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Physics 2102 Exam 2: Review Session Chapters 24.9-28.8 / HW04-06 Physics 2102 Jonathan Dowling Some links on exam stress:

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Presentation on theme: "Physics 2102 Exam 2: Review Session Chapters 24.9-28.8 / HW04-06 Physics 2102 Jonathan Dowling Some links on exam stress:"— Presentation transcript:

1 Physics 2102 Exam 2: Review Session Chapters 24.9-28.8 / HW04-06 Physics 2102 Jonathan Dowling Some links on exam stress: http://appl003.lsu.edu/slas/cas.nsf/$Content/Stress+Management+Tip+1 http://wso.williams.edu/orgs/peerh/stress/exams.html http://www.thecalmzone.net/Home/ExamStress.php http://www.staithes.demon.co.uk/exams.html

2 Exam 2 (Ch24) Sec.11 (Electric Potential Energy of a System of Point Charges); Sec.12 (Potential of Charged Isolated Conductor) (Ch 25) Capacitors: capacitance and capacitors; caps in parallel and in series, dielectrics; energy, field and potential in capacitors. Current and Resistance(Ch 26) Current and Resistance: current, current density and drift velocity; resistance and resistivity; Ohm’s law. (Ch 27) Circuits: emf devices, loop and junction rules; resistances in series and parallel; DC single and multiloop circuits, power; RC circuits. (Ch 28) Magnetic Fields: F=vxB, Right Hand Rule, Circular Motion, Force on Wire, Magnetic Dipole.

3 Potential V of Continuous Charge Distributions Straight Line Charge: dq= dx =Q/L Curved Line Charge: dq= ds =Q/2  R Surface Charge: dq=  dA  =Q/  R 2 dA=2  R’dR’ Straight Line Charge: dq= dx =Q/L Curved Line Charge: dq= ds =Q/2  R Surface Charge: dq=  dA  =Q/  R 2 dA=2  R’dR’

4 Potential V of Continuous Charge Distributions Straight Line Charge: dq= dx =Q/L Straight Line Charge: dq= dx =Q/L Curved Line Charge: dq= ds =Q/2  R Curved Line Charge: dq= ds =Q/2  R

5 Potential V of Continuous Charge Distributions Surface Charge: dq=  dA  =Q/  R 2 dA=R’dR’d  Straight Line Charge: dq= dx =bx is given to you. Straight Line Charge: dq= dx =bx is given to you.

6 Capacitors E =  /  0 = q/A  0 E = V d q = C V C =   0 A/d C =  0 A/d C=  0 ab/(b-a) Connected to Battery: V=Constant Disconnected: Q=Constant Connected to Battery: V=Constant Disconnected: Q=Constant

7 Current and Resistance i = dq/dt V = i R E = J   =  0 (1+  (T  T 0 )) R =  L/A Junction rule

8 DC Circuits Single loop Multiloop V = iR P = iV Loop rule

9 Resistors and Capacitors Resistors Capacitors Key formula: V=iR Q=CV In series: same current same charge R eq =∑R j 1/C eq = ∑1/C j In parallel: same voltage same voltage 1/R eq = ∑1/R j C eq =∑C j

10 Capacitors and Resistors in Series and in Parallel What’s the equivalent resistance (capacitance)? What’s the current (charge) in each resistor (capacitor)? What’s the potential across each resistor (capacitor)? What’s the current (charge) delivered by the battery?

11 RC Circuits Time constant: RC i(t)=dq/dt

12 Capacitors: Checkpoints, Questions

13 Problem 25-21 When switch S is thrown to the left, the plates of capacitor 1 acquire a potential V 0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q 1, q 2, and q 3 on the capacitors? Series

14 Seri-Q

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16 Current and Resistance: Checkpoints, Questions

17 Problem 26-56 A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5 x 10 -5  m. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?

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19 Circuits: Checkpoints, Questions

20 Problem: 27.P.018. [406649] Figure 27-33 shows five 5.00 resistors. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Fig. 27-33 (a) Find the equivalent resistance between points F and H. (b) Find the equivalent resistance between points F and G.

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22 Problem: 27.P.046. [406629] In an RC series circuit, E = 17.0 V, R = 1.50 M , and C = 1.80 µF. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 10.0 µC?

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24 Magnetic Forces and Torques L v F

25 C C Top view Side view (28-13)

26 (28-14)

27 Ch 28: Checkpoints and Questions

28 Circular Motion: Since magnetic force is perpendicular to motion, the movement of charges is circular. B into blackboard. v F In general, path is a helix (component of v parallel to field is unchanged). In general, path is a helix (component of v parallel to field is unchanged). r

29 .. electron C r Radius of Circlcular Orbit Angular Frequency: Independent of v Angular Frequency: Independent of v Period of Orbit: Independent of v Period of Orbit: Independent of v Orbital Frequency: Independent of v Orbital Frequency: Independent of v

30 Problem: 28.P.024. [566302] In the figure below, a charged particle moves into a region of uniform magnetic field, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 160 ns in the region. (a) What is the magnitude of B? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 3.00 times its previous kinetic energy, how much time does it spend in the field during this trip?

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