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Lecture 20R: MON 13 OCT Exam 2: Review Session CH 23.1–9, 24.1–12, 25.1–6,8 26.1–9 Physics2113 Jonathan Dowling Some links on exam stress:

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Presentation on theme: "Lecture 20R: MON 13 OCT Exam 2: Review Session CH 23.1–9, 24.1–12, 25.1–6,8 26.1–9 Physics2113 Jonathan Dowling Some links on exam stress:"— Presentation transcript:

1 Lecture 20R: MON 13 OCT Exam 2: Review Session CH 23.1–9, 24.1–12, 25.1–6,8 26.1–9 Physics2113 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 (Ch23) Gauss’s Law (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.

3 Gauss ’ law:  =q/  0. Given the field, what is the charge enclosed? Given the charges, what is the flux? Use it to deduce formulas for electric field. 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.

4 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 = -38 k N/C, and on the bottom face of the cube E = +11 k N/C. Determine the net charge contained within the cube. [-2.29e-09] C Gauss’ law

5 Gauss’s Law: Cylinder, Plane, Sphere

6 Problem: Gauss ’ Law to Find E

7 Gauss ’ law A long, non conducting, solid cylinder of radius 4.1 cm has a nonuniform volume charge density that is a function of the radial distance r from the axis of the cylinder, as given by  = Ar 2, with A = 2.3 µC/m 5. (a)What is the magnitude of the electric field at a radial distance of 3.1 cm from the axis of the cylinder? (b)What is the magnitude of the electric field at a radial distance of 5.1 cm from the axis of the cylinder?

8 The figure shows conducting plates with area A=1m 2, and the potential on each plate. Assume you are far from the edges of the plates. What is the electric field between the plates in each case? What (and where) is the charge density on the plates in case (1)? What happens to an electron released midway between the plates in case (1)?

9 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?

10 Derive an expression in terms of q 2 /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/m 2 )x 2 - (4.0 V/m 2 )y 2. What are the magnitude and direction of the electric field at point (3.0 m, 3.0 m)?

11 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 +Q Do this from scratch! Don ’ t memorize the formula in the book! We will change the numbers!!!

12 Potential Energy of A System of Charges: Solution No energy needed to bring in first charge: U 1 =0 Energy needed to bring in 2nd charge: Energy needed to bring in 3rd charge = Energy needed to bring in 4th charge = +Q Total potential energy is sum of all the individual terms shown on left hand side = So, final kinetic energy of each charge =

13 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’

14 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

15 Potential V of Continuous Charge Distributions Surface Charge: dq=  dA  =Q/  R 2 dA=2πR’dR‘ Straight Line Charge: dq=  dx  =bx is given to you. Straight Line Charge: dq=  dx  =bx is given to you.

16 Capacitors E =  /  0 = q/A  0 E = V d q = C V C plate =   0 A/d C plate =  0 A/d C sphere =  0 ab/(b-a) Connected to Battery: V=Constant Disconnected: Q=Constant Connected to Battery: V=Constant Disconnected: Q=Constant

17 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 –Q Q is fixed! d increases! C decreases (=  0 A/d) V=Q/C; V increases.

18 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. +Q –Q V is fixed constant by battery! C decreases (=  0 A/d) Q=CV; Q decreases E = σ/  0 = Q/  0 A decreases Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease? Battery does work on capacitor to maintain constant V!

19 Capacitors Capacitors Q=CV In series: same charge 1/C eq = ∑1/C j In parallel: same voltage C eq =∑C j

20 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?

21 Capacitors: Checkpoints, Questions

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

25 Current and Resistance

26 Current and Resistance: Checkpoints, Questions

27 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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