Lectures 7 & 8 Electrostatics: Dielectrics

Lectures 7 & 8 Electrostatics: Dielectrics

Introduction Dielectrics represent a class of materials which, although insulators, exhibit a number of effects when placed in an E-field. A good example is their effect on capacitors. A capacitor has capacitance C0 when the space between its two conductors is a vacuum, filling this space with a dielectric increases the capacitance to a new value Cm. The ratio Cm/C0=r is known as the relative permittivity of the dielectric. The plates of a capacitor are charged to some value Q and then isolated. Because of the relationship Q=CV, if C increases when a dielectric is placed between the conductors, as Qmust remain constant V must decrease. Because V is related to the E-field by E=-V, the introduction of the dielectric must decrease Ebyr. The aim of this lecture is to understand why this occurs.

LIH dielectrics • In order to simplify our treatment we will consider a sub-class of dielectrics known as LIH ones. Such dielectrics exhibit the following properties. • Linearity (LIH) All properties are proportional to the magnitude of any applied E-field. This implies that r is a constant and is independent of applied voltage or E-field. • Isotropy (LIH) ris independent of the orientation of the material i.e. it is independent of the direction of the E-field. • Homogeneity (LIH) r has the same value at all points in the dielectric.

For the majority of dielectrics r is found to vary with temperature and the frequency (if any) at which the applied E-field is modulated. The value of r in a steady E-field is known as the static relative permittivity. It is this value that is relevant to electrostatic situations.

The effect of an E-field on a dielectric (qualitative treatment) Because dielectrics are insulators they contain no free charges (charges which are able to move around freely within the material when an E-field is applied). However although unable to move far, the positive and negative charges may be able to move slightly (but in opposite directions) when an E-field is applied.

Consider an atom which consists of negatively charged electrons orbiting the positively charged nucleus. In general the centroids of the positive and negative charges coincide and the atom produces no external E-field. Neither the electrons or the nucleus is free to move over large distances when an E-field is applied.

However the application of an E-field will cause the electrons to be moved, on average, a small distance in one direction and (possibly) the nucleus to move in the opposite direction. The centroids of the positive and negative charges now no longer coincide and we have an electric dipole which, from Lecture 4, we know produces an external E-field. In this case the atoms are said to be polarised.

Referring to the above diagram. For no external field the atoms are unpolarised (centroids of q coincide) and hence they produce no external E-field. When an external field (Eext) is applied the atoms are polarised and form small electric dipoles. In the present case the field moves the positive charges upwards and the negative ones downwards. As their centroids no longer coincide they produce an external field E*.

Because E-fields point from positive to negative charges, in the present case E* will point downwards and hence will oppose Eext. The total E-field within the dielectric will hence be less than the external field (Eext-E*) explaining our previous observation that the E-field between the conductors of a capacitor decreases when a dielectric is introduced.

The effect of an E-field on a dielectric (quantitative treatment) Polarisation. An applied E-field causes the atoms or molecules in a dielectric to become small electric dipoles. The polarisation P at any point in the dielectric is defined as the electric dipole moment per unit volume at that point. Hence if we have a small volume d it will contain a total dipole moment dp of Pd. This is the definition of the polarisation P(units Cm-2).

Consider now a small volume element of length dL and cross-sectional area dS. The application of an E-field causes the displacement of positive charge (+dq) out of one end of the element and an equal but negative charge (–dq) out of the other end This results in a dipole moment (p=sQ) of dp=dq.dL.

If instead of dq we use a surface charge density dq=.dS and hence dp=.dS.dL But dS.dL is simply d, the volume of the element and from the definition of P P=dp/dP= So polarisation charge (dq) crossing area dS is PdS. More generally if the area dS is not normal to the polarisation vectorP then Polarisation charge crossing dS: =PdS(A) This represents an alternative definition of P

Surface polarisation charges Consider a slab of dielectric in an applied E-field. Along the field direction we can consider the dielectric in terms of a sequence of small elements. If the dielectric is homogeneous (P is the same at all points) then the positive polarisation charge at the end of an element is balanced by an equal but opposite negative charge at the end of the neighbouring element. Only at the surface of the dielectric is there an unbalanced charge.

From above if Pn is the normal component of P at the surface then a surface polarisation charge density (P) exists given by P=Pn. If the material is not homogeneous then there will also be a volume polarisation charge.

Gauss’s Law in dielectrics, D-field Consider a general situation in which we have both free charges (labelled Qf) and charges due to the polarisation of one or more dielectrics (labelled QP). Gauss’s law can still be applied but we must now take account of both the free and polarisation charges. Applying Gauss’s law to the surface S

After some manipulation we arrive at because the term 0E+P occurs very frequently it is given a special name, the electric displacement or D-field, symbol D D=0E+P The units of D are Cm-2.

Gauss’s law in terms of D is now (integral form) In words this states ‘the outward flux of D over any closed surface Sequals the algebraic sum of the free charges enclosed by S’. The differential form of Gauss’s law for D is where fis the volume density of free charges.

From the previous equations we see that while both free and polarisation charges are sources of E, only free charges are sources of D. Because D is given by a modified form of Gauss’s law techniques used to find E for a given charge distribution can also be used to find D. In free space (no dielectric) P=0 and hence D=0E. Example. The D-field a distance r from a point charge Q is

Worked Example • Calculate the D- and E-fields1cm from a +1nC point charge • In vacuum • In a dielectric with r=10

Relationships between E, D and P Electric susceptibility. If the E-field at a point within a dielectric is E and the polarisation at that point is P then the electric susceptibility (e) is defined via the equation P=0eE For an LIH dielectric e is independent of the size and direction of E and position within the dielectric. Because D=0E+P D=0(1+e)E (B)

Relationship between e and r. Consider a system of free charges which produce an E-field and hence a D-field. If we now introduce a dielectric of relative permittivity r we know that D can not change and hence (B) above tells us that within the dielectric E must decrease to 1/(1+e) of its value in vacuum. However from our definition of r in terms of capacitance we know that introducing a dielectric results in a decrease in E by a factor r. Hence we must have the general result r=1+e

And hence we can also write D=r0E and P=0(r-1)E Some times =r0 is used where is the permittivity of the dielectric.

Summary of E, D and P The above diagram summarises the results. A parallel plate capacitor carries charge densities of f. A dielectric partially fills the volume between the plates.

D has the same value in the dielectric and in the vacuum. • The polarisation P is only non-zero in the dielectric. • The polarisation P results in surface polarisation charges P on the top and bottom surfaces of the dielectric.

The polarisation surface charges produce an additional E-field, Ep which opposes the vacuum field Evac to give a reduced total field Em within the dielectric. • Lines of D pass through the dielectric • Some lines of E terminate at the polarisation surface charges.

Worked example A parallel plate capacitor has A=20cm2, d=2mm and is charged to 5nC A dielectric with r=5 and thickness 1mm is placed between the plates, touching one plate. Find the E, D and P fields, the surface charge on the dielectric and the voltage between the plates.

Effect of dielectrics on the E-field circuital law in vacuum. Charges make no contribution to the integral of E around a closed path and this is true for both free charges and polarisation charges. Hence this result is still valid in a dielectric or in a vacuum or dielectric (but not when there is a time varying magnetic field!!).

General electrostatic equations in the presence of dielectrics In general all the previously derived equations for V and E can be used if 0 is replaced by 0r. (This is only strictly true for LIH dielectrics) For example if a point charge Q is placed in a dielectric then the electric potential V, at a distance r is given by

The only equation which can not be modified in this manner is Coulomb's equation for electrostatic forces. This is because to place a charge in a dielectric we would have to make a small hole in which to place the charge. Hence the charge would not strictly be in the dielectric. The solution of this problem is not trivial but fortunately we are generally interested in E-fields not forces.

Worked example A spherical capacitor has conductors of radii a and 4a. The region between a and 2a is filled with a dielectric of relative permittivity 5. What is the capacitance of the system?

Boundary conditions for D and E These are required for problems where we are concerned with what happens at the interface between two different media (e.g. vacuum and a dielectric, two different dielectrics etc). Consider the interface between two media 1 and 2 (these can be vacuum, conductors or insulators). For generality we assume there is a free charge density f at the interface.

For D take a Gaussian surface of the cylinder shown. If height is made infinitesimally small then need only consider the flux through the two ends. Applying Gauss’s law for D. D1dS-D2dS=f dS In the limit dS0 D1n-D2n=Dn=f Where Dn is the component of D normal to the surface. This result shows that the normal component of the D-field is discontinuous by f across any surface. If f=0 then the normal component of D is continuous.

For the E-field take the rectangular path as the closed loop. If its height is made infinitesimally small then the only contribution to the line integral is along the top and bottom edges Where Etis the tangential component of E. Hence E1t=E2t; the tangential component of E is continuous across any interface.

Now consider a general case where an E-field (and hence D-field) passes from a first dielectric to a second. If there is no surface charge at the interface then we must have the following (E-tangential) (C) (D-normal) (D) but D1=0rE1 and D2=0rE2 hence we can write (E) finally dividing (E) by (C)

Stored energy in terms of E and D In the absence of dielectrics it was shown in Lectures 2 and 3 that electrical potential energy (U) could be written in terms of the E-field (E) as This equation is modified in the presence of dielectrics.

For a parallel plate capacitor filled with a dielectric having a relative permittivity rwe have the following hence but hence

This can be shown to be a general result. Hence for situations where E (and D) are not uniform The dot product of D and E is required because for non-LIH dielectrics D and E may not be parallel to each other.

Conclusions • Effects of a dielectric on E, Vand C • LIH dielectrics • Qualitative explanation of dielectrics • Quantitative explanation of dielectrics, definitions of P • Surface polarisation charge • Electric displacement (D-field) • Gauss's law for D-fields • Relationships between E, D and P (r, e) • Electrostatics in the presence of dielectrics • Boundary conditions for D and E • Electrical potential energy in terms of D and E

Lectures 7 & 8 Electrostatics: Dielectrics