Electric Potential Equipotentials and Energy

Transcription

1 Electic Potentil Equipotentils nd Enegy Tody: Mini-quiz hints fo HWK U = qv

2 Should lightening ods hve smll o lge dius of cuvtue? V is popotionl to /R. If you wnt high voltge to pss though the od then use smll dius of cuvtue. Ai is nomlly n insulto, howeve fo lge E fields (E>3 x 6 V/m) it stts conducting. Empie Stte Building, NYC

3 Review: Electicl Potentil W = wok done y foce in going fom to long pth. W U V = = U = V F dl = U V = = W U q qe dl = qe dl U = potentil enegy = U U q W = q F = E dl θ dl Potentil diffeence is minus the wok done pe unit chge y the electic field s the chge moves fom to. Only chnges in V e impotnt; cn choose the zeo t ny point. Let V = t = infinity nd V V, then: V = E dl V = electic potentil llows us to clculte V eveywhee if we know E

4 Potentil fom chged spheicl shell E-field (fom Guss' Lw) < R: E = V q R q >R: E = q 2 R R Potentil R > R: V ( ) = = E dl = E ( d = ) = q Q < R: = R V( ) = E dl = E (d ) = E (d ) Qq E (d ) = = R R

5 ELECTRIC POTENTIAL fo Chged Sphee (Y&F, ex.23.8) Suppose we hve chged metl sphee with chge q. q Wht is the electic potentil s function of dius? N.B. V is continuous ut E is not.

6 Clicke execise A point chge Q is fixed t the cente of n unchged conducting spheicl shell of inne dius nd oute dius. Wht is the vlue of the potentil V t the inne sufce of the spheicl shell? Q () V = () V = 4 πε Q (c) V = Q

7 Clicke execise A point chge Q is fixed t the cente of n unchged conducting spheicl shell of inne dius nd oute dius. Wht is the vlue of the potentil V t the inne sufce of the spheicl shell? Q E out () V = () V = Q (c) V = Q V How to stt?? The only thing we know out the potentil is its definition: V V V = E dl To clculte V, we need to know the electic field E Outside the spheicl shell: Q ˆ Apply Guss Lw to sphee: E = 2 Inside the spheicl shell: E = = Q Q E d l E d l V = V =

8 Peflight 6: Two spheicl conductos e septed y lge distnce. They ech cy the sme positive chge Q. Conducto A hs lge dius thn conducto B. A B 2) Compe the potentil t the sufce of conducto A with the potentil t the sufce of conducto B. ) V A > V B ) V A = V B c) V A < V B

9 Electicl Potentil Two wys to find V t ny point in spce: Use electic field: = V E dl Sum o Integte ove chges: q q P V = i q i i dq q 3 P V = dq Exmples of integting ove distiution of chge: line of chge (eview this one) ing of chge disk of chge You should e le to do these.

10 Infinite line chge o conducting cylinde. Line chge density λ E = 2 λ πε λ λ = = = 2πε 2πε V V Ed ln( ) Suppose we set to infinity, potentil is infinite Insted, set = nd = t some fixed dius.

11 Cove homewok hints

12 Potentil fom chged sphee V( ) = q V( ) (whee ) E Equipotentil The electic field of the chged sphee hs spheicl symmety. The potentil depends only on the distnce fom the cente of the sphee, s is expected fom spheicl symmety. Theefoe, the potentil is constnt on sphee which is concentic with the chged sphee. These sufces e clled equipotentils. Notice tht the electic field is pependicul to the equipotentil sufce t ll points.

13 Equipotentils Defined s: The locus of points with the sme potentil. Exmple: fo point chge, the equipotentils e sphees centeed on the chge. Why?? Along the sufce, thee is NO chnge in V (it s n equipotentil!) Theefoe, V V We cn conclude then, tht B A The electic field is lwys pependicul to n equipotentil sufce! B A E dl = B = E dl A V = E dl is zeo. If the dot poduct of the field vecto nd the displcement vecto is zeo, then these two vectos e pependicul, o the electic field is lwys pependicul to the equipotentil sufce.

14 EXAMPLES of Equipotentil Lines

15 Conductos Clim The sufce of conducto is lwys n equipotentil sufce (in fct, the entie conducto is n equipotentil). Why?? If sufce wee not equipotentil, thee would e n electic field component pllel to the sufce nd the chges would move!!

16 Peflight 6: A B 3) The two conductos e now connected y wie. How do the potentils t the conducto sufces compe now? ) V A > V B ) V A = V B c) V A < V B 4) Wht hppens to the chge on conducto A fte it is connected to conducto B? ) Q A inceses ) Q A deceses c) Q A doesn t chnge

17 Chge on Conductos? How is chge distiuted on the sufce of conducto? KEY: Must poduce E= inside the conducto nd E noml to the sufce. Spheicl exmple (with little off-cente chge): q E= inside conducting shell. chge density induced on inne sufce non-unifom. chge density induced on oute sufce unifom E outside hs spheicl symmety centeed on spheicl conducting shell.

18 Equipotentil Exmple Field lines moe closely spced ne end with most cuvtue highe E-field Field lines to sufce ne the sufce (since sufce is equipotentil). Ne the sufce, equipotentils hve simil shpe s sufce. Equipotentils will look moe cicul (spheicl) t lge.

19 Consevtion of Enegy The Coulom foce is CONSERVATIVE foce (i.e., the wok done y it on pticle which moves ound closed pth etuning to its initil position is ZERO.) Theefoe, pticle moving unde the influence of the Coulom foce is sid to hve n electic potentil enegy defined y: U = qv this q is the test chge in othe exmples... The totl enegy (kinetic electic potentil) is then conseved fo chged pticle moving unde the influence of the Coulom foce.

20 Lectue 6, ACT 3 3A Two test chges e ought septely to the vicinity of positive chge Q. chge q is ought to pt A, distnce fom Q. chge 2q is ought to pt B, distnce 2 fom Q. Compe the potentil enegy of q (U A ) to tht of 2q (U B ): Q Q A q 2 B 2q () U A < U B () U A = U B (c) U A > U B 3B Suppose chge 2q hs mss m nd is elesed fom est fom the ove position ( distnce 2 fom Q). Wht is its velocity v f s it ppoches =? () v f Qq m = () v f 2πε Qq m = (c) v = f

21 Clicke Polem 3A Two test chges e ought septely to the vicinity of positive chge Q. chge q is ought to pt A, distnce fom Q. chge 2q is ought to pt B, distnce 2 fom Q. Compe the potentil enegy of q (U A ) to tht of 2q (U B ): () U A < U B () U A = U B (c) U A > U B Q Q q A 2 2q B The potentil enegy of q is popotionl to Qq/. The potentil enegy of 2q is popotionl to Q(2q)/(2). Theefoe, the potentil enegies U A nd U B e EQUAL!!!

22 Clicke Polem 3B Suppose chge 2q hs mss m nd is elesed fom est fom the ove position ( distnce 2 fom Q). Wht is its velocity v f s it ppoches =? () v f Qq m = () v f 2πε Qq m = (c) v = f Wht we hve hee is little comintion of 7 nd 272. The pinciple t wok hee is CONSERVATION OF ENERGY. Initilly: The chge hs no kinetic enegy since it is t est. The chge does hve potentil enegy (electic) = U B. Finlly: The chge hs no potentil enegy (U /R) The chge does hve kinetic enegy = KE U B =KE Q ( 2q ) = 2 2 mv 2 f 2 v f = 2πε Qq m

23 How to otin vecto E field fom V Detemine V fom E: Detemining E fom V: V dv = V V Fo n infinitesiml step: = E dl = V = E dl = Edl = E cosθ Cses: θ = : dv = E dl (mximum) θ = 9 o : dv = θ = 8 o : dv = -E dl Cn wite: dv = E dl = ( E = ( E x dx E y iˆ E x dy E z y dz) ˆj E z dv dl Exmple: V due to spheicl chge distiution. E kˆ) ( dxiˆ dy θ dl diectionl deivtive dv depends on diection ˆj dzkˆ)

24 Potentil Gdient Tke step in x diection: (dy = dz = ) Similly: And: dv E = ( E dx E dy E dz) x E y x dv = dx V = y y, z const. y E z V = x V = z V V V E = ( iˆ ˆj x y z ˆ ( ˆ = i j kˆ) x y z z kˆ) = E = V x dx gdient opeto Gdient of V points in the diection tht V inceses the fstest with espect to chnge in x, y, nd z. E points in the diection tht V deceses the fstest. E pependicul to equilpotentil lines. ptil deivtive

25 Potentil Gdient Exmple: chge in unifom E field E U = qey V = U/q = Ey whee V is tken s t y =. = V = (iˆ = ( iˆ x E ˆj kˆ) y ˆj = E ˆj kˆ) Ey z Given E o V in some egion of spce, cn find the othe. Cylindicl nd spheicl symmety cses: Fo E dil cse nd is distnce fom point (spheicl) o xis (cylindidl): E V = E o y q Exmple: E of point chge: V q E = = ( ) q = ( )( ) = 2 q 2

26 ˆ) ˆ 3 ˆ ( 2 ˆ) 2 ˆ 6 ˆ (2 )} 3 ( ˆ){ ˆ ˆ ( 2) 2 2 zk yj xi A Azk j Ay Axi z y x A k z j y i x V E = = = =

27 UI7PF7: Clicke Polem This gph shows the electic potentil t vious points long the x-xis. 2) At which point(s) is the electic field zeo? A B C D

28 UI7ACT: Clicke Polem The electic potentil in egion of spce is given y V ( ) 2 3 x = 3x x The x-component of the electic field E x t x = 2 is () E x = () E x > (c) E x <

29 UI7ACT: Clicke Polem The electic potentil in egion of spce is given y V 2 3 ( x) = 3x x The x-component of the electic field E x t x = 2 is () E x = () E x > (c) E x < We know V(x) eveywhee To otin E x eveywhee, use E = V E x V = x E x E x = 6x 3x ( 2 ) = 2 2 = 2

30 The Bottom Line/Tke Home Messge If we know the electic field E eveywhee, B W AB V V VB V A = E dl B A q A llows us to clculte the potentil function V eveywhee (keep in mind, we often define V A = t some convenient plce) If we know the potentil function V eveywhee, E = V llows us to clculte the electic field E eveywhee Units fo Potentil! Joule/Coul = VOLT (ut ememe we mesue potentil diffeences with voltmete)