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Exam 2 covers Ch. 27-32, Lecture, Discussion, HW, Lab Exam 2 is Wed. Mar. 26, 5:30-7 pm, 2103 Ch: Adam(301,310), Eli(302,311), Stephen(303,306), 180 Science Hall: Amanda(305,307), Mike(304,309), Ye(308) Chapter 27: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field Chapter 28: Current & Conductivity Chapter 31: Circuits Chapter 32: Magnetic fields & forces (exclude 32.6,32.8,32.10) 1 Electric flux Suppose surface make angle surface normal E E|| sˆ E nˆ A Anˆ nˆ Component || surface Component surface Only component ‘goes through’ surface E = EA cos E =0 if E parallel A E = EA (max) if E A sˆ Flux SI units are N·m2/C E E A 2 Gauss’ law net electric flux through closed surface = charge enclosed / E E dA Qenclosed o 3 Field outside uniformly-charged sphere Field direction: radially out from charge Gaussian surface: Sphere of radius r Surface area where E dA 0 : Value of E dA on this area: 2 4 r E Flux thru Gaussian surface: 2 E 4 r Charge enclosed: Q Q Gauss’ law: E 4 r Q/o E 2 4o r 2 1 4 Electric potential energy Work is Force x distance (taking into account cosθ between 2 vectors!) Whand U >0 since they repel! potential energy increases If opposte charges they attract => W <0 and potential energy decreases 5 Electric Potential Electric potential energy per unit charge units of Joules/Coulomb = Volts Example: charge q interacting with charge Q Qq Electric potential energy UQq ke r UQq Q ke Electric potential of charge Q VQ r q r Q source of the electric it potential, q ‘experiences’ 6 Example: Electric Potential Calculate the electric potential at B q q VB k 0 d d Calculate the electric potential at A q q 2q VA k k d1 3d1 3d1 B y x d -12 C d2=4 m - +12 C A + d1=3 m 3m 3m Calculate the work YOU must do to move a Q=+5 mC charge from A to B. 2qQ WYou U UB UA Q(VB VA ) k 3d1 7 Work and electrostatic potential energy Question: How much work would it take YOU to assemble 3 negative charges? A. W = +19.8 mJ B. W = -19.8 mJ C. = 0 Likes repel, so YOU will still do positive work! q3 W1 0 W2 k 6 C 6 q1q2 110 2 10 9 10 9 3.6mJ r12 5 qq qq W 3 k 1 3 k 2 3 16.2mJ r13 r23 W tot k q1q2 qq qq k 1 3 k 2 3 19.8mJ r12 r13 r23 UE 19.8mJ 5m q1 C 5m q2 5m C electric potential energy of the system increases 8 Potential from electric field dV E d Electric field can be used to find changes V V o in potential Potential changes largest in direction of E-field. Smallest (zero) perpendicular to E-field d d E d V Vo E d V=Vo V Vo E d 9 Electric Potential and Field • Uniform electric field of E = 4i+3j N/C • Points A at 2m and B at 5m on the x axis. • What is the potential difference VA - VB? A(2,0) 0 E = 4i N/C A) -12V B) +12V C) -24V D) +24V dV E V E x dx VA VB 4 3 12V B(5,0) B A dV E x B A x(m) dx 10 Capacitors V Q /C Conductor: electric potential proportional to charge: C = capacitance: depends on geometry of conductor(s) +Q Example: parallel plate capacitor V Q /C C -Q o A Area A d d Energy stored in a capacitor: Q2 1 1 2 U CV QV 2C 2 2 V Stored energy Isolated charged capacitor Plate separation increased The stored energy A) 1) Increases B) 2) Decreases C) 3) Does not change q2 U 2C Cini 0 A d C fin 0 A D q unchanged because C isolated q is the same E is the same = q/(Aε0) ΔV increases = Ed C decreases U increases C fin Cini U fin U ini 12 Conductors, charges, electric fields Electrostatic equilibrium No charges moving No electric fields inside conductor. Electric potential is constant everywhere Charges on surface of conductors. Not equilibrium Charges moving (electric current) Electric fields inside conductors -> forces on charges. Electric potential decreases around ‘circuit’ 13 L Electric current Average current: Instantaneous value: SI unit: ampere 1 A = 1 C / s n = number of electrons/volume n x AL electrons travel distance L = vd Δt Iav = Q/ t = neAL vd /L Current density J= I/A = nqvd (direction of + charge carriers) Resistance and resistivity V = R I (J = E or E = ρ J V = EL and E = J /A = V/L R = ρL/A Resistance in ohms () Ohm’s Law: 15 I2 Current conservation Iin I1 I3 I1=I2+I3 I1 I3 Iout Iout = Iin I2 I1+I2=I3 16 Resistors in Series and parallel Series I1 = I2 = I Req = R1+R2 Parallel V1 = V2 = V Req = (R1-1+R2-1)-1 I1+I2 I R1 R1+R2 R2 = I I1 R1 I R2 I2 = 1 2 resistors in series: RL Like summing lengths 1 1 R1 R2 L R A 17 Quick Quiz How does brightness of bulb B compare to that of A? A. B brighter than A B. B dimmer than A C. Both the same Battery maintain constant potential difference Extra bulb makes extra resistance -> less current 18 Quick Quiz What happens to the brightness of bulb B when the switch is closed? A. Gets dimmer B. Gets brighter C. Stays same D. Something else Battery is constant voltage, not constant current 19 Quick Quiz What happens to the brightness of bulb A when the switch is closed? A. Gets dimmer B. Gets brighter C. Stays same D. Something else 20 Capacitors as circuit elements Voltage difference depends on charge Q=CV Current in circuit Q on capacitor changes with time Voltage across cap changes with time 21 R RC Circuits C C Start w/uncharged C Close switch at t=0 q(t) C(1 e I(t) e R t / RC ) t / RC R Vcap t 1 et / RC Start w/charged C Close switch at t=0 qt qoet / RC qo /C t / RC It e R Vcap t qo /Cet / RC 22 Capacitors in parallel and series ΔV1 = ΔV2 = ΔV Qtotal = Q1 + Q2 Ceq = C1 + C2 Q1=Q2 =Q • ΔV = ΔV1+ΔV2 •1/Ceq = 1/C1 + 1/C2 23 Calculate the equivalent Capacitance C1 = 10 F C2 = 20 F C3 = 30 F C4 = 40 F V = 50 Volts 1 1 1 1 Ceq 6.9F Ceq C1 C2 C3 C4 C1 C2 V V V1 V23 V4 V23 V2 V3 Q Q1 Q23 Q4 Q23 Q2 Q3 Q Q Q Q V Ceq C1 C2 C3 C4 C3 C4 parallel C1, C23, C4 in series 24 RC Circuits What is the value of the time constant of this circuit? A) 6 ms B) 12 ms C) 25 ms D) 30 ms 25 Magnetic fields and forces I FB q vBsin F qv B Magnetic force on currentcarrying wire B FB Ids B Magnetic force on moving charged particle I B Magnetic torque on current loop 26 Effect of uniform magnetic field Effect of uniform B-field on charged particle If charged particle is not moving - no effect If particle is moving: force exerted perpendicular to both field and velocity F qv B vector ‘cross product’ 27 Lorentz force Electron moves in plane of screen the page. B- field is in the plane of screen to the right. Direction of instantaneous magnetic force on electron is A) toward the top of the page B B) into the page C) toward the right edge of the page F v D) out of the page electron 12/09/2002 U. Wisconsin, Physics 208, Fall 2006 28 Trajectory in Constant B Field • Charge enters B field with velocity shown. (vB) • Force is always to velocity and to B. x x x x x x x x x x x x x x x x x x x x x x x vx B x x x x x x x x x x x x q v F F Path is a circle. mv Radius determined by velocity: R qB 29 Current loops & magnetic dipoles Current loop produces magnetic dipole field. Magnetic dipole moment: IA current Area of loop magnitude direction Effect of uniform magnetic field Magnetic field exerts torque B, B sin Torque rotates loop to align with B 30 Question on torque Which of these loop orientations has the largest magnitude torque? (A) a (B) b (C) c μ τ a μ τ b c Answer: (c). all loops enclose same area and carry same current magnitude of μ is the same for all. (c) μ upwards, μ B and τ = μB. (a), = 0 (b) = Bsin 12/09/2002 U. Wisconsin, Physics 208, Fall 2006 31