Electrostatic Fields in Matter

3 Polarisation MechanismsThere are 3 mechanisms involved in the polarisation of atoms andmolecules:1. Electronic polarization - displacement of electrons from their normalposition in molecules.2. Ionic polarization (or atomic polarization) - the displacement ofions from their normal positions within molecules.3. Dipolar polarization - alignment of permanent molecular dipolesby the field.The size of the effect is largest for 3, then 2 then 1. However effects3 and 2 are not always present.4 Frequency Dependence (non-examinable)The polarisation generated by an alternating field varies with frequencyand temperature.• Dipole polarisation occurs only at relatively low frequencies (upto high radio frequencies) and disappears at low temperatures.• Atomic polarisation extends to the infrared.• Electronic polarisation occurs for virtually all frequencies.The refractive index is due to electronic polarisation.3

5 Parallel Plate Capacitor.A parallel plate capacitor is a useful model for demonstrating someof the basic ideas about dielectrics.The figure shows two parallel conducting plates. A charge Q hasbeen transferred from the bottom plate to the top and the plates arenow isolated.+QAdE-QVFrom Gauss’s law ⃗ E = σ ɛ 0= Q/Aɛ 0(5)V =∫⃗E · ⃗ds = Ed (6)C = Q V = ɛ 0Ad(7)These equations are based on the assumption that the field betweenthe plates is constant, which is reasonable if the plates areclose together.What happens when a dielectric is placed between the plates?The original field, produced by the charges on the plates, pushesthe +ve charges in the dielectric down and -ve charges up. Thedielectric becomes polarised.4

This leaves a layer of negative charge on the top surface of thedielectric and a positive layer at the bottom.The original charge Q, which was placed on the capacitor plates,is referred to as free charge.The charge induced on the dielectric surface is called bound charge.The bound charge generates a new field, in the opposite directionto the original field. This new field is called the depolarisation field.The total field in the dielectric is the sum of the original field (fromthe free charges) and the depolarisation field - it will be smaller thanthe original field.Bound chargesFree chargesDepolarisation fieldOriginal fieldLooking at equations (6) and (7), we see that when a dielectricis placed between charged plates which are not connected, then Qremains constant while V decreases; hence C increases. The factorby which C increases is called the dielectric constant K (or therelative permittivity, ɛ r ).-5

6 Bound Charges• Detailed derivation: Griffiths section 4.2.1, page 166 (non-examinable)• Simpler derivation: Griffiths section 4.2.2, page 170.The following is similar to Griffith’s simpler derivation.When a dielectric is polarised, the movement of charge createsregions in the dielectric which can have a net charge. We will deriveexpressions for the densities of these bound charges.The electric field produced by a polarised dielectric is calculatedsimply by calculating the field due to the bound charges.6.1 Surface Bound Charge DensityThe diagram shows a small area da at the surface of a polariseddielectric. This area is represented by a vector da, ⃗ normal to thesurface. The polarisation at this point is at some angle θ to thesurface normal.In each atom the positive charges have been displaced by d ⃗ relativeto the negative, so ⃗p = Qd ⃗ (and remember P ⃗ = n⃗p).PθdatThe charges move in the direction of ⃗ P . At the surface this leavesa layer of uncompensated charge, with a thickness t = d cos θ.6

The amount of charge in the element in the diagram is the productof the element’s volume and the charge per unit volume:i.e. the charge is: (da×t)×(nQ) = nQd cos θda = n⃗p· ⃗da = ⃗ P · ⃗da(8)The surface charge density (i.e. the charge per unit area) is thus:σ b = ⃗ P · ˆn (9)where ˆn is a unit vector, perpendicular to the surface and pointingout of the surface.6.2 Volume Bound Charge DensityWe saw above, that the net charge flowing across a surface when adielectric is polarised is P ⃗ · ⃗da. If the volume is enclosed by a surface,then the net charge which flows out of that volume is given by :∮Q out = ⃗P · ⃗da (10)areaThe charge left behind in the volume is just −Q out .∮∮−Q out = − ⃗P · ⃗da = − ∇ · ⃗P dτ (11)areaThe last step uses Gauss’ theorem.What this tells us is that the net charge density inside the volumeelement is :volρ b = −∇ · ⃗P (12)Note that since the volume charge density depends on spatialderivatives of the polarisation, it will be zero inside a uniformly polariseddielectric.7

7 Permittivity and SusceptibilityThe permittivity (and dielectric constant) are related to the susceptibility.For simplicity we consider the case of a constant field, as foundbetween parallel plates. We can then omit the vector nature of thefield.From Gauss’s lawUsing (9) and (4):E = σ f − Pɛ 0E = σ f − σ bɛ 0(13)= σ f − ɛ 0 χ e Eɛ 0giving E = σ fɛ 011 + χ e(14)The field is reduced by the factor χ e + 1. The capacitance isincreased by the same factor. Keeping in mind the definition of thedielectric constant, we have:K = ɛ r = χ e + 1 (15)The permittivity of a dielectric is : ɛ = ɛ r ɛ 0Equation (14) could be written in a variety of ways:E =σ fɛ 0 (1 + χ e ) = σ f= σ fɛ r ɛ 0 ɛ(16)Equation (16) illustrates an important point. In dielectrics it isusually possible to use the normal equations for calulating electricfields due to free charges by replacing ɛ 0 by ɛ.8

This could all get confusing!Remember• ɛ will be a very small number (since ɛ 0 = 8.84 × 10 −12 ).• Relative permittivity and dielectric constant are the same thing,with values usually between 1 and 100 (although 50,000 is possible).• χ e is one less than ɛ r . (So what is its value for free space?).8 Electric DisplacementThe differential form of Gauss’s law is:∇ · ⃗E = ρ ɛ 0(17)We now think of the charge as two parts: ρ = ρ f + ρ b .This means that ∇ · ⃗E = ρ f + ρ bɛ 0= ρ f − ∇ · ⃗Pɛ 0and this gives us ∇ · (ɛ 0⃗ E + ⃗ P ) = ρf (18)We now define a new quantity, the electric displacement vector⃗D, defined as⃗D = ɛ 0⃗ E + ⃗ P (19)(This also means that ⃗ D = ɛ ⃗ E) (20)We then have ∇ · ⃗D = ρ f (21)We can think of equation (21) as Gauss’s law in a dielectric.9

¨¨© ©© ©¨⃗D is convenient because it depends only on the free charge. Youcan calculate ⃗ D in the same way as you calculate ⃗ E, except thatyou only need include the free charge and there will be an ɛ 0 missingsomewhere.The integral form of equation (21) would be:∮⃗D · ⃗da = Q fenc (22)In a dielectric then, we have two equations:∇ · ⃗D = ρ f and ∇ × ⃗ E = 0 (23)ExampleA parallel plate capacitor with plates 5 mm apart is filled withpolythene, which has ɛ r = 2.26. The capacitor is connected to a 9 Vbattery.1. Calculate ⃗ E, ⃗ P , σ b , ⃗ D and σ f in the dielectric.2. Calculate the field due to the free charge and the depolarisingfield.3. What extra information would you need if you wished to calculate⃗p, the atomic dipole moment?¢ ¢ £ £ £ ¢¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡E P D¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥¤¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤ ¤¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥ ¥¤¦ ¦ ¦ § § § § ¦10

SolutionSince in this simple geometry, all the vector quantities are in thesame direction, we will ignore their vector nature.1.E = V d =P = ɛ 0 (K − 1)E9 V5 mm = 1.8 × 103 Vm −1= 8.85 × 10 −12 (2.26 − 1)1.8 × 10 3= 2.01 × 10 −8 Cm −2σ b = P = 2.01 × 10 −8 Cm −2D = ɛ 0 E + P= 8.85 × 10 −12 × 1.8 × 10 3 + 2.01 × 10 −8= 3.60 × 10 −8 Cm −2σ f = D = 3.60 × 10 −8 Cm −211

2. For a sheet of charge, with charge per unit area of σ, Gauss’slaw gives the field as σ/ɛ 0 .So E free = σ fɛ 0= 4.07 × 10 3 Vm −1and E bound = σ bɛ 0= 2.27 × 10 3 Vm −13. The polarisation ⃗ P is defined as the total dipole moment per unitvolume. In order to calculate ⃗p, the moment of an individualdipole, we would need to know the number of dipoles per unitvolume.12

9 Boundary Conditions.What happens at the junction of two dielectrics?In the figure, the dielectric above the horizontal line has permittivityɛ 1 , while below the line it is ɛ 2 .ε1ε2Dn1Dn2Et1Et2Take the Gaussian pill-box on the left in the figure and imagineits height goes to zero. If there is no free charge at the boundary,there is no net flux of ⃗ D out of the pill-box. Hence the boundarycondition for ⃗ D is:D n1 = D n2 (24)We also know that the line integral of the electric field aroundthe rectangle on the right must be zero (since ∇ × ⃗ E = 0). As weimagine the rectangle shrinking to an infinitesimal size, we see that:E t1 = E t2 (25)i.e. across the boundary of 2 dielectrics, the normal component of⃗D is continuous and the tangential component of ⃗ E is continuous.13

10 Force on a DielectricIn general, an electric field will exert a torque on a dipole which isnot parallel to the field and a force on a dipole if the two centres ofcharge are at points where the field is different.We will not look at this in detail, although it is covered in Griffiths.However there is an interesting example, which shows how lateralthinking can sometimes get answers out of apparently difficultsituations.Consider the situation shown below.Lx+QA parallel plate capacitor with a slab of dielectric partly insertedbetween the plates. The plates are isolated, so Q, the free charge onthe plates, is constant.Our task is to calculate the force on the dielectric slab. The situationis complex. The field is not constant and hence the polarisationwill also vary with position.The method used is to consider the energy. If the slab moves adistance dx into the capacitor, the energy changes since the capacitancechanges. From conservation of energy we know that the workdone by the force moving the dielectric must be equal to the changein the stored energy.( 1 Q 2 ) (i.e. F dx = ∆ = − 1 Q 2 )dC =(− 1 )2 C 2 C 2 2 V 2 dC (26)14-Q

Take the length of the plates to be L and the distance the dielectricintrudes into the space between the plates to be x. The totalcapacitance is thus:C = ɛ 0A L − xd L+ Kɛ 0A xd L = ɛ 0Ad+ ɛ 0A (K − 1)xd LIf x changes by dx, then the change in C will be:(27)dC = ɛ 0A (K − 1)dx (28)d LCombining equations (26), (27) and (28) gives finally:F = − 1 2 V 2ɛ 0A (K − 1)(29)d LThe expression itself is not particularly important. The interestingpoint is that it is possible to calculate the force in a complex situationin such a simple way.15

11 Energy in a Dielectric.The following represents a simple derivation, although the result obtainedis quite general.We know that the energy stored in a charged capacitor isW = 1 2 CV 2 (30)For a parallel plate capacitor C = Kɛ o A/d and V = Ed.Thus W = 1 2 Kɛ oAdE 2 (31)The energy per unit volume stored in the capacitor is thus:U = 1 2 Kɛ 0E 2 = 1 2 ɛE2 = 1 2 ⃗ D · ⃗E (32)The energy stored in a dielectric, for a given field, is K times largerthen the energy stored in a vacuum. The extra energy comes fromthe polarization of the dielectric.16

12 SummaryThe state of a dielectric is characterised by three vector quantities:The electric field ⃗ EThe polarisation ⃗ P (= total dipole moment per unit volume).The displacement ⃗ D defined by ⃗ D = ɛ 0⃗ E + ⃗ P .There are relations between these quantities.⃗D = ɛ ⃗ E = ɛ r ɛ 0⃗ E and ⃗ P = ɛ0 χ e⃗ EThe polarisation gives rise to bound charges:σ b = ⃗ P · ˆn and ρ b = −∇ · ⃗PThe displacement D ⃗ is determined by the free charges.∮∇ · ⃗D = ρ f (in integral form this is ⃗D · ⃗da = Q fenc )The field E ⃗ is determined by ALL the charges (i.e.bound)free plusAt a dielectric boundary the normal component of ⃗ D and thetangential component of ⃗ E are continuous.The energy density in a polarised dielectric is 1 2 ⃗ D · ⃗E.17

ExampleA hollow dielectric sphere, with inner radius R 1 and outer radiusR 2 has susceptibility χ e . A charge Q is spread uniformly over theinner surface of the sphere’1. Find ⃗ D, ⃗ E and ⃗ P in the three regions r < R1 , R 1 ≤ r ≤ R 2and r > R 2 .2. Calculate the bound charges.++++ ++++ + + 18+++R+ + ++1R+++2+++