Electric potential energy Electrostatic force does work on a particle : Potential energy (: i initial state f : final state):

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1 Electc ptental enegy Electstatc fce des wk n a patcle : v v v v W = F s = E s. Ptental enegy (: ntal state f : fnal state): Δ U = U U = W. f ΔU Electc ptental : Δ : ptental enegy pe unt chag e. J ( Jule) N N m J Unt : ( vlt) =, E = = = =. C( Cul) C C m C m m e = e = J f E

2 Wk dne by appled fce W = -W. app suce E W < 0 Δ U > 0 W > 0 Δ U < 0 W > 0 W < 0 app test app suce E W > 0 Δ U < 0 W app < 0 test W < 0 Δ U > 0 W app > 0

3 Electc Ptental Enegy : In chapte 8 we defned the change n ptental enegy asscated wth a cnsevatve fce as the negatve value f the wk W that the fce must d n a patcle t take t fm an ntal pstn x t a fnal pstn x. Δ U = U f U = W = F( x) dx x x f f ΔU... O Δ U x f = F( x) dx x F(x) x x x f x (4 - )

4 Cnsde an electc chage mvng fm an ntal pstn at pnt A t a fnal pstn at pnt B unde the nfluence f a knwn electc feld E. The fce exeted n the chage s: F = E f f f Δ U = F ds = E ds Δ U = E ds A B (4 - )

5 P = 1 R SI Unts f : Defntn f vltage : W Δ = Unts f : J/C knwn as the "lt" Ptental due t a pnt chage Cnsde a pnt chage placed at the gn. We wll use the defntn gven n the pevus O page t detemne the ptental at pnt P a dstnce R f m O. P (4-4)

6 P = 1 R R = E ds = Ed cs0 = Ed P R The electc feld geneated by s: R E = P = d d x R = 1 x 1 1 P = = R R O (4-4)

7 Eu-ptental suface ΔU W Δ = = = 0 v E Euptental suface ( If nt, E 0. tan tan tan E = F F S = W 0 Δ 0) 0 E v v E = 0 = 0 Calculatng fm E: v v v w dw = F ds = E ds f v v f v v W = dw = E ds = E ds W Δ = f v f = E ds = = Ecsθ ds f

8 EX. E + _ ds 1 d Δ = 1 1 v v = E ds = E 1 ds = Ed < 0. ( <, W > 0) 1 E + _ d s 1 d Δ = 1 v v = E ds 1 1 = Ecs180 ds = Eds = E ds = Ed > 1 1 ( >, W < 0) 1 0

9 EX. v=? e - + ~0 + ~0k 1 W = Δ U =ΔK Δ = mv 19 1 ( ) 0k = mev 7 1 v= m/ s c 4

10 Ptental f gup f pnt chages =? R 1 R R3 k = R Ptental f cntnuus chage dstbutn Q ΔQ Ο =? k0δq Δ = k0δq = k0δq k0dq Δ 0 = lm = Δ 0 Q Q

11 Ptental due t a gup f pnt chages Cnsde the gup f thee pnt chages shwn n the fgue. The ptental geneated by ths gup at any pnt P s calculated usng the pncple f supepstn 1. We detemne the ptentals,, and geneated 1 3 by each chage at pnt P =, =, = P 3 = (4-5)

12 . We add the thee tems: = = The pevus euatn can be genealzed f n chages as fllws: n 1 n = = 1 1 n 1 1 P 3 3 = (4-5)

13 EX. Ptental f unfmly chaged thn ng Q dq R dφ 0 ( Z ) =? z kdq Q Q =, dq = Rdϕ = dϕ Q πr π π k Q 0 = d ϕ 0 π kq 0 π = 0 d ϕ π kq 0 = kq 0 = R + z

14 p A B C Example : Ptental due t an electc dple Cnsde the electc dple shwn n the fgue. We wll detemne the electc ptental ceated at pnt P by the tw chages f the dple usng supepstn. Pnt P s at a dstance fm the cente O f the dple. Lne OP makes an angle θ wth the dple axs 1 = ( + ) + ( ) = = ( ) ( + ) ( + ) ( ) ( ) ( + ) (4-6)

15 We assume that? d whee d s the p chage sepaatn. Fm tangle ABC we have: ( ) ( + ) dcsθ Als: dcsθ 1 pcsθ = ( ) ( + ) whee p= d = the electc dple mment. A C = 1 pcsθ B (4-6)

16 Detemnng E fm : ρ ρ ρ ρ dw = F ds = E ds dw ρ ρ = E ds = d ( In geneal, ρ E = E ˆ + E E x x y ˆj + E kˆ z d =, E y =, E z =, x y z ˆ ˆ ρ + j + kˆ, del) x y z E =. E ρ ds ρ f ( E ds, f E = = d ds E ds Fundamental pncple f feld paallel t path Calculus.) ρ ˆ ˆ ( ˆ) (ˆ ˆ j k j kˆ E = + + = + + ). x y z x y z EX. Check feld f sngle chage k = 0 d d k0 E = = = ds d ρ E = k0 ˆ EX. Check fled f unfmly chaged thn ng Q knwn E x = = 0 E y = = 0 x y E z = z = d dz =...

17 Example : O A d Ptental ceated by a lne f chage f length L and unfm lnea chage densty λ at pnt P. Cnsde the chage element d = λdx at pnt A, a dstance x fm O. Fm tangle OAP we have: = + Hee s the dstance OP d x The ptental d ceated by d at P s: d (4-8)

18 O A d d = 1 d 1 λdx = = d + x λ dx L 0 ( ) x d x = ln + + λ = ln x+ d + x d + x λ = ln + d dx + x ( ) ( ) L L x L 0 + lnd (4-8)

19 Lets assume that the electc feld E fms an angle θ wth the path Δ. The wk dne by the electc feld s: W = F Δ = FΔ csθ = EΔcsθ We als knw that W = 0. Thus: EΔ csθ = 0 0, E 0, Δ 0 Thus: csθ = 0 θ = 90 The cect pctue s shwn n the fgue belw E S (4-11)

20 Examples f euptental sufaces and the cespndng electc feld lnes Unfm electc feld Islated pnt chage Electc dple (4-1)

21 Calculatng the electc feld E fm the ptental Nw we wll tackle the evese pblem.e. detemne E f we knw. Cnsde tw euptental sufaces that cspnd t the values and + d sepaated by a dstance ds as shwn n the fgue. Cnsde an abtay dectn epesented by the vect ds. We wll allw the electc feld t mve a chage fm the euptenbtal suface t the suface + d. A B +d (4-13)

22 The wk dne by the electc feld s gven by: W = d (es.1) als W = Fdscsθ = E dscs θ (es.) If we cmpae these tw euatns we have: d E dscsθ = d E csθ = ds Fm tangle PAB we see that E cs θ s the cmpnent E f E alng the dectn s. Thus: E s s = s E s = s A B +d (4-13)