Field due to a collection of N discrete point charges: r is in the direction from

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1 Physcs 46 Fomula Shee Exam Coulomb s Law qq Felec = k ˆ (Fo example, f F s he elecc foce ha q exes on q, hen ˆ s a un veco n he decon fom q o q.) Elecc Feld elaed o he elecc foce by: Felec = qe (elecc foce on a chage q n a feld E ) Feld of a pon chage Q : Q E = k ˆ ( ˆ s n he decon fom Q o he feld pon : he pon a whch he feld s beng evaluaed.) Feld due o a collecon of dscee pon chages: Q E = k ˆ = ( s he dsance fom Q o he feld pon. ˆ Feld due o a connuous chage dsbuon: s n he decon fom Q o he feld pon.) dq E = k ˆ (Inegal s aken ove he whole chage dsbuon. ˆ s n he decon fom dq o he feld pon.) Feld due o a unfomly chaged ng of neglgble hckness and adus a locaed n he y-z plane wh cene a ogn: E = kq x ˆ x + a Q s he oal chage on he ng. ( ) 3/ Elecc Flux Φ E = E da (Inegal s aken ove he ene suface.) Fo a unfom feld and a fla suface, hs smplfes o: Φ E = E A Do poduc: If A = A ˆ ˆ ˆ x + Ay j+ Az k and B = B ˆ ˆ ˆ x + By j+ Bz k, hen AB = AB+ AB + AB = ABcosθ x x y y z z θ s he (smalles) angle beween A and B.

2 Fomula Shee Exam Page Gauss s law E da= Q encl ε 0 Elecc Poenal Geneal defnon (poenal as a funcon of poson n some egon whee hee s an E-feld E ): ( ) V = E d ( d s an nfnesmal veco elemen of ac lengh along he pah ove whch he negal s evaluaed. oe ha hs defnon akes he poenal o be zeo a nfny.) Poenal due o a pon chage Q : Q V = k Poenal due o a collecon of dscee pon chages: Q V = k = ( s he dsance fom Q o he pon a whch he poenal s beng evaluaed.) Poenal due o a connuous chage dsbuon: dq V = k (Inegal s aken ove he whole chage dsbuon.) Poenal dffeence (volage) beween wo pons a and b : Fndng E fom V : Δ V = Vb Va = E d E = V b a V V V Ex =, Ey =, Ez = x y z

3 Fomula Shee Exam Page 3 Elecc Poenal Enegy elaon o he Elecc Poenal, V and Poenal Dffeence, Δ V : U V = U = qv Δ U = qδ V q Poenal Enegy of a Collecon of Dscee Pon Chages: Capacance qq U = k = j= < j j s he dsance beween q and j j q j. Q = CV A Paallel-plae capaco: C = ε0. d πε0l Cylndcal capaco: C = b ln a a b Sphecal capaco: C = 4πε0 b a Combnaons of Capacos Paallel: C = C+ C + + C Sees: = C C C C Fo wo caps n sees: CC C = (poduc ove he sum) C+ C Fo ual caps n sees, each of capacance C: Capacos Wh Deleccs C = KC 0 C C =.

4 Fomula Shee Exam Page 4 Enegy Soed n a Capaco Q U = CV = = QV C Enegy densy (enegy pe un volume): Elecc Cuen Iav ΔQ (aveage cuen fo me Δ ) Δ ue = ε0e dq I (nsananeous cuen) d Mcoscopc Model of Cuen j = nqvd I j = A I = nqv A d j = σ E (Ohm s law) essance and Tempeaue L = ρ A ρ = σ ρ = ρ + α( T T ) 0 0 = + α ( T T ) 0 0 V = I (Ohm s law) Eleccal Enegy and Powe P = IV (powe suppled by a souce o delveed o a load) If load s a esso (.e., obeys Ohm s law), hen: V P = I =

5 Fomula Shee Exam Page 5 Combnaons of essos Sees: = Paallel: = Fo wo essos n paallel: = + (poduc ove he sum) Kchhoff s ules Fo ual essos n paallel, each of essance : =. Kchhoff s Cuen ule (KC): In = Iou Kchhoff s Volage ule (KV): Δ V = 0 C Ccus Chagng: closed loop q () Q e = C max () = 0 Dschagng: I e C q () = Qe C 0 () = 0 I e C τ = C (C me consan) Magnec Foces Magnec foce on a pacle wh chage q movng wh velocy v n magnec feld B : F = qv B mag e elecomagnec foce on chaged pacle movng n egon of E and B felds: F = q E+ v B ne ( ) (Loenz foce law)

6 Fomula Shee Exam Page 6 Magnec foce on a sagh cuen-cayng conduco n a B-feld: F = IL B mag Coss Poduc If A = A ˆ+ A ˆj+ A kˆ x y z and B = B ˆ+ B ˆj+ B kˆ, hen: x y z ˆ ˆj kˆ A B = A A A = A B A B A B A B j+ A B A B kˆ ( ) ˆ ( ) ˆ ( ) x y z y z z y x z z x x y y x B B B x y z A B = ABsnφ. φ s he angle beween A and B. Moon of Chaged Pacle n Unfom B-feld adus of pah: Velocy Seleco E v = B = mv qb Cycloon Fuency ω = qb m f qb = π m Thomson s e/m Expemen e E = m VB

7 Fomula Shee Exam Page 7 Physcal Consans C ε0 = m (pemvy of fee space) 9 m k = k = C 4πε 0 9 e =.60 0 C (magnude of chage on elecon) 3 m e = 9. 0 kg (mass of elecon) 7 =.67 0 kg (mass of poon) m p Aeas & Volumes Suface aea of sphee, adus : A = 4π 4 3 Volume of sphee, adus : V = π 3 Aea of laeal suface of a gh ccula cylnde, adus, lengh : A = π Volume of gh ccula cylnde, adus, lengh : V = π

5-1. We apply Newton s second law (specifically, Eq. 5-2). F = ma = ma sin 20.0 = 1.0 kg 2.00 m/s sin 20.0 = 0.684N. ( ) ( )

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