Examples of Dielectric Problems and the Electric Susceptability Lecture 10 1 A Dielectric Filled Parallel Plate Capacitor Suppose an infinite, parallel plate capacitor filled with a dielectric of dielectric constant ǫ. The field will be perpendicular to the plates and to the dielectric surfaces. We use Gauss Law as previously to find the field between the plates. For a cylindrical Gaussian surface through a plate we write; D d A = qfree So that; E = σ free /ǫ The potential on the plates is; 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 capacitor can store additional charge due to the polarization of the dielectric which reduces the field between the plates. 2 A Dielectric Sphere in a Uniform Electric Field. In a previous lecture we considered a conducting sphere in a uniform electric field. The field caused charge to move so that there was no E component parallel to the surface and no field inside the conductor. In the case of a dielectric sphere, figure 1, charge cannot move but will be polarized, and we expect to find a field within the dielectric. There is no free charge on the sphere, so we apply Laplace s equation with appropriate boundry conditions. 2 V = ρ/ǫ We solve this equation using separation of variables with the boundary conditions; 1
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 V = E 0 z = E 0 r cos(θ) as r E is finite as r = 0 E = (ǫ/ǫ 0 ) E, and; E = E at r = a In spherical coordinates the solution with azmuthal symmetry has the form; For r < a For r > a V = (κ/r ) A l ( r r ) l P l (x) V = (κ/r) B l ( r r )l P l (x) In the above, x = cos(θ). Apply the boundry 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 All other cofficients must vanish to satisfy the boundary conditions as r and r 0. Then match the field as r = a with E = ǫ V 2
ǫ V r in = ǫ 0 V r out Then we have ǫ r = ǫ/ǫ 0 which gives; ǫ r Al a l 1 P l = B l (l + 1) a (l+2) P l E 0 cos(θ) The requirement that tangential E is continous 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; 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 (ǫ r + 2) r cos(θ) r > a V = E 0 rcos(θ) + (ǫ r 1) (ǫ r + 2) E 0 (a/r) 3 r cos(θ) 3
σ E z θ E Figure 2: The geometry used to find the field at the center of the polarized sphere 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 charge density is given by ρ = P, but the field within the sphere is constant. The Polarization is given by; P = (ǫ ǫ 0 ) E Thus the volume charge density vanishes. 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 4
charge and in fact forms a dipole field. We calculate this field at the center of the sphere 2. The field due to the small element as shown in the figure is; de z = κ P cos(θ) r 2 cos(θ) r 2 dω Integrate this over the solid angle dω; E z = P 3ǫ0 This field is the same for all points in the sphere as we have found that the field is constant. Note that it is directed opposite to the applied field. 4 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 ploarized spherical dielectric. We then combine the Polarization to the applied field to obtain; E = ǫ 0 E + P = D = ǫ E We then combine the Polarization to the applied field to obtain; E = ǫ 0 E + P = D = ǫ E In this case the polarization is independent of surface effects. Assume the applied field is E 0 and E the field due to the polarized material. The total field inside the dielectric is the superposition of these fields. It is the total field that causes the polarization ie the polarization field also causes polarization so the effect is non-linear (self interaction). The field in the dielectric is, E total. E total = E 0 + E pol The field at the center of the dielectric sphere as solved above is; E z = P 3ǫ 0 The dipole moment of the sphere is obtained from the atomic polarizability, α. 5
p = α E 0 = α( E total + P 3ǫ0 ) 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 = Nα 1 P/(3ǫ 0 ) E volume The electric susceptibility χ e is; χ e = Nα (1 P/3ǫ 0 ) Then the field inside the sphere from the previous solution is 3E 0 ǫ r + 2. The applied field is E 0, so that neglecting vectors as all fields are in the ẑ direction; E pol = 3E 0 ǫ r + 2 E 0 = ǫ r 1 ǫ r + 2 E 0 Note to first order the polarization is, P 1 ; P 1 = ǫχ e E 0 = (ǫ r 1)ǫ 0 E 0 This does not equal the polarization above, P P 1 so that the polarization acts to create new polarization (ie a non-linear effect). We can iterate the first order result obtaining a first order polarization field, E 1 E 1 = P 1 3ǫ = (ǫ r 1) 0 3 E 0 This creates an incremental polarization, P 2 ; P 2 = (ǫ r 1)ǫ 0 E 1 = (ǫ r 1) 2 3 ǫ 0 E o and this creates an additional E field; E 2 = ( ǫ r 1 3 ) 2 E 0 Continuing the iterations; P = 3 n ( ǫ r 1 3 ) n ǫ 0E0 = 3(ǫ r 1) ǫ r + 2 ǫ 0E 0 To summarize, we have found the solution for a dielectric in a uniform electric field in the interior of the sphere to have the form; 6
E in = 3E 0 ǫ r + 2 From the definition of the electric displacement, D = ǫ 0 E + P so that; P = ǫ 0 (ǫ r 1) E For the dielectric sphere, E z = P 3ǫ 0, which when subsituted for E z above; P = ǫ 0 (ǫ r 1) 3E 0 ǫ r + 2 E 0 This is the expected result. 5 Energy in a Dielectric For a parallel plate capacitor filled with a dielectric constant, ǫ, and plate separation, d. The capacitance is ; C = Q/V V = Ed C = Q/Ed Use Gauss s Law to get E; D d A = Qfree E = σ/ǫ C = ǫ(area)/d Without the dielectric the capacitance is C 0 = ǫ 0 (Area)/d Therefore; C = ǫ r C 0 = ǫ 0E t + P E t C 0 7
In the above, E t is the total field in the capacitor. From this we can obtain; ǫ r = (ǫ 0 + P/E t ) = ǫ 0 (1 + χ e ) The above equation connects the permittivity (dielectric constant) to the susceptibility. The energy of a parallel plate capacitor is obained 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 filled capacitor. Look at this additional energy. The differential energy to align a dipole, as previously obtained is ; dw = E dp Then use the dipole density, N, to obtain the potential energy per unit volume due to the polarization, P = N p. 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 which gives the expected result (ǫ/2)e 2. Thus the additional energy 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 moves up into the capacitor. The final position of the liquid can be determined by minimizing the system energy. The geometry is shown in figure 3. 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. As in the figure, the system consists of 2 capacitors in parallel. Assume the width of the capacitor plates is w. The capacitance values for each capacitor are;. 8
ε l d h Figure 3: A parallel plate capacitor dipped into a dielectric liquid C 1 = ǫ 0w(l h) d C 2 = ǫwh h So that the system capacitance is C t = C 1 + C 2. C t = C 0 [1 + (h/l)(ǫ r 1)] Where we have used C 0 = ǫ 0lw d. The energy stored in the capacitor as a function of h is; 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. The minimum in the energy is then found to obtain the equilibrum position. 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 equilibrum position. This results in a cubic equation for h. 9
z ε εo x q a y Figure 4: The geometry of a problem with a point charge q placed at the center of a spherical tank of water h 3 + 2 l (ǫ r 1) h2 + l 2 (ǫ r 1) 2 2Q 2 l 4ρw 2 ǫ 0 (ǫ r 1)C 0 = 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) Which will be the case in physical situations. 7 Point Charge placed at the center of a Spherical Tank of Water. The geometry of the problem is shown in figure 4. We use Gauss law to get the electric displacement in the water. The electric displacement (and electric field) is radial and independent of angle. Thus; D d A = Qfree = q Thus because of symmetry; 10
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 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 There is an inner induced charge symmetrically placed about q at some 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] We put this dipole inside a small spherical volume of radius, b, in the center of the tank in order to keep the solution appropriately bounded as r 0. Thus the boundry conditions at r = b are; ǫe r (water) = ǫ 0 E r (vacuum) E (water) = E (vacuum) Therefore inside the water; E(water) = p r 3 [2ǫ r cos(θ) ˆr + sin(θ) ˆθ] 11
We 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. 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 = κp 3ǫ r [[ ǫ r + 2 r 2 + 2(ǫ r 1)r b 3 ] cos(θ) ˆr + [ ǫ r + 2 r 3 + 2(ǫ r 1) b 3 ] sin(θ) ˆθ] The polrization 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)] 12
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 boundry conditions at r = a must now be carefully done. As the field does not 0 as r but has a dipole form, with the potential given by; V = 2κp(ǫ r 1) b 3 r cos(θ) This potential should be subtracted from the dipole potential and comes from the problem of defining a dipole as a point. 13