result (ǫ/2)e 2. Thus the missing energy in the field is stored in the polarization of the material. 6 Example of Energy in a Dielectric Suppose a charged parallel plate capacitor is dipped into a dielectric liquid. The liquid is pulled up into the capacitor. The final position of the liquid can be determined by minimizing the system energy. The geometry is shown in Figure 4. In this problem the voltage is disconnected from the capacitor so the charge remains constant, but the voltage changes as the liquid fills the volume between the plates. On the other hand, if the voltage supply remains connected to the capacitor, then the voltage remains constant, but the charge changes. In this case the battery charging the plates continues to insert energy into the system, so the total energy in the capacitor is not constant. From the figure, the system can be considered as 2 capacitors connected in parallel. Assume the width of the capacitor plates is w. The capacitance values for each capacitor are; C 1 = ǫ 0w(l h) d C 2 = ǫwh h 9
10 So that the system capacitance is C t = C 1 + C 2. C t = C 0 [1 + (h/l)(ǫ r 1)] Where C 0 = ǫ 0lw is used. Write the energy stored in the capacitor as a function of h in d terms of the stored charge, W = (1/2)Q 2 /C. W = Q 2 2C 0 [1 + (h/l)(ǫ r 1)] Since ǫ r > 1 the energy decreases as h increases. The difference in the energy goes into raising the liquid. The system energy is then; W S = W + mg(h/2) = W + g(h/2)ρ(wdh) In the above ρ is the mass density and the second term on the left represents the potential energy of the raised liquid. The minimum in the energy determining the equilibrium position is then found. W S h = 0 Q 2 l(ǫ r 1) 2C 0 [l + h(ǫ r 1)] 2 + 2ρwd(h/2) = 0 Solve for h to find the equilibrium position. This results in a cubic equation for h. h 3 + (ǫ 2 l r 1) h2 + l 2 2Q (ǫ r 1) 2 h 2 l 4ρw 2 = 0 ǫ 0 (ǫ r 1)C 0 There is one real root of the equation if q 3 + r 2 > 0 where; q = (1/3)[ l ǫ r 1 ]2 r = (1/9)( l 2Q (ǫ r 1) )3 (3/4) 2 l ρw 2 gǫ 0 (ǫ r 1) This will be the case in all physical situations. The solution is obtained as follows. a 2 = 2l ǫ r 1 s 1 = [r + (q 3 + r 2 ) 1/2 ] 1/3 s 2 = [r + (q 3 r 2 ) 1/2 ] 1/3 10
11 z ε εo x q a y Figure 5: The geometry of a problem with a point charge q placed at the center of a spherical tank of water h = (s 1 + s 2 a 2 /3 7 Point Charge placed at the center of a Spherical Tank of Water. The geometry of the problem is shown in figure 5. Use Gauss law to get the electric displacement in the water. The electric displacement (and electric field) is radial and independent of angle. Thus assume a small spherical shell centered on the charge. Because the field is radial, the electric displacement, D equals the electric displacement in the water, D. This means; D d A = Qfree = q Thus because of symmetry; D = 1 4π q r 2 ˆr and D = ǫ E. Then the polarization is, P = ǫ 0 (ǫ r 1) E = ǫ 0 (ǫ r 1) 1 4πǫ q r 2 ˆr The volume charge density is; ρ = P = 1 r 2 r [r2 P r ] = 0 11
12 Thus there is no volume charge density. The surface charge density at r = a is; σ = P ˆr = ǫ 0 (ǫ r 1) E = ǫ 0 (ǫ r 1) 1 4πǫ q a 2 Inside there is an induced charge symmetrically placed about q at a finite radius so that the total induced charge sums to zero. Outside the water tank the field is the same as the field from a point charge q in the vacuum. 8 Dipole placed at the center of a Spherical Tank of Water This problem is similar to the problem in the last section, but the point charge is replaced by a dipole aligned along the ẑ axis. The field of the dipole in vacuum is; E d = κ p r 3 [2 cos(θ)ˆr + sin(θ) ˆθ] Put this dipole inside a small spherical volume of radius, b, in the center of the tank. This keeps the solution appropriately bounded as r 0. Thus the boundary conditions at r = b are; ǫe r (water) = ǫ 0 E r (vacuum) E (water) = E (vacuum) Therefore inside the water; E(water) = κp r 3 [2(ǫ 0 /epsilon) cos(θ) ˆr + sin(θ) ˆθ] Solve for the potential in the water using separation of variables. The solution has the forms; r > a V = A l r (l+1) P l (x) b < r < a V = [B l r (l+1) + C l r l ]P l (x) Now match the boundary conditions at r = b. 12
13 ǫ[2b 1 /b 3 C 1 ]cos(θ) = 2ǫ 0 κp/b 3 cos(θ) (1/b)[B 1 /b 2 + C 1 b]sin(θ) = κp/b 3 sin(θ) All other coefficients vanish. Solve for B 1 and C 1. B 1 = κp 3ǫ r [ǫ r + 2] C 1 = 2κp 3ǫ r b 3 [ǫ r 1] This gives the potential; V = 3ǫ κp [ ǫ r + 2 r r 3 + 2(ǫ r 1) b 3 r]cos(θ) From this one gets the field; E = 3ǫ κp [ [ ǫ r + 2 r r 2 + 2(ǫ r 1)r ] cos(θ) ˆr + b 3 [ ǫ r + 2 r 3 + 2(ǫ r 1) b 3 ] sin(θ) ˆθ] The polarization is P = ǫ 0 (ǫ r 1) E. So that the volume charge density is; ρ = (ǫ 0 (ǫ r 1) E) ρ = ǫ 0(ǫ r 1) κp 3ǫ r r 2 [[ 2(ǫ r + 2) r 2 2(ǫ r 1) b 3 ] cos(θ) + [ ǫ r + 2 r 3 + 2(ǫ r 1) b 3 ]cos(θ)] ρ = ǫ 0(ǫ r 1) κp 3ǫ r r 2 [3(ǫ r + 2) + 6(ǫ r 1)(r/b) 3 ] cos(θ) The surface charge density is; r = b σ = (ǫ ǫ 0 ) 8κp 3ǫ r b 3 cos(θ) r = a σ = (ǫ ǫ 0 ) 4κp 3ǫ r a 3 [ǫ r (1 ǫ r (a/b) 3 ) + 2(a/b) 3 ] cos(θ) Matching the boundary conditions at r = a must now be carefully done. The field does not 0 as r but has a dipole form, with the potential given by; 13
14 z d ε ε 2 a x y Figure 6: The geometry used to find the capacitance of a parallel plate capacitor filled with 2 dielectrics V = 2κp(ǫ r 1) b 3 r cos(θ) This potential should be subtracted from the dipole potential. There still remains a problem in defining a dipole as a point. 9 Examples 9.1 Parallel plate capacitor filled with 2 dielectric materials A parallel plate capacitor is filled with 2 dielectric materials of dielectric constants ǫ 1, and ǫ 2 as shown in figure 6. The plates have area, A, and are separated by a distance, d. The thickness of dielectric, ǫ 2 is a. Find the capacitance. Place a potential, V 0, between the plates. By symmetry, the potential is assumed to be dependent only on z as the plate lengths are >> d. Thus the electric field is perpendicular to the plates. The solution to Laplace s equation, 2 V = 0, must have the form; V = Az + B In the above A and B are constants which are used to satisfy the boundary conditions. Thus; V 2 = A 2 z + B 2 for 0 < z < d a) V 1 = A 1 z + B 1 for d a < z < d Then when Z = 0 V 2 = B 2 = 0 14
15 and when z = d V = A 1 d + B 1 = V 0 Solving for the constants the potentials become; V 1 = [ V 0 B 1 ]z + B d 1 V 2 = A 2 z The fields are; E 1 = V 1 z = [V 0 B 1 ] d E 1 = V 2 z = A 2 The remaining constants, A 2 and B 1, are determined at the dielectric boundary, z = a, where; Also require; D 1 = ǫ 1 E 1 = D 2 = ǫ 2 E 2 A 2 = (ǫ 1 /ǫ 2 )[ V 0 B 1 ] d 0 d d l E = V 0. Then solve the 2 above equations for the constants. V B 1 = 0 a(ǫ 1 ǫ 2 ) [ǫ 1 a + ǫ 2 (d a)] A 2 = ǫ 1 V 0 [ǫ 1 a + ǫ 2 (d a)] This yields the potentials; V 1 = V 0[ǫ 2 z + (ǫ 1 ǫ 2 )a] [ǫ 1 a + ǫ 2 (d a)] V 2 = ǫ 1 V 0 [ǫ 1 a + ǫ 2 (d a)] z The fields are; E 2 = E 1 = ǫ 1 [ǫ 1 a + ǫ 2 (d a)] V 0z ǫ 2 [ǫ 1 a + ǫ 2 (d a)] V 0z 15
16 These solutions should be checked for the limiting cases when a = 0, a = d, and ǫ 1 = ǫ 2 Now the charges on the plates are obtained from Gauss law, D d A = Qfree. Bottom Plate; σ F = ǫ 2 E 2 = ǫ 1 ǫ 2 ǫ 1 a + ǫ 2 (d a) V 0 Top Plate; σ F = ǫ 1 E 1 = ǫ 1 ǫ 2 ǫ 1 a + ǫ 2 (d a) V 0 So the free charge on the top and bottom plates are equal as they should be. Then using the free charge, the capacitance per unit area is; C/A = σ F /V 0 = ǫ 1 E 1 = ǫ 1 ǫ 2 ǫ 1 a + ǫ 2 (d a) V The field of a polarized sphere Suppose a sphere of radius, R, composed of polarized material. The polarization has only and angular dependence given by; P = P 0 cos(θ) ˆr There is a surface charge density; σ = P ˆr = P 0 cos(θ) The volume charge density is; ρ = P = 2P 0 cos(θ) r Then find the potential outside the sphere, Figure 7. The volume contribution to the potential has the form;. V = κ dω r 2 ρ r r Substitute for the volume charge density and use the addition theorem to expand 1 r r 16
17 r P o cos ( q ) r Figure 7: Geometry to obtain the potential of a polarized sphere 1 r r = l,m 4π 2l + 1 (r l /r (l+1) ) Yl m (θ, φ ) Yl m (θ, φ). Note, cos(θ) = 4π/3Yl 0. In the following use the orthogonality of the spherical harmonics. V = 8πκ(P 0 /r) 1 2l + 1 V = 8πκ( P 0R 3 3r 3 ) cos(θ) [ dω cos(θ, φ ) Y m l The integral for the surface proceeds in a similar way. V = κ R 2 dω P 0 cos(θ) r r V = 4πκ P 0R 3 cos(θ) 3r 3 (θ, φ )][ dr (r l+2 /r l )] Y m l The potential is the sum of the potentials from the volume and surface charge densities. 17
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