Lecture 22: WED 14 OCT

1. Physics2113 Jonathan Dowling Lecture 22: WED 14 OCT Exam 2: Review Session CH 24–27.3 & HW 05–07 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 • (Ch 23) Gauss’s Law: Flux, Sphere, Plane, Cylinder • (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. • (Ch 26) Current and Resistance. Current Density. Ohm’s Law. Power in a Resistor.

3. Gauss’ law At each point on the surface of the cube shown in Fig. 24-26, the electric field is in the z direction. The length of each edge of the cube is 2.3 m. On the top surface of the cube E = -38k N/C, and on the bottom face of the cube E = +11k N/C. Determine the net charge contained within the cube.[-2.29e-09] C

4. Gauss’s Law: Cylinder, Plane, Sphere

5. Problem: Gauss’ Law to Find E

6. Two Insulating Sheets

7. Two Conducting Sheets E does not pass through a conductor Formula for E different by Factor of 2 7.68 4.86 4.86 7.68 12.54

8. Electric Fields With Insulating Sphere

10. Electric potential: • What is the potential produced by a system of charges? (Several point charges, or a continuous distribution) • Electric field lines, equipotential surfaces: lines go from +ve to –ve charges; lines are perpendicular to equipotentials; lines (and equipotentials) never cross each other… • Electric potential, work and potential energy: work to bring a charge somewhere is W = –qV (signs!). Potential energy of a system = negative work done to build it. • Conductors: field and potential inside conductors, and on the surface. • Shell theorem: systems with spherical symmetry can be thought of as a single point charge (but how much charge?) • Symmetry, and “infinite” systems.

11. Electric potential, electric potential energy, work In Fig. 25-39, point P is at the center of the rectangle. With V = 0 at infinity, what is the net electric potential in terms of q/d at P due to the six charged particles?

12. Derive an expression in terms of q2/a for the work required to set up the four-charge configuration of Fig. 25-50, assuming the charges are initially infinitely far apart. The electric potential at points in an xy plane is given by V = (2.0 V/m2)x2 - (4.0 V/m2)y2. What are the magnitude and direction of the electric field at point (3.0 m, 3.0 m)?

13. +Q +Q +Q +Q Potential Energy of A System of Charges • 4 point charges (each +Q) are connected by strings, forming a square of side L • If all four strings suddenly snap, what is the kinetic energy of each charge when they are very far apart? • Use conservation of energy: • Final kinetic energy of all four charges = initial potential energy stored = energy required to assemble the system of charges Do this from scratch! Don’t memorize the formula in the book! We will change the numbers!!!

14. +Q +Q +Q +Q Potential Energy of A System of Charges: Solution • No energy needed to bring in first charge: U1=0 • Energy needed to bring in 2nd charge: • Energy needed to bring in 3rd charge = Total potential energy is sum of all the individual terms shown on left hand side = • Energy needed to bring in 4th charge = So, final kinetic energy of each charge =

15. Potential V of Continuous Charge Distributions Straight Line Charge: dq=λdx λ=Q/L Curved Line Charge: dq=λds λ=Q/2πR Surface Charge: dq=σdA σ=Q/πR2 dA=2πR’dR’

16. Potential V of Continuous Charge Distributions Curved Line Charge: dq=λds λ=Q/2πR Straight Line Charge: dq=λdx λ=Q/L

17. Potential V of Continuous Charge Distributions Surface Charge: dq=σdA σ=Q/πR2 dA=2πR’dR‘ Straight Line Charge: dq=λdx λ=bx is given to you.

18. ✔ ✔ ✔ ✔

19. –Q +V +Q 0 –V + + + + + + + + + – – – – – – – – ICPP: Consider a positive and a negative charge, freely moving in a uniform electric field. True or false? • Positive charge moves to points with lower potential voltage. • Negative charge moves to points with lower potential voltage. • Positive charge moves to a lower potential energy. • Negative charge moves to a lower potential energy. (a) True (b) False (c) True (d) True

20. electron ✔ ✔ ✔ ✔

21. downhill

22. No vectors! Just add with sign. One over distance. Since all charges same and all distances same all potentials same.

24. (a) Since Δx is the same, only |ΔV| matters! |ΔV1| =200, |ΔV2| =220, |ΔV3| =200 |E2| > |E3| = |E1| The bigger the voltage drop the stronger the field. Δx (b) = 3 (c) F = qE = ma accelerate leftward

25. Cplate = κε0A/d Capacitors E = σ/ε0 = q/Aε0 E = V d q = C V Cplate = ε0A/d Connected to Battery: V=Constant Disconnected: Q=Constant Csphere=ε0ab/(b-a)

26. –Q +Q Isolated Parallel Plate Capacitor: ICPP • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential voltage difference = V. • Battery is then disconnected. If the plate separation is INCREASED, does the capacitance C: (a) Increase? (b) Remain the same? (c) Decrease? If the plate separation is INCREASED, does the Voltage V: (a) Increase? (b) Remain the same? (c) Decrease? • Q is fixed! • d increases! • C decreases (= ε0A/d) • V=Q/C; V increases.

27. –Q +Q Parallel Plate Capacitor & Battery: ICPP • A parallel plate capacitor of capacitance C is charged using a battery. • Charge = Q, potential difference = V. • Plate separation is INCREASED while battery remains connected. • V is fixed constant by battery! • C decreases (=ε0A/d) • Q=CV; Q decreases • E = σ/ε0 = Q/ε0A decreases Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease? Battery does work on capacitor to maintain constant V!

28. Capacitors Capacitors Q=CV In series: same charge 1/Ceq=∑1/Cj In parallel: same voltage Ceq=∑Cj

29. Capacitors in Series and in Parallel • What’s the equivalent capacitance? • What’s the charge in each capacitor? • What’s the potential across each capacitor? • What’s the charge delivered by the battery?

32. Capacitors: Checkpoints, Questions

33. Parallel plates: C = ε0 A/dSpherical: • Cylindrical: C = 2πε0 L/ln(b/a)] Hooked to battery V is constant. Q=VC (a) d increases -> C decreases -> Q decreases (b) a inc -> separation d=b-a dec. -> C inc. -> Q increases (c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases

34. PARALLEL: • V is same for all capacitors • Total charge = sum of Q • SERIES: • Q is same for all capacitors • Total potential difference = • sum of V • Parallel: Voltage is same V on each but charge is q/2 on each since • q/2+q/2=q. (b) Series: Charge is same q on each but voltage is V/2 on eachsince V/2+V/2=V.

35. dielectric slab Example: Battery Connected — Voltage V is Constant but Charge Q Changes • Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E; • Battery remains connected • V is FIXED; Vnew = V (same) • Cnew =κC(increases) • Qnew = (κC)V = κQ(increases). • Since Vnew = V, Enew = V/d=E (same) Energy stored? u=ε0E2/2 => u=κε0E2/2 = εE2/2 increases

36. Example: Battery Disconnected — Voltage V Changes but Charge Q is Constant • Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E; • Battery remains disconnected • Q is FIXED; Qnew = Q (same) • Cnew =κC(increases) • Vnew = Q/Cnew= Q/(κC)(decreases). • Since Vnew < V, Enew = Vnew/d = E/κ (decreases) dielectric slab Energy stored?

37. V = i R E = Jρ Current and Resistance i = dq/dt R = ρL/A Junction rule r = ρ0(1+a(T-T0))

38. Current and Resistance

39. A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x10-5Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?

41. Current and Resistance: Checkpoints, Questions

42. 26.2: Electric Current, Conservation of Charge, and Direction of Current: Fill in the blanks. Think water in hose!

43. All quantities defined in terms of + charge movement! (a) right (c) right (b) right (d) right

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

45. Resistors vs Capacitors ResistorsCapacitors Key formula: V=iR Q=CV In series: same current same charge Req=∑Rj 1/Ceq=∑1/Cj In parallel: same voltage same voltage 1/Req=∑1/Rj Ceq=∑Cj

46. Resistorsin Series and in Parallel • What’s the equivalent resistance? • What’s the current in each resistor? • What’s the potential across each resistor? • What’s the current delivered by the battery? • What’s the power dissipated by each resisitor?

47. Series and Parallel

48. Series and Parallel

49. Series and Parallel

50. 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. Slide Rules: You may bend the wires but not break them. You may slide any circuit element along a wire so long as you don’t slide it past a three (or more) point junction or another circuit element.