in Static Electric

™ No free charge in dielectrics to make interior 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 material

¾ a. Polar : permanent moments -30 ex) H2O (two or more dissimilar ) Æ P ~ 10 (C·m) )Individual are randomly oriented Æ macroscopically no net dipole )Some have a permanent dipole moment even in the absence of external field Æ

¾ b. nonpolar molecules : no permanent dipole moments

Radio Technology Laboratory 1 Equivalent Charge Distribution of Polarized Dielectrics

™ Define vector , P nvΔ p ∑ k P = limk =1 [C m2 ] : volume density of 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' ′ επε 44πε 00RR Polarization charge density ∴ ρ ps = Pni ˆ ρ p =−∇′iP or bound charge density

Equivalent polarization Equivalent polarization 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 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 ): 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

™ of dielectric material ¾ linear and isotropic dielectric media :

) Polarization is directlyyp prop ortional to the electric field PE=εχ0 e

where χe : (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 εχ000 e =+ (1 ) E 0 εε ε e ==0 r EE εχ ε re=+ 1 = : or dielectric constant of the medium ε0

ε = εε0 r : absoult e permitti vit 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 . ⎡⎤ ⎡⎤ Dxx⎡⎤ε εεE ⎢⎥⎢⎥εεε11 12 13 ⎢⎥ ⎢⎥DEyy= ⎢⎥21 22 23 ⎢⎥ ⎢ ⎥⎢⎥⎢⎥εεε ⎣⎦DEzz⎣⎦31 32 33 ⎣⎦

™ 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 , επ επε πε 44πε 01bb 02 4 01 b 4 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

ρρ = −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 )ssor 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. EE sinαα= sin 2211 ™ Normal D should be continuous at boudary.

εαεα22EE cos 2= 11 cos 1 22 22 ∴=EEE222tn + = ( E 2 sinαα222 )+ (E cos ) 1 αααα22⎡ ⎤ 2 22⎛⎞εε11 ⎛⎞ =+ (EEE11 sin )⎜⎟ 1 cos 1 =+ 1⎢ sin 1 ⎜⎟ cos ⎥ ⎝⎠εε22⎣⎢ ⎝⎠⎦⎥ 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.

9 The radius of the inner conductor : 0 .4 cm ε 9 Concentric layers of rubber ε rr = 3.2

9 Polystyrene ,rp = 2.6

¾ Design a cable that is to at a 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 → C onsider only Er component

6 ρl ⎛⎞1 maxEr =××= 0.25 25 10 ⎜ ⎟ 23.2πε0 ⎝ ri ⎠ ρ ⎛⎞1 maxE =××= 0.25 20 106 l p ⎜⎟ 22.6πε0 ⎝⎠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⎜⎟ ε εεπε ε 222.63.2πε ε 00⎣⎦⎝⎠⎢⎥rp⎝⎠⎝⎠rr rr 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] 2 πε0

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 = ⋅ ⋅ επε 2πε 00 2.6rrir 2 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 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 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 bound 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 μ C/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 πε p 1/2 E =+⎡⎤ 1 3cos2 θ 3 ⎣⎦ 4 0 r

Radio Technology Laboratory 14