Download presentation

Presentation is loading. Please wait.

Published byMegan Newton Modified over 2 years ago

1
Microscopic picture – change in charge density when field is applied 3). Dielectric phenomena E (r) Change in electronic charge density Note dipolar character r No E field E field on - + (r) Electronic charge density

2
Dipole Moments of Atoms Total electronic charge per atom Z = atomic number Total nuclear charge per atom Centre of mass of electric or nuclear charge distribution Dipole moment Zea

3
Electrostatic potential of point dipole +/- charges, equal magnitude, q, separation a axially symmetric potential (z axis) a/2 r+r+ r-r- r q+q+ q-q- x z p

4
Equipotential lines: dipole Contours on which electric potential is constant Equipotential lines perp. to field lines

5
Field lines: point dipole Generated from r i j k ← NB not a point dipole

6
Insulators vs metals Insulator –Localised wave functions Metal –Delocalised wave functions No E field E field on

7
Polarisation P = dipole moment p per unit volume Cm/m 3 = Cm -2 Mesoscopic averaging: P is a constant vector field for a uniformly polarised medium Macroscopic charges p in a uniformly polarised medium Polarisation P - + p E P E E dSdS E p = ___?

8
Depolarising electric field Depolarising electric field E Dep in uniformly polarised ∞ slab Macroscopic electric field E Mac = E + E Dep E Dep - + E Mac - + EE E Dep = P o + P o E Dep = -P/ o E Mac = E - P/ o

9
Relative Permittivity and Susceptibility E Mac = E – P/ o = ( plates – P)/ o in magnitude o E = o E Mac + PP = o E E Mac o E = o E Mac + o E E Mac = o (1 + E )E Mac = o E Mac –E Mac = E / –E = E Mac Dielectric constant (relative permittivity) = 1 + E –Typical values: silicon 11.8, diamond 5.6, vacuum 1 Dielectric susceptibilty E

10
Polar dielectrics molecules possess permanent dipole moment in the absence of electric field, dipoles randomly oriented by thermal motion hence, no polarisation. e.g. HCl and H 2 O …..but not CS 2 no net dipole moment + + + + _ __ _ E zero field, random preferential alignment net P=0 but P Np

11
Effect of orientation on net field Effect of alignment is to reduce the net field Tendency to align is opposed by thermal effects Balance is determined by Boltzmann statistics Key factor is ratio of the potential energy of the dipole (U) to the temperature (T), which enters as exp(-U/kT) E appl + _ _ + E dip

12
Potential energy of dipole in E field Potential energy U (U c.f. W from before) when charge density of molecule ( ) is in slowly spatially varying external potential (No factor of ½ c.f. W) E _ + If q = 0, leading term is –p.E = -pEcos

13
Number Distribution function Angular distribution function: N( ) (no E field), N’( ) (E field) Number of dipoles oriented between and +d : N( )d Total number of dipoles N dd E||z no E field E field on

14
Total number of dipoles N Number Distribution function

15
Susceptibility of polar dielectric Molecules acquire induced dipole moment through: - reorientation - polarisation of molecular charge (polar or nonpolar molecules) EE 1/T

16
The Langevin Equation When U/kT is not small, integration of N( )d yields: Plotting P vs pE/kT shows two distinct regimes: (1)High E, low T: all dipoles aligned: (2)Low E, high T: small U/kT approximation: pE/kT NpNp P

17
Clausius-Mossotti equation Relationship between r and polarisability density N including local fields Neglected local field for polar dielectrics (dilute gases) Each molecule, atom, etc. located in spherical cavity C-M local field is external field + field due to polarisation charges on cavity surface E loc = E + E pol pol = P.dS dd R P, E dSdS - ---- - + ++++ + cos - - cos P.dS= - P dS cos ring area element = 2 Rsin Rd P dSdS

18
Clausius-Mossotti equation Charge on ring area element - P dS cos - o E E 2 Rsin Rd cos Contribution to field at centre of cavity from pol on ring o E E 2 Rsin Rd cos 4 o R 2 ) = E E sin cos d 2 Field || P due to all charge on cavity surface E pol = E E/3 Local field E loc = E + E pol = (1+ E /3)E P = o N E loc = o N (1+ E /3)E (in cavity) P = o E E (in bulk) N (1+ E /3) = E N = E / (1+ E /3) N /3 = ( r – 1)/( r + 2) since r = 1 + E

19
Non-uniform polarisation Uniform polarisation surface charges only Non-uniform L polarisation bulk charges also Displacements of positive charges Accumulated charges P - + P E ++--

20
Non-uniform polarisation Box with origin of local axes at (x,y,z), volume x y z Charge crossing area dS = P.dS (x,y,z) ( x+ x,y,z ) Charge entering LH yz face Charge exiting RH yz face Net charge entering box Total charge including zx and xy pairs of faces

21
Electric displacement D What happens when a charge is added to a neutral dielectric ? Two types of charge: Those due to polarisation (bound charges) Those due to extra charges (free charges) (charge injection by electrode, etc) Total charge Added (free) charge Polarisation (bound) charge response of dielectric to added charge

22
Electric displacement D Gauss’s Law Displacement: a vector whose div equals free charge density Units: C·m -2 (same as P) D relates E and P D = o E + P is a constitutive relation Can solve for D field and implicitly include E and P fields

23
Validity of expressions Always valid:Gauss’ Law for E, P and D relation D = o E + P Limited validity: Expressions involving r and E Have assumed that E is a simple number: P = o E E only true in LIH media: Linear: E independent of magnitude of E interesting media “non-linear”: P = E o E + 2 E o EE + …. Isotropic: E independent of direction of E interesting media “anisotropic”: E is a tensor (generates vector) Homogeneous: uniform medium (spatially varying r )

24
Boundary conditions on D and E Simplest example – charged capacitor with dielectric D is continuous ┴ boundaries (no free charges there) E is discontinuous ┴ boundaries - + D = o E D = o E + P = o r E mac D = o E - + E E mac =E/ r E

25
Boundary conditions on D We know that Absence of free charges at boundary D 1 cos 1 S – D 2 cos 2 S = 0 D 1 cos 1 = D 2 cos 2 D 1 ┴ = D 2 ┴ Perpendicular component of D is continuous Presence of free charges at boundary D 1 cos 1 S – D 2 cos 2 S = S f D 1 ┴ = D 2 ┴ + f Discontinuity in perpendicular component of D is free charge areal density 1 2 (E1,D1)(E1,D1) (E2,D2)(E2,D2) 22 11 S

26
Boundary Conditions on E We know that for an electrostatic E field E and D are constant along the horizontal sides of C in regions 1 or 2 Sides of C thin enough to make no contribution Parallel component of E is continuous across boundary 1 2 (E2,D2)(E2,D2) (E1,D1)(E1,D1) 22 11 dℓ1dℓ1 dℓ2dℓ2 C AB

27
Interface between 2 LIH media LIH D = r o E E and D bend at interface 1 2 11 22

28
Energy of free charges in dielectric In vacuum Assembling free charges in a dielectric

29
Method of Images Derives from Uniqueness Theorem: “only one potential Satisfies Poisson’s Equation and given boundary conditions” Can replace parts of system with simpler “image” charge arrangements, as long as same boundary conditions satisfied Method exploits: (1)Symmetry (2)Gauss’s Law or specified Image charges reproduce BC

30
Basic Image Charge Example Consider a point charge near an infinite, grounded, conducting plate: induced -ve charge on plate; potential zero at plate surface Complex field pattern, combining radial (point charge +Q) and planar (conducting plate) symmetries, can also be viewed as half of pattern of 2 point charges (+Q and -Q) of equal magnitude and opposite sign! +Q -Q

31
Basic Image Charge continued Arrangement is equivalent because it keeps the same boundary condition (potential zero on plate and zero potential on the median line). Point charge -Q is located same distance behind, like an image in a plane mirror. The resulting field is easy to calculate (vector sum of fields of 2 point charges of equal and opposite sign) Field lines must be normal to surface of conductor Also easy to calculate the induced -ve charge on plate!

32
Distribution of induced charge Induced charge is related to the outward E field at the surface: Find E using image charge (-ve and varying with r) D +Q-Q E E+E+ E-E- r

33
Total induced charge Introduce parameter s ind has azimuthal symmetry: consider elemental annulus, radius s, thickness ds ds r D ss

34
Total induced charge: implications 2 conclusions from the result: (1)Induced charge equals the negative of original point charge - trivially true in this case only! (2)Induced charge equals the image charge - generally true! Consider Gauss’s Law, concept of enclosed charge Must not try to determine E in the region of image charge! In this case (behind infinite conductor) it is zero, which is not the answer the image charge would yield

35
Point charge near grounded conducting sphere By comparison with previous example: (1)Distance D to centre of symmetry, radius a (2)Image (charge) location (3)-ve induced charge predominantly on side facing +Q (4)Boundary condition, zero potential on sphere surface Expect image charge will be a point charge on centre line, left of centre of sphere, magnitude not equal to Q, call it Q D +Q a - - - -

36
Point charge near grounded conducting sphere Q distance b from centre, =0 at symmetry points P 1 and P 2 Q D P2P2 P1P1 Q b

37
Point charge near floating conducting sphere On its own, floating the sphere at V relative to ground results in uniform +ve charge density over the surface. In the presence of Q, induced -ve charge predominantly on left; this complex system easily solved by 2 image charges: Q V QQQ

38
Point charge near isolated conducting sphere With no connection to ground, the sphere is at an unknown non-zero potential ; easily solved by same 2 image charges: the potential is still determined by Q but in this case, the sphere is overall neutral: Q+Q =0 (same potential as if sphere was absent!) Q QQQ

39
Point charge near LIH dielectric block Q polarises dielectric and produces bound surface charge b,ind b,ind = P.n = o ( r -1)E ni E ni normal component of E inside b,ind negative if Q positive Q D s r E ni E no b,ind =1 = r

40
Point charge near LIH dielectric block Image charge for E no Image charge for E ni +Q r E E+E+ E-E- r r E E+E+ r

41
Point charge near LIH dielectric block (see Lorrain,Corson & Lorrain pp 212-217) Outside: remove dielectric block and locate image charge Q a distance D behind Inside: remove dielectric block and replace original point charge Q by Q

42
Point charge near LIH dielectric block contrasting E field patterns conductor dielectric note dielectric distorted outside but radial inside

Similar presentations

Presentation is loading. Please wait....

OK

Application of Gauss’ Law to calculate Electric field:

Application of Gauss’ Law to calculate Electric field:

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on computer languages and platforms Ppt on sources of energy for class 8th science Ppt on water pollution problems and solutions Ppt on tyres manufacturing process Ppt on basic concepts of cost accounting Ppt on social contract theory definition Ppt on barack obama leadership academy Ppt on environmental issues and their solutions Ppt on national education day 2017 Ppt on agriculture system in india