College Physics B - PHY2054C. Final Review 12/03/2014. My Office Hours: Tuesday 10:00 AM - Noon 206 Keen Building. College Physics B.

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1 College - PHY2054C Final Review 12/03/2014 My Office Hours: Tuesday 10:00 AM - Noon 206 Keen Building

2 PHY2054C 1 Final Exam: Wednesday, 10:00 AM - Noon, UPL 101 Take conceptual questions seriously! 2 Some scores are available on Blackboard! Check your scores this week and talk to me and/or your recitation instructor/lab TA in case you have questions.

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4 Question 1 Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces? A C B D

5 Two uniformly charged spheres are firmly fastened to and electrically insulated from frictionless pucks on an air table. The charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows the magnitude and direction of the electrostatic forces? A C B D

6 F = k q 1 q 2 r 2 Charles-Augustin de Coulomb (14 June August 1806) k = 1 4πǫ 0 = N m 2 /C 2 ǫ 0 is called permittivity of free space

7 Superposition of Forces When there are more than two charges in a problem, the Superposition Principle must be used. 1 Find the forces on the charge of interest due to all the other forces. 2 Add the forces as vectors. F = F1 + F 2

8 The is defined as the force exerted on a tiny positive test charge at that point divided by the magnitude of the test charge: Consider a point in space where the electric field is E. If a charge q is placed at the point, the force is given by F = q E. Unit of the electric field is N/C: F = k Q q r 2 = q E E = k Q r 2

9 Energy The change in electric potential energy is PE elec = W = F x = q E x The change in potential energy depends only on the endpoints of the motion. A positive amount of energy can be stored in a system that is composed of the charge and the electric field. Stored energy can be taken out of the system: This energy may show up as an increase in the kinetic energy of the particle.

10 Two Point Charges From : F = k q 1q 2 r 2 The electric potential energy is given by: PE elec = k q 1q 2 r = q 1q 2 4πǫ 0 r Note that PE elec varies as 1/r while the force varies as 1/r 2. PE elec approaches zero when the two charges are very far apart. The electric force also approaches zero in this limit.

11 : Voltage Electric potential energy is a property of a system of charges or of a charge in an electric field, it is not a property of a single charge alone. Electric potential energy can be treated in terms of a test charge, similar to the treatment of the electric field produced by a charge: V = PE elec q Units are the Volt or [V]: 1 V = 1 J/C. (Named in honor of Alessandro Volta) The unit of the electric field can also be given in terms of the Volt: 1 N/m = 1 V/m.

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13 Question 2 In the United States and Canada, the standard line voltage is VRMS = 110 V. In much of the world (Europe, Australia, Asia), the standard line voltage is VRMS = 220 V. The light output of a 60 Watt Tallahassee light bulb if used in Europe would A be twice as bright. B be four times as bright. C be half as bright. D be one fourth as bright. E remain the same brightness.

14 In the United States and Canada, the standard line voltage is VRMS = 110 V. In much of the world (Europe, Australia, Asia), the standard line voltage is VRMS = 220 V. The light output of a 60 Watt Tallahassee light bulb if used in Europe would A be twice as bright. B be four times as bright. C be half as bright. D be one fourth as bright. E remain the same brightness. P = V I = V 2 / R It must get brighter.

15 In the United States and Canada, the standard line voltage is VRMS = 110 V. In much of the world (Europe, Australia, Asia), the standard line voltage is VRMS = 220 V. The light output of a 60 Watt Tallahassee light bulb if used in Europe would B be four times as bright. How much brighter? P = V I = V 2 / R P Europe V 2 Europe = 1 R = P USA V 2 USA P Europa = P USA V 2 Europe V 2 USA P Europa = 4 P USA

16 Energy in a Resistor The test charge gained energy when it passed through the battery. It lost energy as it passed through the resistor. Energy is converted into heat energy inside the resistor: The energy is dissipated as heat. It shows up as a temperature increase of the resistor and its surroundings. P(Power) = energy transformed time = Q V t = I V P = I V = I 2 R = V 2 / R Reminder: : R = V / I

17 Question 3 Two light bulbs, A and B, are connected in series to a constant voltage source. When a wire is connected across B as shown, bulb A A burns more brightly. B burns as brightly. C burns more dimly. D goes out.

18 Two light bulbs, A and B, are connected in series to a constant voltage source. When a wire is connected across B as shown, bulb A A burns more brightly. B burns as brightly. C burns more dimly. D goes out.

19 When current passes through one resistor and then another, the resistors are said to be in series: E I R 1 I R 2 = 0 Kirchhoff s Loop Rule Any number of resistors can be connected in series. The resistors will be equivalent to a single resistor with: R equiv = R 1 + R 2 + R

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21 Question 4 A current in a long, straight wire produces a magnetic field. The magnetic field lines A go out from the wire to infinity. B come in from infinity to the wire. C form circles that pass through the wire. D form circles that go around the wire.

22 A current in a long, straight wire produces a magnetic field. The magnetic field lines A go out from the wire to infinity. B come in from infinity to the wire. C form circles that pass through the wire. D form circles that go around the wire.

23 Point the thumb of your right hand in the direction of the current: Your thumb will be parallel to the wire. Curling the fingers of your right hand around the wire gives the direction of the magnetic field.

24 Question 5 Two current-carrying wires are parallel as shown below; the current is the same in both wires. The current in both wires is flowing to the right. At a point midway between the wires, the direction of the net magnetic field is A to the right B to the left C into the screen D out of the screen E The field is zero. P

25 Two current-carrying wires are parallel as shown below; the current is the same in both wires. The current in both wires is flowing to the right. At a point midway between the wires, the direction of the net magnetic field is A to the right B to the left C into the screen D out of the screen E The field is zero.

26 Question 6 Two current-carrying wires are parallel as shown below; the currents are now in the opposite directions. At a point midway between the wires (point A), the direction of the net magnetic field is A to the right B to the left C into the screen D out of the screen E The field is zero.

27 Two current-carrying wires are parallel as shown below; the currents are now in the opposite directions. At a point midway between the wires (point A), the direction of the net magnetic field is A to the right B to the left C into the screen D out of the screen E The field is zero.

28 Question 7 The current-carrying wire as shown below is bent into a loop. At any point in the wire loop, the direction of the net magnetic field is: A to the right B to the left C into the screen D out of the screen E The field is zero.

29 The current-carrying wire as shown below is bent into a loop. At any point in the wire loop, the direction of the net magnetic field is: A to the right B to the left C into the screen D out of the screen E The field is zero.

30 Treat the loop as many small pieces of wire: Apply the right-hand rule to find the field from each piece of wire. Applying superposition gives the overall pattern shown on the right. Current Loop At the center of the loop: B = µ 0 I 2R

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32 Question 8 A double loop of wire (making 2 turns) is in the x-y plane centered at the origin. A uniform magnetic field is increasing at constant rate in the negative z direction. In which direction is the induced magnetic field in the loop? A In the positive z direction. B In the negative z direction. C There is no induced field because of the double loop. D There is no induced field because the rate of change of the magnetic field is constant.

33 Example A Assume a metal loop in which the magnetic field passes upward through it. B Assume the magnetic flux increases with time. C The magnetic field produced by the induced emf must oppose the change in flux. The induced magnetic field must be downward and the induced current will be clockwise (right-hand rule).

34 Another Example A Assume a metal loop in which the magnetic field passes upward through it. B Assume the magnetic flux decreases with time. C The magnetic field produced by the induced emf must oppose the change in flux. The induced magnetic field must be downward and the induced current will be counterclockwise (right-hand rule).

35 A double loop of wire (making 2 turns) is in the x-y plane centered at the origin. A uniform magnetic field is increasing at constant rate in the negative z direction. In which direction is the induced magnetic field in the loop? A In the positive z direction. B In the negative z direction. C There is no induced field because of the double loop. D There is no induced field because the rate of change of the magnetic field is constant.

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37 Question 9 If an electric field wave oscillates north and south (horizontally), and the electromagnetic wave is traveling vertically straight up, then what direction does the magnetic field wave oscillate? A It does not oscillate: the situation is impossible. B East and west (horizontally) C North and south (horizontally) D Up and down (vertically)

38 Electromagnetic Waves If an electric field wave oscillates north and south (horizontally), and the electromagnetic wave is traveling vertically straight up, then what direction does the magnetic field wave oscillate? A It does not oscillate: the situation is impossible. B East and west (horizontally) C North and south (horizontally) D Up and down (vertically)

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40 Question 10 A fish swims below the surface of the water at P. An observer at O sees the fish at A a greater depth than it really is. B the same depth. C a smaller depth than it really is.

41 Refraction A fish swims below the surface of the water at P. An observer at O sees the fish at A a greater depth than it really is. B the same depth. C a smaller depth than it really is.

42 The ratio c/v is called index of refraction and is denoted by n: n = c/v sinθ 1 = n sinθ 2 A more general statement can be applied to any two materials with indices of refraction n 1 and n 2 : n 1 sinθ 1 = n 2 sinθ 2 Snell s

43 Question 11 A convex lens has a focal length of magnitude F. At which of the following distances from this lens would an object give an upright virtual image? A F/2 B 2F C Any value greater than 2F D This cannot be done with a convex lens.

44 Image Formation A convex lens has a focal length of magnitude F. At which of the following distances from this lens would an object give an upright virtual image? A F/2 B 2F C Any value greater than 2F D This cannot be done with a convex lens.

45 Question 12 A convex lens has a focal length of magnitude F. At which of the following distances from this lens would an object give an inverted virtual image? A F/2 B 2F C Any value greater than 2F D This cannot be done with a convex lens.

46 Image Formation A convex lens has a focal length of magnitude F. At which of the following distances from this lens would an object give an inverted virtual image? A F/2 B 2F C Any value greater than 2F D This cannot be done with a convex lens.