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Exam 2 covers Ch. 28-33, Lecture, Discussion, HW, Lab. Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 Ch. Chapter 28: Electric flux & Gauss’ law Chapter 29: Electric potential & work Chapter 30: Electric potential & field (exclude 30.7) Chapter 31: Current & Conductivity Chapter 32: Circuits
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Exam 2 covers Ch. 28-33,Lecture, Discussion, HW, Lab Exam 2 is Tue. Oct. 28, 5:30-7 pm, 2103 Ch • Chapter 28: Electric flux & Gauss’ law • Chapter 29: Electric potential & work • Chapter 30: Electric potential & field • (exclude 30.7) • Chapter 31: Current & Conductivity • Chapter 32: Circuits • (exclude 32.8) • Chapter 33: Magnetic fields & forces • (exclude 33.3, 33.6, 32.10, Hall effect)
Electric flux • Suppose surface make angle surface normal Component || surface Component surface Only component‘goes through’ surface • E = EA cos • E =0 if E parallel A • E = EA (max) if E A • Flux SI units are N·m2/C
Gauss’ law • net electric flux through closed surface = charge enclosed /
Properties of conductors - - + + - + - + - - + + • everywhere inside a conductor • Charge in conductor is only on the surface • surface of conductor
Gauss’ law example: Charges on parallel-plate capacitor • Apply Gauss’ law: • E=0 inside metal • E=0 to left • No charge on outer surface • Apply Gauss’ law: • E=0 inside metal • E=Q/Ao in middle -Q Q • Determine fields by superposition Area A=Length X Width
Electric potential: general • Electric field usually created by some charge distribution. • V(r) is electric potential of that charge distribution • V has units of Joules / Coulomb = Volts Electric potential energy difference U proportional to charge q that work is done on Electric potential difference Depends only on charges that create E-fields
Electric Potential Electric potential energy per unit chargeunits of Joules/Coulomb = Volts Example: charge q interacting with charge Q Electric potential energy Electric potential of charge Q Q source of the electric potential, q ‘experiences’ it
Example: Electric Potential y Calculate the electric potential at B B x d d2=4 m -12 μC +12 μC A - + Calculate the electric potential at A d1=3 m 3 m 3 m Calculate the work YOU must do to move a Q=+5 mC charge from A to B. Work done by electric fields
Work and electrostatic potential energy electric potential energy of the system increases −3μC −1μC −2μC Question: How much work would it take YOU to assemble 3 negative charges? Likes repel, so YOU will still do positive work! A. W = +19.8 mJ B. W = -19.8 mJ C. W= 0 q3 5 m 5 m q2 q1 5 m
Potential from electric field • Electric field can be used to find changes in potential • Potential changes largest in direction of E-field. • Smallest (zero) perpendicular to E-field V=Vo
Electric Potential and Field y A 5m B 2m x 2m 5m Uniform electric field of What is the electric potential difference VA-VB? A) -12V B) +12V C) -24V D) +24V
Capacitors +Q -Q Area A d Conductor: electric potential proportional to charge: C = capacitance: depends on geometry of conductor(s) Example: parallel plate capacitor Energy stored in a capacitor:
Isolated charged capacitor • Plate separation increased • The stored energy • Increases • Decreases • Does not change Stored energy A) B) C) q unchanged because C isolated q is the same E is the same = q/(Aε0) ΔV increases = Ed C decreases U increases
Spherical capacitor Gaussian surface to find E + + + + + + + + + Path to find V Charge Q moved from outer to inner sphere Gauss’ law says E=kQ/r2until second sphere Potential difference Along path shown
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’
Electric current L • Average current: • Instantaneous value: • SI unit: ampere1 A = 1 C / s n = number of electrons/volume n x AL electrons travel distance L = vd Δt Iav = ΔQ/ Δt = neAL vd /L • Current density J= I/A = nqvd (direction of + charge carriers)
Resistance and resistivity • Ohm’s Law: ΔV = R I (J = σE or E = ρJ) • ΔV = EL and E = ρ J => ρ I/A = ΔV/L • R = ρL/A Resistance in ohms (Ω)
Current conservation I2 I1 I3 I1=I2+I3 I1 I3 I2 I1+I2=I3 Iin Iout Iout = Iin
Resistors in Series and parallel R1 R2 • Parallel • V1 = V2 = V • Req = (R1-1+R2-1)-1 • Series • I1 = I2 = I • Req = R1+R2 I1+I2 I R1 R1+R2 I = = I1 I2 I R2 2 resistors in series: R L Like summing lengths
Quick Quiz What happens to the brightness of bulb A when the switch is closed? Gets dimmer Gets brighter Stays same Something else
Quick Quiz R1=200Ω R4=100Ω 9V R1=200Ω R3=100Ω 6V 3V Req=100Ω 9V Req=50Ω What is the current through resistor R1? 5 mA 10 mA 20 mA 30 mA 60 mA
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
Capacitors in parallel and series • ΔV1 = ΔV2 = ΔV • Qtotal = Q1 + Q2 Q1=Q2 =Q ΔV = ΔV1+ΔV2 1/Ceq = 1/C1 + 1/C2 Ceq = C1 + C2 Series Parallel
Example: Equivalent Capacitance C1 C2 C3 V C4 C1 = 30 μF C2 = 15 μF C3 = 15 μF C4 = 30 μF in series Parallel combinationCeq=C1||C2
R R RC Circuits C C ε Charge Discharge Time constant Start w/uncharged CClose switch at t=0 Start w/charged CClose switch at t=0
Question R1=100Ω 10V C=1µF R2=100Ω What is the current through R1 Immediately after the switch is closed? A. 10A B. 1 A C. 0.1A D. 0.05A E. 0.01A
Question R1=100Ω 10V C=1µF R2=100Ω What is the charge on the capacitor a long time after the switch is closed? A. 0.05µC B. 0.1µC C. 1µC D. 5µC E. 10µC
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
FB on a Charge Moving in a Magnetic Field, Formula FB = q v x B • FB is the magnetic force • q is the charge • v is the velocity of the moving charge • B is the magnetic field • SI unit of magnetic field: tesla (T) • CGS unit: gauss (G): 1 T = 104 G (Earth surface 0.5 G)
Magnetic Force on a Current S I • Force on each charge • Force on length of wire • Force on straight section of wire, length L Current N Magnetic force Magnetic field
Magnetic field from long straight wire:Direction r = distance from wire = permeability of free space y • What direction is the magnetic field from an infinitely-long straight wire? x I
Current loops & magnetic dipoles • Current loop produces magnetic dipole field. • Magnetic dipole moment: Area of loop current direction magnitude In a uniform magnetic field Magnetic field exerts torqueTorque rotates loop to align with