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## Exam 2 covers Ch. 28-33, Lecture, Discussion, HW, Lab

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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