Electrostatics (RECAP) Electric Current

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What is the constant polaization?
Outwad flux of the electic field is the net?
How is the outwad flux of the electic field divided by ε 0?

1 Electostatics (ECAP) Electic Cuent

2 Key ideas Electic chage: conseved and quantized Electic field: foce pe unit chage, field lines, adding vectos Flux: amount of field passing though an aea Electic potential: enegy pe unit chage, integal of field new unit: electon volt (ev) Dipole moment: paied + and - chages Capacitance: device to stoe chage and enegy Dielectics: polaization, dielectic constant

3 Coulomb s Law: Gauss s Law: Fundamental Laws kq elations between potential and field: F E V kqq / kq The outwad flux of the electic field though any closed suface equals the net enclosed chage divided by ε 0. The potential diffeence between A and B is the wok equied to cay a unit positive chage fom A to B. V E x dx E x / / dv dx etc.

4 Teminology Wods whose pecise definitions you must know: Field Flux Potential Potential diffeence Dipole moment Capacitance Dielectic constant And of couse the SI units fo all these things.

5 Q. 5-3 C πε 0 log( b / a) In the text, this fomula is deived fo the capacitance pe unit length of a long cylindical capacito, such as a coaxial cable. In this deivation, the potential diffeence was calculated by means of an integal ove the electic field. What was that integal? (1) d d d () d (3) (4)

6 Q.5-3 What was the integal needed fo the capacitance of a coaxial cable? 1) d ) d 3) (1/) d 4) (1/ ) d

7 Long Line of Chage

8 Q. 5-3 Fig. 5-6: Gauss s Law: Potential diffeence: (1) E V k Ed d d d () d (3) (4) a b λ / k λ a b d

9 eview: The electon-volt One ev is the enegy to move an electon though a potential diffeence of one volt. Note this is a unit of enegy, not potential. This is not an SI unit but is used in all pocesses involving electons, atoms, etc. U qv So if V1 and qe then: U 1 ev e V ( C ) (1V ) J

10 Cathode ay Tube Electon gun: potential V gives electon enegy in ev. 1 K mv qv So if V 500 volts, electon enegy is K 500 ev

11 High enegies Fo X-ay machines, acceleate electons with potentials of thousands of volts, so we speak of kinetic enegies in kev. In nuclea physics, acceleatos poduce beams of paticles with enegies in MeV. In elementay-paticle physics, high-enegy paticles beams have enegies measued in giga-volts: GeV ev ev ev

12 Q. 5-4 Which diection will the electic field lines point in this electon gun? (1) To the ight (3) Upwad () To the left (4) Downwad

13 Q.5-4 Which way does E point in the acceleating egion? 1) To the ight. ) To the left. 3) Upwad. 4) Downwad.

14 Q. 5-4 Which diection will the electic field lines point in this electon gun? F qe ee So fo foce to the ight, we need field to the left! (1) To the ight (3) Upwad () To the left (4) Downad

15 Text poblem 3-49 Nonconducting sphee of adius a with chage q unifomly distibuted Concentic metal shell of inne adius b, oute adius c, with total chage q. (a) Find field eveywhee. (b) How is chage distibuted on shell?

16 Text poblem 49(a) Fo <a, have pevious solution: ρ Fo a<<b, shell theoem gives: E E Fo b<<c, we e inside metal, so E 0. 3ε 0 kq Fo c<, shell theoem gives E 0. In all cases, field is adially outwad.

17 Text poblem 49(b) Daw gaussian sphee of adius with b<<c. Because we e inside a metal, E 0. Theefoe flux 0. Theefoe enclosed chage 0. Theefoe thee is q on inne suface of shell. Theefoe thee is no chage on oute suface.

18 Electic Dipole The combination of two chages of equal but opposite sign is called a dipole. If the chages +q and q ae sepaated by a distance d, then the dipole moment p is defined as a vecto pointing fom q to +q of magnitude p qd. q p + q

19 Electic Field Due to a Dipole

20 Potential due to a dipole Exact potential at P: V kq / + kq / + + ( d / ) d cosθ + ( d / ) + d cosθ Appox. potential at P, >>d: V θ kpcos /

21 esult: the dipole potential So we have found that fo lage, the potential poduced by a dipole is: V ( ) kp cos / θ p θ Note this can also be witten: V ( ) k p 3

22 Toque on a Dipole in a Field τ d F ( sinθ ) qe d sinθ pe τ p E sinθ

23 Enegy of dipole in given field E θ p U Q( E cos )( d / ) pe cos p E θ θ So dipole tends to align with an applied field.

24 Inteaction of two dipoles p p 1 U qv ( + d / ) qv ( d / ) ( kqp + d / ) ( kqp d / ) k p 1 3 p Attactive potential: wok equied to pull apat.

25 TODAY: Cuent and esistance Chaptes 6, 7 Ohm s Law Ideal Metes Souces of Voltage and Powe esistos in seies and paallel

26 Cuent is ate of flow of chage If you watch closely at a fixed point as cuent is flowing along a wie, the cuent is the amount of chage that passes by pe unit time. SI unit is the Ampee: 1A 1C/s. Example: In a cathode-ay tube, n 3x10 15 electons leave the electon gun pe second. What is the cuent of this electon beam? Solution: ne C / s 0. 5 ma

27 Cuent and esistance When a cuent flows though a pefect conducto (e.g. coppe wie) thee is no change in potential. When cuent flows though an impefect conducto (esisto) thee must be a field E to make it flow and thus a change in potential V Ed fo a distance d. If the cuent I is popotional to the field E which is causing it, then it will be popotional to the potential change V. This is Ohm s Law and the popotionality constant is called the esistance. V I

28 Notes on Ohm s Law This is not a fundamental law of natue like Gauss s Law. It s just a popotionality which is appoximately tue fo some mateials unde some conditions. The entie electonics industy is based on mateials which violate Ohm s Law. Ohm s Law should eally be witten as a potential dop V I because if you follow the cuent the potential deceases.

29 Micoscopic Fom E V / L I / L JA/ L Jρ So we define esistivity ρ: ρ L / A and cuent density j: J I / A So the micoscopic fom of Ohm s Law is E Jρ

30 esistance So fo a given object (esisto) we can measue its esistance V/I. The SI unit of esistance is the ohm (Ω). Clealy 1Ω 1V/A (ohm volt/amp). Thus esistivity ρ has units Ω-m. But emembe Ohm s Law is only an appoximation. Fo example, esistance nomally changes with tempeatue.

31 Example: Poblem 6-15 A wie is m long, with a diamete of 1 mm. If its esistance is 50 mω, what is the esistivity of the mateial? ρ ρl / A A/ l 10 8 ( 3 ) ( Ω m) π Ωm (Note this is consistent with table on page 689.)

32 Voltage and Powe Souces When a battey o any othe voltage souce delives a cuent i at a potential diffeence V it is supplying powe P iv. U qv P t t iv When a cuent i flows though a esisto with a voltage dop V, a fiction-like pocess called Joule heating convets this powe PiV into heat.

33 Note on P I V This is the easiest equation in the wold to emembe, IF you know what the thee quantities mean: enegy chage enegy voltage cuent chage time time powe

34 Ideal Metes V Ideal ammete Measues cuent and has zeo potential dop (zeo ) a A b Ideal voltmete Measues V ab V a -V b but daws zeo cuent (infinite ) Ideal wies V is constant along any wie (zeo )

35 Example Mete A eads i A flowing upwad as shown, and mete V eads V 6V. a A V b (a) What is esistance? V / i 6V / A 3Ω (b) Which point is at the highe potential, A o B? Point A because potential dops by 6 volts.

36 EMF Electomotive foce (emf) is not a foce: it s the potential diffeence povided by a powe supply. Fo a battey poviding a cuent i, the teminal voltage V is less than the emf E because thee is an intenal esistance : i E i V E i V

37 Powe in a Simple Cicuit i i + ε i 0 V loop + ε i( ) E ε Dy cell Light bulb P chem P i ε heating battey i + + i P heating filament

38 Example Fo a battey with intenal esistance 5 Ω, what load esisto will get maximum powe? ( ) + + i P load ε ε ( ) 0 ) ( ) ( ) ( d d Ω + 5 ) ( ) ( + +

39 esistos in Seies Voltage dops add Cuents ae equal. V V ε i + + b a i i 1 3 ε But we want i eq + + eq 1 3

40 Cuents add esistos in Paallel Voltage dops ae equal. i i + i 1 ε V eq But ε i 3 V + V 3 3 V V V 1 3 1/ 1/ + + eq 1/ 1/ 1 3

41 Example: Poblem 7-30 ε 6.0V Ω 75Ω 100Ω (a) Find the equivalent esistance of the netwok. (b) Find the cuent in each esisto.

42 Poblem 7-30 (pat a) ε 6.0V Ω 75Ω 100Ω (a) Find the equivalent esistance of the netwok. 1/ So 34 1/ + 1/ + 1/ 1/ 50+ 1/ 50+ 1/ / /16 19Ω 3 4

43 Poblem 7-30 (pat a cont d) ε 6.0V Ω 75Ω 100Ω (a) Find the equivalent esistance of the netwok. Now eq Ω+ 19Ω 119Ω 1 and ae in seies so

44 Poblem 7-30 (pat b) ε 6.0V (b) Find the cuent in each esisto Ω 75Ω Fist get the total cuent fom the battey, which is also the cuent though 1 : 100Ω ε i / eq 6.0 / A 50 1 ma

45 Poblem 7-30 (pat b cont d) V V i A 19Ω 0.95V (b) Find the cuent in each esisto. i V /.95 / ma i i 3 4 V / V / / / ma 1mA

46 Poblem 7-30 (check) i 1mA Check by adding the cuents in the thee banches: i i 4 i 1 i 3 50 ma 19 ma i + i + i ma i 1

The Electric Potential, Electric Potential Energy and Energy Conservation. V = U/q 0. V = U/q 0 = -W/q 0 1V [Volt] =1 Nm/C

The Electric Potential, Electric Potential Energy and Energy Conservation. V = U/q 0. V = U/q 0 = -W/q 0 1V [Volt] =1 Nm/C Geneal Physics - PH Winte 6 Bjoen Seipel The Electic Potential, Electic Potential Enegy and Enegy Consevation Electic Potential Enegy U is the enegy of a chaged object in an extenal electic field (Unit

Electrostatics (RECAP) Electric Current

What is the constant polaization?

Outwad flux of the electic field is the net?

How is the outwad flux of the electic field divided by ε 0?