Examples of Dielectric Problems and the Electric Susceptability 1 A ...

Examples of Dielectric Problems and the ElectricSusceptabilityLecture 101 A Dielectric Filled Parallel Plate CapacitorSuppose 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 surfacethrough a plate we write;∮ ⃗D · d ⃗ A = qfreeSo that;E = σ free /ǫThe potential on the plates is;V = ∫ ⃗ E · d ⃗ l = σd/ǫwhere d is the plate separation. The capacitance is;C = Q/V = (Area)ǫ/dFor a given value of V , the capacitor can store additional charge due to the polarizationof 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. Thefield caused charge to move so that there was no E ⃗ component parallel to the surface and nofield inside the conductor. In the case of a dielectric sphere, figure 1, charge cannot movebut will be polarized, and we expect to find a field within the dielectric. There is no freecharge on the sphere, so we apply Laplace’s equation with appropriate boundry conditions.∇ 2 V = ρ/ǫWe solve this equation using separation of variables with the boundary conditions;1

ε oE = Eo r cos( θ ) z^E = Eo r cos( θ ) z^εE = Eo r cos( θ ) z^E = Eo r cos( θ ) z^Figure 1: The geometry of a dielectric sphere placed in a uniform fieldV = E 0 z = E 0 r cos(θ) as r → ∞E is finite as r = 0E ⊥ = (ǫ/ǫ 0 ) E ′ ⊥ , and;E ‖ = E ′ ‖ at r = aIn spherical coordinates the solution with azmuthal symmetry has the form;For r < aFor r > aV = (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 < aV = (κ/r)[B 0 + B 1 /r cos(θ) − V 0 rcos(θ)] for r > aAll other cofficients must vanish to satisfy the boundary conditions as r → ∞ and r → 0.Then match the field as r = a with ⃗ E = −ǫ ⃗ ∇V2

ǫ ∂V∂r in = ǫ 0 ∂V∂r outThen 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 thepotential.∑Al a l P l = ∑ B l a −(l+1) P l − E 0 P lEquate the constants;A 0 = B 0 /a and B 0 a −2 = 0A 1 a = B 1 a −2 − E 0 a and −ǫ r a 1 = 2B 1 a −3 + E 0A 2 a 2 = B 2 a −3 and −ǫ r aA 2 = −3B 2 a −4This means that ;B 0 = 0; A 0 = 0A 1 = B 1 /a 3 − E 0All other values of A l and B l are zero. Finally;B 1 = (ǫ r − 1)(ǫ r + 2) a3 E 0A 1 = − 3(ǫ r + 2) E 0The potential is then;r < aV = − 3(ǫ r + 2) r cos(θ)r > aV = −E 0 rcos(θ) + (ǫ r − 1)(ǫ r + 2) E 0 (a/r) 3 r cos(θ)3

σE zθ EFigure 2: The geometry used to find the field at the center of the polarized sphere3 Polarization of the Dielectric Sphere in a UniformElectric Field.The field inside the dielectric sphere as obtained in the last section is;⃗E = − ⃗ ∇Vwith V in = − 3E 0(ǫ r + 2) r cos(θ)Thus ⃗ E = 3E 0(ǫ r + 2) ẑThe field is uniform and in the ẑ direction. The volume charge density is given byρ = − ⃗ ∇ · ⃗P, but the field within the sphere is constant. The Polarization is given by;⃗P = (ǫ − ǫ 0 ) ⃗ EThus 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 surface4

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 2cos(θ) r 2 dΩIntegrate this over the solid angle dΩ;E z = − ⃗ P3ǫ0This field is the same for all points in the sphere as we have found that the field isconstant. Note that it is directed opposite to the applied field.4 Connection between the Electric Susceptibility andAtomic PolarizabilityAn applied field induces a polarization in a dielectric material. To better understand thisprocess consider the polarization at the center of a ploarized spherical dielectric. We thencombine the Polarization to the applied field to obtain;⃗E = ǫ 0⃗ E + ⃗ P = ⃗ D = ǫ ⃗ EWe then combine the Polarization to the applied field to obtain;⃗E = ǫ 0⃗ E + ⃗ P = ⃗ D = ǫ ⃗ EIn this case the polarization is independent of surface effects. Assume the applied fieldis ⃗ E 0 and ⃗ E ′ the field due to the polarized material. The total field inside the dielectricis the superposition of these fields. It is the total field that causes the polarization ie thepolarization field also causes polarization so the effect is non-linear (self interaction). Thefield in the dielectric is, ⃗ E total .E total = ⃗ E 0 + ⃗ E polThe field at the center of the dielectric sphere as solved above is;E z = − P3ǫ 0The dipole moment of the sphere is obtained from the atomic polarizability, α.5

⃗p = α ⃗ E 0 = α( ⃗ E total + ⃗ P3ǫ0)The polarization is the dipole moment per unit volume, ⃗ P = N⃗p, where N is the numberdensity of dipoles. Now solve for the polarization.⃗P =Nα1 − P/(3ǫ 0 ) ⃗ E volumeThe electric susceptibility χ e is;χ e =Nα(1 − P/3ǫ 0 )Then the field inside the sphere from the previous solution is3E 0ǫ r + 2. The applied fieldis E 0 , so that neglecting vectors as all fields are in the ẑ direction;E pol = 3E 0ǫ r + 2 − E 0 = − ǫ r − 1ǫ r + 2 E 0Note to first order the polarization is, P 1 ;P 1 = ǫχ e E 0 = (ǫ r − 1)ǫ 0 E 0This does not equal the polarization above, P ≠ P 1 so that the polarization acts to createnew polarization (ie a non-linear effect). We can iterate the first order result obtaininga first order polarization field, E 1E 1 = P 13ǫ= − (ǫ r − 1)0 3E 0This creates an incremental polarization, P 2 ;P 2 = (ǫ r − 1)ǫ 0 E 1 = (ǫ r − 1) 23ǫ 0 E oand this creates an additional E field;E 2 = −( ǫ r − 13) 2 E 0Continuing the iterations;⃗P = 3 ∑ n(− ǫ r − 13) n ǫ 0E0 ⃗ = 3(ǫ r − 1)ǫ r + 2 ǫ ⃗ 0E 0To summarize, we have found the solution for a dielectric in a uniform electric field inthe interior of the sphere to have the form;6

⃗E in = 3E 0ǫ r + 2From the definition of the electric displacement, ⃗ D = ǫ 0⃗ E + ⃗ P so that;⃗P = ǫ 0 (ǫ r − 1) ⃗ EFor the dielectric sphere, E z = − P3ǫ 0, which when subsituted for E z above;P = ǫ 0 (ǫ r − 1) 3E 0ǫ r + 2 E 0This is the expected result.5 Energy in a DielectricFor a parallel plate capacitor filled with a dielectric constant, ǫ, and plate separation, d. Thecapacitance is ;C = Q/VV = EdC = Q/EdUse Gauss’s Law to get E;∮ ⃗D · d ⃗ A = QfreeE = σ/ǫC = ǫ(Area)/dWithout the dielectric the capacitance isC 0 = ǫ 0 (Area)/dTherefore;C = ǫ r C 0 = ǫ 0E t + PE tC 07

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 2W = (ǫ/2) ∫ dτ E 2When one keeps the same voltage across the capacitor, there is an increase in energyW = ǫ r W 0 in a dielectric filled capacitor. Look at this additional energy. The differentialenergy to align a dipole, as previously obtained is ;dW = E dpThen use the dipole density, N, to obtain the potential energy per unit volume due tothe polarization, ⃗ P = N⃗p.∫dW =∫dP E =∫dE ǫ0 (ǫ r − 1)EW = (1/2)ǫ 0 (ǫ r − 1)E 2This is to be added to the energy density of the vacuum field (ǫ 0 /2)E 2 which gives theexpected result (ǫ/2)E 2 . Thus the additional energy is stored in the polarization of thematerial.6 Example of Energy in a DielectricSuppose a charged parallel plate capacitor is dipped into a dielectric liquid. The liquidmoves up into the capacitor. The final position of the liquid can be determined by minimizingthe system energy. The geometry is shown in figure 3. In this problem the voltageis disconnected from the capacitor so the charge remains constant, but the voltage changesas the liquid fills the volume between the plates. On the other hand, if the voltage supplyremains connected to the capacitor, then the voltage remains constant, but the chargechanges. As in the figure, the system consists of 2 capacitors in parallel. Assume the widthof the capacitor plates is w. The capacitance values for each capacitor are;.8

εldhFigure 3: A parallel plate capacitor dipped into a dielectric liquidC 1 = ǫ 0w(l − h)dC 2 = ǫwhhSo 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 0lwd. The energy stored in the capacitor as a function of h is;W =Q 22C 0 [1 + (h/l)(ǫ r − 1)]Since ǫ r > 1 the energy decreases as h increases. The difference in the energy goes intoraising 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 obtainthe equilibrum position.∂W S∂h = 0Q−2 l(ǫ r − 1)2C 0 [l + h(ǫ r − 1)] 2 + 2ρwd(h/2) = 0Solve for h to find the equilibrum position. This results in a cubic equation for h.9

zεεoxqayFigure 4: The geometry of a problem with a point charge q placed at the center of a sphericaltank of waterh 3 +2 l(ǫ r − 1) h2 +l 2(ǫ r − 1) 2 −2Q 2 l4ρw 2 ǫ 0 (ǫ r − 1)C 0= 0There is one real root of the equation if q 3 + r 2 > 0 where;q = −(1/3)[ lǫ r − 1 ]2r = (1/9)( l2Q(ǫ r − 1) )3 − (3/4)2 lρw 2 gǫ 0 (ǫ r − 1)Which will be the case in physical situations.7 Point Charge placed at the center of a SphericalTank of Water.The geometry of the problem is shown in figure 4. We use Gauss’ law to get the electricdisplacement in the water. The electric displacement (and electric field) is radial andindependent of angle. Thus;∮ ⃗D · d ⃗ A = Qfree = qThus because of symmetry;10

⃗D = 14π q r 2 ˆrand ⃗ D = ǫ ⃗ E. Then the polarization is,⃗P = ǫ 0 (ǫ r − 1) ⃗ E = ǫ 0 (ǫ r − 1) 14πǫ q r 2 ˆrThe Volume charge density is ;ρ = − ⃗ ∇ · ⃗P = 1 r 2 ∂ ∂r [r2 P r ] = 0Thus 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) 14πǫ q a 2There is an inner induced charge symmetrically placed about q at some finite radius sothat the total induced charge sums to zero. Outside the water tank the field is the same asthe field from a point charge q in the vacuum.8 Dipole placed at the center of a Spherical Tank ofWaterThis problem is similar to the problem in the last section, but the point charge is replacedby 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 tankin order to keep the solution appropriately bounded as r → 0. Thus the boundry conditionsat 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 hasthe forms;r > aV = ∑ A l r −(l+1) P l (x)b < r < aV = ∑ [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 = κp3ǫ r[ǫ r + 2]C 1 = 2κp3ǫ r b 3 [ǫ r − 1]This gives the potential;V =3ǫ κp [ ǫ r + 2r r 3 + 2(ǫ r − 1)b 3 r]cos(θ)From this one gets the field;⃗E = κp3ǫ r[[ ǫ r + 2r 2 + 2(ǫ r − 1)rb 3] cos(θ) ˆr +[ ǫ r + 2r 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) ⃗ Eρ = ǫ 0(ǫ r − 1) κp3ǫ r r 2 [[ 2(ǫ r + 2)r 2 − 2(ǫ r − 1)b 3 ] cos(θ) +[ ǫ r + 2r 3 + 2(ǫ r − 1)b 3 ]cos(θ)]12

ρ = ǫ 0(ǫ r − 1) κp3ǫ r r 2 [3(ǫ r + 2) + 6(ǫ r − 1)(r/b) 3 ] cos(θ)The surface charge density is;r = bσ = −(ǫ − ǫ 0 ) 8κp3ǫ r b 3 cos(θ)r = aσ = −(ǫ − ǫ 0 ) 4κp3ǫ 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 doesnot → 0 as r → ∞ but has a dipole form, with the potential given by;V = 2κp(ǫ r − 1)b 3r cos(θ)This potential should be subtracted from the dipole potential and comes from the problemof defining a dipole as a point.13