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Dielectrics in Static Electric Field

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Dielectrics in Static Electric Field Dielectrics in Static Electric Field No free charge in dielectrics to make interior charge density and electric field vanish Dielectrics contain bound charge Æ effect on the electric field E-field Æ small disppplacement of positive and neg gg(ative charges (bound charge) Æ polarize a dielectric material ¾ a. Polar molecules : permanent dipole moments -30 ex) H2O (two or more dissimilar atoms) Æ P ~ 10 (C·m) )Individual dipoles are randomly oriented Æ macroscopically no net dipole )Some have a permanent dipole moment even in the absence of external field Æ electrets ¾ b. nonpolar molecules : no permanent dipole moments Radio Technology Laboratory 1 Equivalent Charge Distribution of Polarized Dielectrics Define polarization vector , P nvΔ p ∑ k P = limk =1 [C m2 ] : volume density of electric dipole moment Δ→v 0 Δv dP of an elemental volume dp= Pdv′ PRi ˆ ⎛⎞PRi ˆ ′ dV = 2 dv ⎜⎟cf) potential due to a dipole V = 2 4πε 0 R ⎝⎠4πε 0 R 1 PRi ˆ ∴ VdvR= ′′, where is the distance from dv to a fixed field point. ∫v' 2 4πε 0 R cf) Rxxyyzz2222=−()()()′′′ +− +− in Cartesian coordinate. 11⎛⎞RRˆ ′′⎛⎞ ∇=∇⎜⎟ ⎜⎟ ==32 RRR⎜⎟222 ⎝⎠ ⎝⎠()()()xx−+−+−′′′ yy zz Rxxxyyyzzz=−(′′′ )ˆˆ +− ( ) +− ( ) ˆ Radio Technology Laboratory 2 Equivalent Charge Distribution of Polarized Dielectrics 11 ⎛⎞ ∴=VPdvi ∇′ ′ ∫v' ⎜⎟ 4πε0 ⎝⎠R ⎛⎞11⎛⎞P cf) ∇=∇+∇⇒∴∇=∇−∇′′′′′′iiiii(f A )f AA f PP⎜⎟ ⎜⎟ ( i ) ⎝⎠R ⎝⎠RR 1 ⎡⎤⎛⎞PP∇′i ∴Vdvdv=∇−⎢⎥′′i⎜⎟ ′ ∫∫vv'' 4πε 0 ⎣⎦⎢⎥⎝⎠RR 1'Pni ˆ 1(−∇′iP ) = ds''+ dv 4πε ∫∫Sv′ RR4πε ' 0 0 surface volume cf) potential due to surface and volume charge 11ρρ VdvVds==vs′′ (3-61), (3-62) ∫∫vS' ′ 44πε00RR πε Polarization charge density ∴ ρ ps = Pni ˆ ρ p =−∇′iP or bound charge density Equivalent polarization Equivalent polarization surface charge density volume charge density Radio Technology Laboratory 3 Equivalent Charge Distribution of Polarized Dielectrics cf) Imagi inary el emental surf ace Δs of a nonpoldililar dielectric, net ch arge crossing the surface Δs is, ΔΔΔQdQ =⋅nq ()d nsˆ Δ , whithbfllhere n is the number of molecules per unit vo lume nqd : dipole moment per unit volume ⇒ polarization vector P ΔQ Δ=QPnsiiˆˆ( Δ ) ⇒ ρ ps = = Pn Δ s outward normal cf) For a surface SV bounding a volume , the net total charge flowing out of V = negative of the net charge remai ning within the volume V QPndsP=−iiˆ = −( ∇ ) dv= ρ dv ∫S ∫∫vvp Radio Technology Laboratory 4 Electric Flux density and Dielectric Constant 1 ∇=iE ()ρρ + ε p 0 equivalent volume charge ρ p =−∇iP density of polarization ∴ ∇i()ε 0 EP+=ρ Define new fundamental qqyuantity 2 DEP=+ε0 (C/m ): electric flux density or electric displacement ∇=iD ρ (C/m3 ) valid everywhere free charge density note no ε 0 appear cf) ∇×E =0 Two of static maxwell equations ⎛⎞ Q ∴ Ddsii=⇒ Q Gauss' sslaw law Eds = ∫∫s ⎜⎟ ⎝⎠ε0 Radio Technology Laboratory 5 Electric Flux density and Dielectric Constant Permittivity of dielectric material ¾ linear and isotropic dielectric media : ) Polarization is directlyyp prop ortional to the electric field intensity PE=εχ0 e where χe : electric susceptibility (dimensionless quantity) cf) medium is linear if χe is independent of E medium is homogeneous if χe is independent of space coordinate. DEPE= ε +=εεχ + E 000 e =+εχ(1 ) E 0 e ==εε0 r EE ε ε εχre=+1 = : relative permittivity or dielectric constant of the medium ε 0 ε = εε0 r : absou lte perm ittivit y ( permitti vit y) Radio Technology Laboratory 6 Electric Flux density and Dielectric Constant ¾ anisotropic medium ) Dielectric constant is different for different directions of the electric field ⇒ DE and vectors generally have different directions. ⇒ permittivity is a tensor. ⎡⎤ ⎡⎤ Dxx⎡⎤ε εεE ⎢⎥⎢⎥11 12 13 ⎢⎥ ⎢⎥DEyy= ⎢⎥εεε21 22 23 ⎢⎥ ⎢ ⎥⎢⎥⎢⎥εεε ⎣⎦DEzz⎣⎦31 32 33 ⎣⎦ Dielectric strength ¾ Maximum electric field intensity that dielectric material can withstand without breakdown. Radio Technology Laboratory 7 Ex) 3-13. Two connected conducting spheres 9 Two spherical conductors with radius bbbb1221 an d(>)d (> ) 9 Connected by a conducting wire 9 Distance is large enough to ignore influence of each sphere on the other 9 Total charge Q is deposited on the sphere sol) Two conductors are at the same potential QQ12 Q 1 Q 2 a) === b) EE12nn22, 44πε01bb πε 02 4 πε 01 b 4 πε 02 b 2 Qb11E1n ⎛⎞ b 2Qb12 = , QQ12+= Q ∴ =⎜⎟ = Qb22 E 2n ⎝⎠ b 1Qb21 bb12 QQQQ12==, bb12++ bb 12 larger curvature ⇒ smaller sphere:higher electric field intensity Radio Technology Laboratory 8 Boundary Conditions for Electrostatic Fields (outward normal to medium 1) ① Δh → 0 Edlii=Δ+−Δ E12 w E i() w ∫abcda =Δ−ΔE12ttwE w= 0 DD ∴ EE= or 12tt= 12tt εε 12 ) the tangential component of an E field is continuous across an interfac e = −nˆ2 ② Cylinder Δ→h 0 Ddsiii=+Δ=−Δ==Δ( D12 nˆˆˆ D n ) S n i ( D 12 D ) S Qρ S ∫s 21 2 s ∴ nDˆˆ21ii(12−= D )ρρssor nD ( 21 −= D ) 2 ˆ i.e., DD12nns−=ρ (C/m ) reference normal is n2 Radio Technology Laboratory 9 Ex) 3 -15 Boundary conditions Tangential E should be continuous at bounda ry. EEsinαα= sin 2211 Normal D should be continuous at boudary. εαεα22EEcos 2= 11 cos 1 22 22 ∴=EEE222tn + =( E 2 sinαα222 )+ (E cos ) 1 22⎡ ⎤ 2 22⎛⎞εε11 ⎛⎞ =+(EEE11 sinαααα )⎜⎟ 1 cos 1 =+ 1⎢sin 1 ⎜⎟ cos ⎥ εε ⎝⎠22⎣⎢ ⎝⎠⎦⎥ note 1 The normal component of D field is discontinuous across an interface where a surface charge exists. note 2 If, ρsnn==0, then DD12 EE12tt= Summary : nˆ1 ⋅()DD21−=ρs Radio Technology Laboratory 10 Ex) 3 -16. coaxial cable 9 The radius of the inner conductor : 0 .4 cm 9 Concentric layers of rubber ε rr = 3.2 9 Polystyrene ,ε rp = 2.6 ¾ Design a cable that is to work at a voltage rating of 20kV. → E field are not to exceed 25% of their dielectric strength. sol) Dielctric strength of rubber : 25 × 106 Vm Dielctric strength of polystyrene : 20 × 106 Vm Cylindrical symmetry → Consider only Er component 6 ρl ⎛⎞1 maxEr =××= 0.25 25 10 ⎜ ⎟ 23.2πε 0 ⎝ ri ⎠ ρ ⎛⎞1 maxE =××= 0.25 20 106 l p ⎜⎟ 22.6πε 0 ⎝⎠rp rrpi==1. 54 0. 616 [cm] Radio Technology Laboratory 11 Ex) 3 -16. coaxial cable Potential difference rr −−=piEdr Edr 20,000 ∫∫rrpr 0. p ρρ⎡⎤⎛⎞1111⎛⎞⎛⎞rrpi 1r 1rp llo⎢⎥−+−=dr dr ⎜⎟ln + ln = 20,000 ⎜⎟⎜⎟∫∫rr op⎜⎟ 222.63.2πε00⎣⎦⎝⎠⎢⎥ εrp⎝⎠⎝⎠rr ε rr πε rr p i rri ==0.4,p 0.616, ρl 6 4 ∴=×××0.25 20 10 2.6rp =×8 10 , ∴ rroi =2.08 = 0.832 [cm] 2πε 0 553.22.6rp cf) =⇒=× rrpi 4 3.2ri 4 2.6 if order is reversed, ρρ11 11 0.25×× 20 1066 =ll ⋅ ⋅ , 0.25 ×× 25 10 = ⋅ ⋅ 2πε00 2.6rrir 2 πε 3.2 4426104442323.2rr .610.4 ==×=<⇒, ∴ rrrrrriiri, Non sense 5 2.6ri 5 3.2 16 Radio Technology Laboratory 12 Homework 3 (Cheng 3 - 6) Two very small conducting spheres, each of a mass 1.0× 10-4 (kg ), are suspended at a common point by very thin nonconducting threads of a length 02(0.2 (m) . A ch arge Q Qi is pl aced on a each sphThltiffhere. The electric force of repul lision separates the spheres, and an equilibrium is reached when the suspending threads make an angle of 10 o. Assuming a gravitational force of 9 .90 (N/kg) and a neglig ible mass of the threads, find Q 3 (Cheng 3 -12) Two infinitely long coaxial cylindrical surfaces, r = a and r = b (b > a), carry suirface charge densities ρsaand ρ sb , respectively. a) Determine E everywhere. b) What must be the relation between a and b in order that E vanishes for r > b? Radio Technology Laboratory 13 Homework 3 (Cheng 3 - 27) What are the boundary conditions that must be satisfied by the electrics potential at an interface between two perfect dielectrics with dielectric constants εr1 and ε r2? 3 A point charge q is enclosed in a linear, isotropic, and homogeneous dielectric medium of infinite extent. Calculate the E field, the D field, the polarization vector P, the bound surface chditharge density ρsb , and the boun d vol ume ch arge d ensit y ρvb . 3 A very thin, finite and uniformly charged line of length 10 m carries a charge of 10 μC/m. Calculate the electric field intensity in a plane bisecting the line at ρ = 5m. 3 Show that the magnitude of the electric field intensity of an electric dipole is p 1/2 E =+⎡⎤1 3cos2 θ 3 ⎣⎦ 4πε 0 r Radio Technology Laboratory 14.
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