Work, Potential Energy, Conservation of Energy. the electric forces are conservative: ur r

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2 Wok, Potentil Enegy, Consevtion of Enegy the electic foces e consevtive: u Fd =

3 Wok, Potentil Enegy, Consevtion of Enegy b b W = u b b Fdl = F()[ d + $ $ dl ] = F() d u Fdl = the electic foces e consevtive

4 Wok, Potentil Enegy, Consevtion of Enegy u W = K K = F dl b b W b = U Ub U U = K K U + K = U + K b b b b b q q U qq qq = U = 4 b πεb nottions : u F q q foce with which chge q cts on q ; W b wok pefomed when chge is moved fom point to point b

5 Wok, Potentil Enegy, infinity s efeence point U qq qq = U = 4 b πεb u U U F dl b = b q q U potentil enegy hs to be mesued with espect to W = b U Ub qq qq = U = 4 b πεb u = = = U U Fq qdl u = = = Ub U b Fq qdl b qq qq b Hee, potentil enegy is mesued with espect to n infinitely emote point U = wok which ws done by chge q when n electic chge q o is deliveed to infinity = wok which ws done (by ) to bing n electic chge q fom infinity

6 Wok, Potentil Enegy, Consevtion of Enegy U = qq U = wok which ws done to bing n electic chge q o fom infinity = the kinetic enegy which this chge will cquie t infinity if it will be elesed u F u el. field = Fq q U U Fextenldl u = = = u u Fel. fielddl = Fel. fielddl = qq Hee, potentil enegy is mesued with espect to n infinitely emote point

7 Hee (!!!), potentil enegy is mesued with espect to n infinitely emote point U qq qq = U = 4 b πεb

8 Potentil Enegy U qq qq = U = 4 b πεb Positon q=e, lph pticle Q=2e

9 Electic Potentil Enegy with Sevel Chges U= wok which ws done to bing n Electic Chge q o fom infinity u N u N u N U = Edl = dl E = E dl = U j j i i j i long long i= 1 i= 1 long i= 1 contou contou contou

10 Electic Potentil Enegy with Sevel Chges U= wok which ws done to bing n electic chge q o fom infinity = the kinetic enegy which this chge will cquie t infinity if it will be elesed

11 potentil enegy U(x,y,z) nd foce F: gdient of the potentil enegy is equl to minus foce U() U() U() Fx = Fy = Fz = x y z u u F = U() 2 u U U Fdl 2 1 = 1 u = i / x + j / y+ k / z The opetion is clled tking gdient ; esult gdient. u F U() U() U() = i + j + k x y z diffeentition of this integl with woks like diffeentition of n odiny integl: u u u Fdl = u F( ) u u u Fdl = F( )

12 u F u E Electic Potentil U() V () = V() U() q u F u E u = U() u = V() Electic field is equl to electic foce pe unit chge Electic potentil is equl to potentil enegy pe unit chge V q () = q CAUTION: potentil enegy hs to be mesued with espect to hee, electic potentil is mesued with espect to n infinitely emote point, while is the distnce fom the loction of chge q

13 Electic Potentil u F u E U ( ) V ( ) V ( ) = U ( ) q u F u E u = U ( ) u = V ( ) Electic potentil is equl to potentil enegy pe unit chge of pobe chge U() qv() q = = q V () = q

14 Units: 1V=1 volt= 1 J/C= 1 joule/coulomb 1V=1 (N/C)m electic field is, thus, mesued in volts pe mete: V/m V ( ) = 1 q ε = C/(Vm) E = Const V ( x) = E 1 9 = 9 1 Vm C x

15 Similly to U(x,y,z), one cn intoduce the electic potentil V(x,y,z) such tht: u V V Edl V() V() V() Ex = Ey = Ez = x y z u u u u u Edl Edl E( ) u u E = V() In electosttics, the electic fields e consevtive: Ed = This implies tht the diffeence between the electic potentils of two points does not depend on the tjectoy connecting these points: 2 1 Ed = [ V ( ) V ( )] 2 1 = = = = u u E = V() if V(x,y,z) is known one cn find the components of the electic field nd vice ves

16 Elements of mth u u E = V() x V() V() V() Ex = Ey = Ez = x y z u V() V() V() E = i+ j+ k x y z ptil deivtive (conside y nd z s constnts) u dv () useful fomul: V ( ) = $ d u dv ()( l ) V( ) = dl is unit vecto long ( )

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18 Moving though potentil diffeence: chnge of the potentil enegy of chge q is equl to the poduct of the electic potentil chnge nd the chge 1eV = (14) 1 17 J

19 Moving though potentil diffeence 1V =1J/C

20 Thee is no electic field inside conducto Net chge cn only eside on the sufce of conducto Any extenl electic field lines e pependicul to the sufce (thee is no component of electic field tht is tngent to the sufce). The electic potentil within conducto is constnt (vlid only in the bsence of cuents)

21 Definition: Owing to the fct tht E = inside the conducto, the diffeence Fo ny two points 1 nd 2 inside the conducto 2 V V Edl 2 1 = the electic potentil diffeence is equl to n integl long line connecting the two points (ny line!) V V = V 1 V = In equilibium (i.e., without cuents): 1) the conducto s sufce is equipotentil. 2) the whole body of the conducto is equipotentil.

22 Equipotentil Sufces An equipotentil sufce is sufce on which the electic potentil V is the sme t evey point. Convesely, the electic field cn do no wok on chge moving long n equipotentil sufce. Electic field must be pependicul to the sufce t evey point so tht the electic foce is lwys pependicul to the displcement of chge moving on the sufce. Field lines nd equipotentil sufces e lwys mutully pependicul.

23 Gdient of the electic potentil is equl to minus electic field. It mens tht long the electic field line the electic potentil goes DOWN, down, down! u u u E() = V() Edl = E( ) 1 Edl= [ V ( ) V ( )] = lim [ U ( ) U ( )] q q So, wht is the use in one moe quntity? Becuse to dw mp of scl quntity epesenting the shpe (elief) of the potentil enegy is much esie thn to dw mp of the vecto field. Look on topogphic mp of mountins! It is mp of the gvittionl potentil: Question: wht is n nlogue of lke?

24 V ( ) = 1 q Exmples E = Const V ( x) = E x + const

25 2 1 Ed = 1 lim [ U( 2) U( 1)] = [ V ( 2) V ( 1)] q q Fo point chge t the oigin (we ledy know the nswe): [ V( ) V( )] Ed q d q = = = V ( ) = dv () V() = $ d u dv () useful fomul: V ( ) = $ d u dv ()( l ) V( ) = dl ( ) is unit vecto pointing fom towd

26 Exmple Nonconducting with homogenously distibuted chge Q 1 VV( ) Q R?

27 R Nonconducting sphee of dius A with homogenously distibuted chge Q E > R E = 1 Q 2 < R E = Q R 3 VV( ) 1 Q R? R V() =? beceful! with espect to wht point? V() V( ) = E( y) dy

28 R Nonconducting sphee of dius A with homogenously distibuted chge Q E > < R R E( ) = E () = 1 Q 2 Q R 3 VV( ) 1 Q R? R V() V( ) = E( y) dy < R V() V( ) = + > R 1 Q V() V( ) = 2 1 Q 3 2 R 2R 2

29 R Nonconducting sphee of dius A with homogenously distibuted chge Q E > < R R E E = = 1 Q 2 Q R 3 R V()V() is like V()V( ), but is shifted down on constnt 3 Q equl to 8πε R V() is pbolic t smll V () V() = E( ydy ) = E( ydy ) 2 < R V() V() = 8πε 1 Q R 3 > R V() V() = R 2 Q R 3

30 Potentil of chged plte E = Const V ( x) = E x + const V ( ) =??? V ( ) =??? +σ Fo point chge t the oigin: [ V( ) V( )] = Ed = q d q = = x

31 Potentil of chged plte +σ σ + S S E = Const V ( x) = E x + const x Cution: it is convenient to plce the initil point, i.e., b, to infinity nd set V(b)=. Neve do it when you del with infinitely lge/extended objects!

32 Two pllel conducting pltes σ S σ V ( ed ) = E x + σ + σ + V ( blue) = E x Field between the pltes is constnt, potentil is line E = σ /ε x

33 Two pllel conducting pltes σ S σ σ +σ σ +σ + Field between the pltes is constnt, potentil is line x E = σ /ε Potentil t infinity is not zeo!

34 V ( ) =??? [ V( ) V( )] = Ed = λ dy λ = = 2πε y 2πε ln y??? V( ) = λ dy λ V( )] = Ed = = ln y 2πε y 2πε Agin the poblem with infinity Resolution: only potentil diffeence mttes, mesue potentil diffeence with espect to n bity point = λ 2πε ln

35 Method of imges: Wht is foce on the point chge ne conducting plte? Equipotentil sufce The tick with imging cn be done not only with flt sufce

36 Additionl mteil 1

37 Physics of Lightings Benjmin Fnklin

38 ioniztion nd coon dischge Thee is mximum potentil to which conducto in i cn be ised becuse of ioniztion. 6 E m = 3 1 V m V m = R E m

39 A lightning od hs shp end so tht lightning bolts will pss though conducting pth in the i tht leds to the od; conducting wie leds fom the lightning od to the gound. The metl mst t the top of the Empie Stte Building cts s lightning od. It is stuck by lightning s mny s 5 times ech ye. Even eltively smll potentils pplied to shp points in i poduce sufficiently high fields just outside the point to ionize the suounding i. Cution: wht cn be misleding in the bove quottion?

40 Even eltively smll potentils pplied to shp points in i poduce sufficiently high fields just outside the point to ionize the suounding i. Thee is mximum potentil to which conducto in i cn be ised becuse of ioniztion. 6 E m = 3 1 V m ioniztion nd coon dischge A shp edge leds to the defomtion of equipotentil lines nd, hence, to high electic field ne the edge The tlle the edge the stonge defomtion of the lines nd, theefoe, the moe pobble dischge by lighting

41 Method of imges: Wht is foce on the point chge ne conducting plte? Equipotentil sufce The tick with imging cn be done not only with flt sufce

42 The foce cting on the positive chge is exctly the sme s it would be with the negtive imge chge insted of the plte. The point chge feels foce towds the plte with mgnitude: F = 1 q 2 2 (2) Cution: 2 the thn!

43 Additionl mteil 2

44 2 1 1 Edl= [ V ( ) V ( )] = lim [ U ( ) U ( )] q q electic potentil V=U/q is simil to U/m in the cse of the Eth gvittion 2 1 u u 1 Fdl= [ U ( 2) U ( 1)] g = U() m u

45 b Edl = [ V( ) V( )] = V( ) + V( b) u b u b dl it is convenient to plce the initil point, i.e., b, to infinity nd set V(b)=. Wht potentil V() cetes positive chge Q? It is positive! Note tht when positive chge is ppoched, is diected ginst, nd theefoe Edl is genelly negtive. Hence V() is positive. E dl V( ) = ( E) dl = [ V( ) V( )] >

46 qv() is equl to wok W needed to pefom in ode to delive the chge q fom infinity to the point. The wok W is done ginst the electic field. u qv ( ) = ( qe) dl = q[ V ( ) V ( )] dl Fo pointlike chge: If qq> this wok is positive; W>. Coespondingly V()=W/q is positive if Q>.

47 u dl qv ( ) = ( qe) dl = q[ V ( ) V ( )] = qedl Now qv() cn be intepeted s wok W which pefoms the electic field of the chge Q when the chge q is moving wy. In this fomultion, the wok W is done by the electic field. Fo pointlike chge. We e moving wy fom positive chge: now nd E hve common diection towd infinity. Hence Edl >, If qq>, this wok is positive; W>. Coespondingly V()=W/q is positive when Q>. dl u dl

48 Additionl mteil 3

49 Clcultion of the electic field ne the conducting sufce is possible, but not the shotest wy, to find foce cting on the metllic plne y u F =? E E + 1 = 1 = q ( + y ) q ( + y ) E = 1 q 2 πε ( + y )

50 Clcultion of the electic field ne the conducting sufce is possible, but not the shotest wy, to find foce cting on the metllic plne E 1 q = 2 πε ( + y ) E π 1 q df = ()! σ2πydy = ydy ε 2 π ( y ) q 1 q 1 q F = ydy = = 4 πε ( + y ) 4 4 πε (2 ) y σ = ε E u F =?

51 Additionl mteil 4

52 In ecoded lectue fom 1961, Richd Feynmn explined to his students why physicists use electon volts to mesue enegy insted of some multiple of the joule: A single tom is such smll thing tht to tlk bout its enegy in joules would be inconvenient. But insted of tking definite unit in the sme system, like 1 2 J, [physicists] hve unfotuntely chosen, bitily, funny unit clled n electonvolt (ev)... I m soy tht we do tht, but tht's the wy it is fo the physicists.

53 Electonvolt The electonvolt (ev) is unit of enegy. By definition, it is equl to the mount of enegy gined by single unbound electon when it cceletes though n electosttic potentil diffeence of one volt. 1 ev = (14) 1 19 J. So n electon volt is 1 volt (1 joule divided by 1 coulomb) multiplied by the electon chge ( (14) 1 19 coulomb). The electonvolt is now ccepted within SI.

54 1eV = (14) 1 19 J 1eV = (14) 1 17 J

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