Physics 169. Luis anchordoqui. Kitt Peak National Observatory. Thursday, February 22, 18

Transcription

1 Physics 169 Kitt Peak National Observatory Luis anchordoqui 1

2 4.1 Capacitors A capacitor is a system of two conductors that carries equal and opposite charges A capacitor stores charge and energy in the form of electro-static field We define capacitance as C = Q V Unit Farad(F) Q = Charge on one plate Note capacitor s C V = Potential difference between plates is a constant that depends only on its shape and material V i.e. If we increase for a capacitor we increase stored Q 2

3 4.2 Calculating Capacitance Parallel-Plate Capacitor Recall ~ E = 0 = Q 0 A 3

4 Recall V = V + V = Z + ~E d~s this integral is independent of path taken path is parallel to ~E -field V = = = Q 0 A Z + ~E d~s = Q 0 A d Z ds + {z } Length of path taken ) C = Q V = 0A d Z + ~E d~s 4

5 4.2.2 Cylindrical Capacitor Consider two concentric cylindrical wires V = Z + ~E d~s of inner and outer radii Length of capacitor is L r 1 and r 2 where r 1 <r 2 L Using Gauss law we determine that ~E -field between conductors is Again we choose path of integration so that ) V = Z r 2 ) C = Q V =2 0 r 1 Edr = Q 2 0 L L ln(r 2 /r 1 ) charge per unit length ~E = r ˆr = Q Lr ˆr r 2 d~s k ˆr k ~ E Z r2 dr r 1 r {z } ln( r 2 r 1 ) 5

6 4.2.3 Spherical Capacitor ~E k ˆr Choose d~s k ˆr For space between two conductors E = Q r 2 V = = Z + Z r2 ~E d~s r Q r 2 dr = Q 4 0 h 1 r 1 1 r 2 i h r1 r i 2 C =4 0 r 2 r 1 r 1 <r<r 2 6

7 4.3 Capacitors in Parallel and Series (a) Capacitors in Parallel potential difference V = V a V b is same across capacitors BUT Charge on each capacitor different Total Charge Q = Q 1 + Q 2 = C 1 V + C 2 V ) For capacitors in parallel Q =(C 1 + C 2 ) {z } V Equivalent capacitance C = C 1 + C 2 7

8 (b) Capacitors in Series charge across capacitors are same BUT Potential difference (P.D.) across capacitors different V 1 = V a V c = Q C 1 P.D. across C 1 V 2 = V c V b = Q C 2 ) Potential difference P.D. across C 2 V = V a V b = V 1 + V 2 1 V = Q + 1 C 1 C 2 = Q C C Equivalent Capacitance ) 1 C = 1 C C 2 8

9 4.4 Energy Storage in Capacitor While charging a capacitor positive charge is being moved from negative plate to positive plate ) NEEDS WORK DONE! dq Suppose we move charge from to plate q +q change in potential energy du = V dq = q C dq Suppose we keep putting in a total charge total potential energy U = Z du = Q Z Q 0 to capacitor q C dq ) U = Q2 2C = 1 2 C V 2 (* Q = C V ) Energy stored in capacitor is stored in electric field between plates 9

10 Note: In parallel-plate capacitor density u = ~E total energy stored -field is constant between plates total volume with E ~ field Recall ) u = U {z} Ad ( 0 A C = d Rectangular volume E = V d ) u = 1 2 ( ) V = Ed C z} { ( V ) 2 0 A d ) ( z} { Ed ) 2 1 Volume z} { 1 Ad Energy per unit volume of electrostatic field u = 1 2 0E 2 10

11 Example Changing capacitance by pulling plates apart Isolated Capacitor Charge on capacitor plates remains constant BUT ) U new = Q2 = 2C new ) Summary In pulling plates apart work done (V = Q C ) ) 1 2 0E 2 = C new = 0A 2d Q! Q V! 2V = 1 2 C old Q 2 2C old /2 =2U old W>0 C! C/2 E! E u! u U! 2U (E = V d ) (U = u volume) 11

12 Electric battery device consisting of 2 or more electrochemical cells that convert stored chemical energy into electrical energy Each cell has: positive terminal (or cathode) negative terminal (or anode) 12

13 Capacitor connected to a battery Potential difference between capacitor plates remains constant U new = 1 2 C new V 2 = C old V 2 = 1 2 U old ) In pulling plates apart work done by battery < 0 Summary Q! Q/2 V! V C! C/2 E! E/2 u! u/4 U! U/2 13

14 4.5 Dielectrics Consider conductor being placed in an external -field E 0 In a conductor charges are free to move inside internal -field set up by these charges satisfies E 0 E 0 = E 0 so that E-field inside conductor =0 For dielectric atoms and molecules behave like dipole in ~E -field 14

15 we can envision this so that in absence of -field direction of dipole in dielectric are randomly distributed ~E Aligned dipoles will generate an induced E 0 -field satisfying E 0 < E 0 We can observe aligned dipoles in form of induced surface charge 15

16 Dielectric Constant When a dielectric is placed in an external -field ~E -field inside a dielectric is induced E 0 ~E = 1 apple ~ E 0 Example apple 1 Vacuum Porcelain Water Perfect conductor Air dielectric constant apple =1 apple =6.5 apple 80 apple!1 apple =

17 Polarization Vector Consider polarized volume with density of ~p 0 s p~p i V Polarization vector ~P defined as ~P = lim V!0 P N i=1 ~p i V [C/m 2 ] N number of molecules in V Macroscopic effects of polarized dielectric material are modeled by which really is average dipole moment per unit volume of material ~P 17

18 Electric susceptibility and permittivity It is customary in electromagnetism to bury effects of bound polarization in materials through electric flux displacement Polarization effects of a dielectric can be accounted for by defining ~D as ~D = 0 ~ E + ~ P [C/m 2 ] ❶ What we desire now is to know ~P in terms of ~E Basically without knowing ~P this theory is not very useful It has been found through experimentation that for many materials with small ~E ~P = 0 e ~ E ❷ e electric susceptibility of material (dimensionless) 18

19 Substituting ❷ into ❶ gives ~D = 0 ~ E + 0 e ~ E = 0 (1 + e ) ~ E We can rewrite this as ~D = ~ E [C/m 2 ] Constant called permittivity of material = apple 0 =(1+ e ) 0 [F/m] with apple called apple =1+ e relative permittivity of material (dimensionless) or dielectric constant 19

20 4.6 Capacitor with Dielectric Case I Charge remains constant after dielectric is inserted apple BUT E new = 1 apple E old ) V = Ed ) V new = 1 apple V old ) C = Q V ) C new = applec old For a parallel-plate capacitor with dielectric C = apple 0A d 20

21 We can also write C = A d in general with = apple 0 0 Recall permittivity of dielectric permittivity of vacuum Energy stored U = Q2 2C ) U new = 1 apple U old <U old ) Work done in inserting dielectric < 0 21

22 Case II Capacitor connected to battery apple Voltage across capacitor plates remains constant after insertion of dielectric ~E -field inside capacitor remains constant (* E = V/d) BUT How can E-field remain constant? ANSWER By having extra charge on capacitor plates 22

23 Recall For conductors But E = 0 ) E = Q 0 A After insertion of dielectric ~E -field remains constant Capacitor E 0 = Q0 apple 0 A ( charge per unit area = Q/A) E 0 = E ) Q0 apple 0 A = Q 0 A ) Q 0 = appleq > Q C = Q/V ) C 0! applec Energy stored U = 1 2 CV 2 ) U 0! appleu U new >U old ) Work done to insert dielectric > 0 23

24 4.7 Gauss Law in Dielectric Gauss law we ve learned is applicable in vacuum only Let s use capacitor as an example to examine Gauss law in dielectric apple Free charge on plates Induced charge on dielectric Gauss Law I S ±Q 0 ~E d ~ A = Q 0 I ) E 0 = Q 0 A but E 0 = E 0 apple S ±Q Q 0 Gauss Law ~E 0 d ~ A = Q Q0 0 ) E 0 = Q Q0 0 A 24

25 From ) Q 0 apple 0 A = Induced charge density Q 0 A Q 0 0 A 0 = Q0 A = 1 1 apple < where is free charge density 0 I S Recall Gauss law in Dielectric ~E 0 d ~ A = Q Q 0 E-field in dielectric free charge induced charge polarization charge I I ( 0 E ~ + P ~ ) d S ~ = S S ~D d ~ S = Q 25

26 Note ) 0 I ) 0 I I apple 1 This goes back to Gauss law in vacuum with 2 Only free charges need to be considered 3 Another way to write I S S S ~E 0 d ~ A = Q Q ~E 0 d ~ A = Q apple ~E 0 d ~ A = Q 0 apple 1 even for dielectric where 1 apple Gauss law in dielectric E = E 0 apple for dielectric there are induced charges S ~E d ~ A = Q 26

27 Energy stored with dielectric Total energy stored U = 1 2 CV 2 With dielectric recall C = apple 0A d V = Ed ) Energy stored per unit volume u = U Ad = 1 2 apple 0E 2 and so u dielectric = appleu vacuum ) More energy is stored per unit volume in dielectric than in vacuum 27

28 4.7 Electrostatic Boundary Conditions Boundary between two media A narrow rectangular contour is used in law of conservation of energy and a coinlike closed surface is used in Gauss law for deriving boundary conditions for vectors ~E and D ~ respectively Boundary condition for tangential components of vector E ~ E 1t = E 2t Boundary condition for normal components of vector D ~ (unit vector normal directed into medium 1) ~D 1 ˆn D2 ~ ˆn = ) D 1n D 2n = 28

29 29

30 30