Chapter 6. Dielectrics and Capacitance

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1 Chapter 6. Dielectrics and Capacitance Hayt; //009; 6- Dielectrics are insulating materials with n free charges. All charges are bund at mlecules by Culmb frce. An applied electric field displaces charges prducing electric diples. This is measured as the dielectric cnstant. 6. The Nature f Dielectric Materials Bund charges are bund at atms r mlecules. They are nt free t mve but displaced under an applied electric field. All dielectric materials can stre electric energy by a relative shift f psitive and negative charges. Plar mlecules : The mlecule has a permanently displaced psitive and negative charges. The permanent diples align t the applied electric field. Nnplar mlecules : N electric diple if there is n applied electric field. The psitive and negative charges are displaced under the applied field. The diple mment is defined as p d A vectr frm the negative t the psitive charge. The plarizatin is defined as diple mment per unit vlume. N P lim pi Δv 0 Δv i The bund-charge density acts like the free vlume-charge density in prducing electric field Diples are prduced by the electric field. p makes an angle θ with Δ. Hypthetical surface in the middle

2 Hayt; //009; 6- Psitive charges belw the surface crss the surface bundary upward. Negative charges abve the surface crss the surface bundary dwnward. The ttal psitive charge abve the bundary is Δ b d cs θ Δ ( n) P Δ Vlume Number density f diple Atms abve and belw f the bundary The net charge within the clsed surface, P d b Using divergence therem ρ dv P dv v b v P ρ b Gauss s law including the bund and the free charges ε E d b + : The electric field frm the bund and the free charges Cmbining tw equatins ε E + P d ( ) D, the electric flux density Using divergence therem D ρ v : D is calculated if the free charge is knwn. The relatin between P and E P εχe e χ e, electric susceptibility The electric flux density is D εe + P ε ( + χe) E εεre εe D εe ε r, relative permittivity r dielectric cnstant ε, permittivity ε, permittivity in vacuum ε is a cnstant in istrpic material but a tensr in anistrpic material Dx εxxex + εxyey + εxzez Dx εex Dy εyxex + εyyey + εyzez Dy 0+ ε Ey + 0 Dz εzxex + εzyey + εzzez Dz ε 3Ez Fr right chice f the crdinates

3 6. Bundary Cnditins fr Perfect Dielectric Materials Hayt; //009; 6-3 Apply Faraday s law fr statics at the bundary E d 0 Etan Δw Etan Δ w 0 E E tan tan In this case the tangential electric flux densities are Dtan εetan Dtan ε D ε E D ε tan tan tan Apply Gauss s law at the bundary D d DNΔ DNΔ ρδ Δ DN DN ρ ρ, surface charge density In mst cases D D N N The electric field ε E ε E N N The ttal D and E The nrmal cmpnent f D shuld be cntinuus, D csθ D csθ The tangential cmpnent f E shuld be cntinuus, D D sinθ sinθ ε ε Cmbine tw eqs. tanθ ε : ince θ > θ, we nte that ε > ε tanθ ε The magnitudes f D and E in terms f D and E ε ε cs θ + sin θ, sin θ + cs θ ε ε D D E E D is larger in the regin f larger permittivity. E is larger in the regin f smaller permittivity.

4 Hayt; //009; 6-4 The bundary cnditin at the interface between a cnductr and a dielectric slab D E 0 inside the cnductr. D 0 at the cnductr surface. t E t D E 0 t t Charge relaxatin time in cnductrs me charges are injected inside a cnductr Cnductinn current will be created. ρρ J ρ σ D t ε t Use J σ E, D εe Use Gauss s law D ρ ε ρ ρ σ t Assuming σ t be independent f ρ ( / ) ρ ρ )t : ε /σ, relaxatin time cnstant. e σ ε D N εe N ρ

5 Hayt; //009; Capacitance Tw charged cnductrs in a hmgeneus dielectric. The equal charge f ppsite sign. The charges shuld be n the cnductr surfaces. The electric fields are nrmal t the surfaces. Each surface is an equiptential surface. The electric flux is frm M t M. M is at the higher ptential than M with ptential difference f V. The capacitance is defined as C V C εe d + + E d An increased charge density by a factr f N Increased electric field by a factr f N Increased ptential by a factr f N The same rati f /V and the same capacitance The capacitance depends nly n the physical dimensin and the permittivity f the dielectric. Its unit is farad [F]. Example An infinite plate parallel capacitr The unifrm electric field frm the unifrm surface charge density ρ is expected. Frm the Gauss s law, D ρ aˆ s z and the electric field is ρs E aˆ z ε The ptential difference lwer 0 s s V E d ρ ρ dz d upper d ε ε s The capacitance is infinite due t the infinite plate and therefre the infinite ttal charge. The real capacitr has a finite area. In this case the ttal charge is ρs ρs V d ε Therefre ε C V d

6 Hayt; //009; everal Capacitance Examples A caxial capacitr f the inner and uter radii f a and b with the length. The ptential difference is frm Eq. ( ) f sectin 4.3 ρ V ab ln ( b/ a ) πε The ttal charge is ρ The capacitance is πε C ln ( b / a ) A spherical capacitr with cncentricc spheres f radii a and b. Fr a given charge the electric field is Er 4πεr The ptential difference is frm Eq. ( ) f sectin 4.3 V ab 4πε a b The capacitance is 4πε C V ab a b

7 Hayt; //009; 6-7 An islated spherical capacitr f radius a. et the uter radius b g t infinity, C 4πεa An islated spherical capacitr with dielectric cating. Dielectric cating f ε ε frm r a t r r Frm Gauss s law Dr 4πr The electric field Er ( a < r < r ) 4πε r ( r > r ) 4πε r The ptential difference is a r Va V dr dr r 4πεr + 4πεr 4π ε a r εr The capacitance is 4π C + ε a r εr A parallel plate capacitr with tw dielectrics () tart frm the ptential difference V Unifrm electric fields E and E in tw regins. The ptential difference is V E d + E d () Using bundary cnditin D N DN εe εe () The surface charge density ρ D εe (3) Insert () and () int (3) V ρ d d + ε ε Using the ttal charge f ρ ρ C V V d d + + ε ε C C

8 () tart frm the ttal charge n the cnductr The surface charge density : ρ Frm Gauss s law : D N DN ρ s Hayt; //009; 6-8 The electric fields : E E D / ε N D / ε N V The ptential differences : V The capacitance is C V + V d d + ε ε E d E d A parallel plate capacitr with tw vertical dielectrics Fr the ptential difference V E E V d / The ttal charge ε E + ε E The capacitance is ε + ε C C + C V d 6.5 Capacitance f a Tw-Wire ine The ptential f a single line charge with zer reference at R ρ R V ln πε R The ptential difference between tw wires ρ ρ ρ ( + ) + R R R x a y V ln ln ln ln πε R R πε R 4πε ( x a) + y (35) Assume R0 R0, zer reference plane is x0 plane et K e πε ρ 4 V / (36) Then K ( ) ( ) x + a + y x a + y

9 Hayt; //009; 6-9 Cllect equal pwers K + x ax + y + a 0 K K + a K x a + y K K h b (37) The equiptential surface f V is a cylinder centered at xh, y0 with radius b, where x0 plane is the zer reference pint. When h, b and V are given, then frm (37) a h b h + h b K Find ρ frm (36) b (The equivalent tw wire-line with a and ρ ) Next, fr a new equiptential surface V V. The new K is calculated frm (36). Insert a and K int (37). Find the capacitance Fr the given ptential difference V K e πε ρ 4 V / 4πεV ρ ln K The capacitance between x0 plane and the cylinder centered at xh, y0 with radius b and length. ρ 4πε πε πε πε C V ln ln K K ln / csh ( h / b h + h b b ) Fr b<<h ln h + h b / b ln h / b and πε C ln h / b [ ] ( ) The slid cylinder is at the ptential 00V while the dtted ne at 50V. The electric field frm the minus gradient f Eq. (35) ρ ( x + a) aˆ ˆ ( ) ˆ ˆ x + yay x a ax + ya y E 4πε ( x + a) + y ( x a) + y and D εe The maximum D r the max surface charge density ρ is at xh-b, y0. The minimum D r the min surface charge density ρ is at xh+b, y0.

10 6.6 Using Field ketches t Estimate Capacitance in Tw-Dimensinal Prblems Hayt; //009; Current Analgies There exists an analgy between J and D The cnductin current in a lssy dielectric J σ Ec E V c c Fields in the lssy dielectric D εed E V d d The ttal current leaving the psitive cnductr I J d σ E d i i The ptential difference V E i d c The resistance V E c c id R I σ E i d c The capacitance ε E d i d C V E id d ince V V and E E RC ε σ c d c d d c c

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