AP Physics 2012 Practice Quiz 4, Conduction & Electric Fields

Transcription

1 Name: Class: Date: ID: A AP Physics 01 Practice Quiz 4, Conduction & Electric Fields Multiple Choice Identify the choice that best completes the statement or answers the question. 1. ( points) A repelling force occurs between two charged objects when the charges are of a. unlike signs. b. like signs. c. equal magnitude. d. unequal magnitude.. ( points) Charge is most easily transferred in a. nonconductors. b. conductors. c. semiconductors. d. insulators. 3. ( points) In the diagram shown above, the circles represent small balls that have electric charges. Ball 1 has a negative charge, and ball is repelled by ball 1. Next, you see that ball repels ball 3 and that ball 3 attracts ball 4. What is the electric charge on ball 4? a. Ball 4 may have either a positive or negative charge. b. Ball 4 has a negative charge. c. Ball 4 has a positive charge. d. It is not possible to determine the charge on ball ( points) A surface charge can be produced on insulators by a. grounding. b. induction. c. polarization. d. contact. 5. ( points) Conductors can be charged by, while insulators cannot. a. grounding b. induction c. polarization d. contact 6. ( points) Which of the following is not a characteristic of electrical potential energy? a. It is a form of mechanical energy. b. It results from a single charge. c. It results from the interaction between charges. d. It is associated with a charge in an electric field. 7. ( points) Two positive point charges are initially separated by a distance of cm. If their separation is increased to 6 cm, the resultant electrical potential energy is equal to what factor multiplied by the initial electrical potential energy? a. 3 b. 9 c. 1 3 d ( points) Charge buildup between the plates of a capacitor stops when a. there is no net charge on the plates. b. unequal amounts of charge accumulate on the plates. c. the potential difference between the plates is equal to the applied potential difference. d. the charge on both plates is the same. 9. ( points) A capacitor consists of two metal plates; is stored on one plate, and is stored on the other. a. negative charge, positive charge b. potential energy, kinetic energy c. potential difference, internal resistance d. residual charge, induced charge 10. ( points) What effect will be produced on a capacitor if the separation between the plates is increased? a. It will increase the charge. b. It will decrease the charge. c. It will increase the capacitance. d. It will decrease the capacitance. 1

2 Name: ID: A 11. ( points) A 0.50 µf capacitor is connected to a 1 V battery. Use the expression PE 1 C( V) to determine how much electrical potential energy is stored in the capacitor. a J b J c J d J 1. ( points) A 1.5 µf capacitor is connected to a 9.0 V battery. Use the expression PE 1 C( V) to determine how much energy is stored in the capacitor. a J b J c J d J 13. ( points) When you flip a switch to turn on a light, the delay time before the light turns on is determined by a. the number of electron collisions per second in the wire. b. the drift speed of the electrons in the wire. c. the speed of the electric field moving in the wire. d. the resistance of the wire. 14. ( points) The energy gained by electrons as they are accelerated by an electric field is a. greater than the average loss in energy due to collisions. b. equal to the average loss in energy due to collisions. c. less than the average loss in energy due to collisions. d. not affected by the gain in energy due to collisions. 15. ( points) An attracting force occurs between two charged objects when the charges are of a. unlike signs. b. like signs. c. equal magnitude. d. unequal magnitude. 16. ( points) Electric charge is a. found only in a conductor. b. conserved. c. found only in insulators. d. not conserved. 17. ( points) If a positively charged glass rod is used to charge a metal bar by induction, the charge on the bar a. will be equal in magnitude to the charge on the glass rod. b. must be negative. c. must be positive. d. will be greater in magnitude than the charge on the glass rod. 18. ( points) Which sentence best describes electrical conductors? a. Electrical conductors have low mass density. b. Electrical conductors have high tensile strength. c. Electrical conductors have electric charges that move freely. d. Electrical conductors are poor heat conductors. 19. ( points) Which statement is the most correct regarding electric insulators? a. Charges within electric insulators do not readily move. b. Electric insulators have high tensile strength. c. Electric charges move freely in electric insulators. d. Electric insulators are good heat conductors. 0. ( points) The process of charging a conductor by bringing it near another charged object and then grounding the conductor is called a. contact charging. b. induction. c. polarization d. neutralization. 1. ( points) Both insulators and conductors can be charged by a. grounding. b. induction. c. polarization. d. contact.. ( points) Which is the most correct statement regarding the drawing of electric field lines? a. Electric field lines always connect from one charge to another. b. Electric field lines always form closed loops. c. Electric field lines can start on a charge of either polarity. d. Electric field lines never cross each other. 3. ( points) Electric field strength depends on a. charge and distance. b. charge and mass. c. Coulomb constant and mass. d. elementary charge and radius. 4. ( points) What occurs when two charges are moved closer together? a. The electric field doubles. b. Coulomb s law takes effect. c. The total charge increases. d. The force between the charges increases. 5. ( points) Resultant force on a charge is the sum of the individual forces on that charge. a. scalar b. vector c. individual d. negative

3 Name: ID: A Problem 6. (5 points) What is the electric force between an electron and a proton that are separated by a distance of m? Is the force attractive or repulsive? (e C, N m /C ) 7. (5 points) A proton (q C) moves 4 cm on a path parallel to the direction of a uniform electric field of strength 5.0 N/C. What is the change in electrical potential energy? 8. (5 points) What is the electric potential at a distance of 0.15 m from a point charge of 6.0 µc? ( N m /C ) 9. (5 points) A current of 8.0 A flows through a resistor when a potential difference of 0.0 V is applied to the resistor. What is the resistance of the resistor? 30. (5 points) A toaster is connected across a 13 V outlet and dissipates 0.95 kw in the form of electromagnetic radiation and heat. What is the resistance of the toaster? 31. (5 points) An alpha particle (charge e) is sent at high speed toward a gold nucleus. The electric force acting on the alpha particle is 91.0 N when it is m away from the gold nucleus. What is the charge on the gold nucleus, as a whole number multiple of e? (e C, N m /C ) 3. (5 points) Two point charges having charge values of.0 µc and 4.0 µc, respectively, are separated by 1.5 cm. What is the value of the mutual force between them? ( N m /C ) 33. (5 points) Charge A and charge B are. m apart. Charge A is 1.0 C, and charge B is.0 C. Charge C, which is.0 C, is located between them and is in electrostatic equilibrium. How far from charge A is charge C? 34. (5 points) The diagram above shows four charges A, B, C, and D at the corners of a square. Charges A and D, on opposite corners, have equal charge, whereas both B and C have a charge of 1.0 C. If the force on B is zero, what is the charge on A? 35. (5 points) Two equal but oppositely charged points are 1.0 m apart in a vacuum. The electric field intensity at the midpoint between the charges is N/C. What is the magnitude of each charge? ( N m /C ) 3

4 AP Physics 01 Practice Quiz 4, Conduction & Electric Fields Answer Section MULTIPLE CHOICE 1. ANS: B PTS: DIF: I OBJ: ANS: B PTS: DIF: I OBJ: ANS: C PTS: DIF: II OBJ: ANS: C PTS: DIF: I OBJ: ANS: B PTS: DIF: I OBJ: ANS: B PTS: DIF: I OBJ: ANS: C PTS: DIF: II OBJ: ANS: C PTS: DIF: II OBJ: ANS: A PTS: DIF: I OBJ: ANS: D PTS: DIF: I OBJ: ANS: D C 0.50 µf 0.50 x 10 6 F V 1 V PE electric 1 C( V) 1 ( F)(1 V) J PTS: DIF: IIIA OBJ: ANS: B C 1.5 µf F V 9.0 V PE electric 1 C( V) 1 ( F)(9.0 V) J PTS: DIF: IIIA OBJ: ANS: C PTS: DIF: II OBJ: ANS: A PTS: DIF: I OBJ: ANS: A PTS: DIF: I OBJ: ANS: B PTS: DIF: I OBJ: ANS: B PTS: DIF: II OBJ: ANS: C PTS: DIF: I OBJ: ANS: A PTS: DIF: I OBJ: ANS: B PTS: DIF: I OBJ: ANS: D PTS: DIF: I OBJ: ANS: D PTS: DIF: II OBJ: ANS: A PTS: DIF: I OBJ: ANS: D PTS: DIF: I OBJ: ANS: B PTS: DIF: I OBJ:

5 PROBLEM 6. ANS: N; attractive q e e C q p +e C r m N m /C Ê Ê q e q p ˆÊ C Ê Nm /C ˆ ˆ ˆ C r Ê ˆ m N PTS: 5 DIF: IIIA OBJ: ANS: J E 5.0 N/C q C d 4 cm m PE electric qed PE electric ( C)(5.0 N/C)( m) PE electric J PTS: 5 DIF: IIIB OBJ: 17-1.

6 8. ANS: V d 0.15 m q 6.0 µc C N m /C r d 0.15 m q Ê V r ( Nm/C ˆ C ) 0.15 m V PTS: 5 DIF: IIIB OBJ: ANS:.5 Ω I 8.0 A V 0.0 V V I R Rearrange to solve for R. R V I 0.0 V 8.0 A.5 Ω PTS: 5 DIF: IIIA OBJ: ANS: 16 Ω V 13 V P 0.95 kw Ê P 0.95 kw P ( V) R W 1 kw Rearrange to solve for R. ˆ W R ( V) P (13 V) W 1519 V W 16 Ω PTS: 5 DIF: IIIB OBJ:

7 31. ANS: 79e e C q α e C 91.0 N r m N m /C q α q Gold r Rearrange to solve for q Gold. q Gold q Gold Ê Felectric ˆr (91.0 N)( m) ( )q α ( Nm /C )( C) C C q e C 79.4 The charge on the gold nucleus must be an integer multiple of e. Integer (79.4)e 79e PTS: 5 DIF: IIIB OBJ: ANS: N C q C r 1.5 cm m N m /C Ê Ê q ˆ C Ê Nm /C ˆ Ê C ˆ ˆ r ( m) N PTS: 5 DIF: IIIB OBJ:

8 33. ANS: 0.91 m r A,B. m r C,A d r C,B. m d 1.0 C.0 C q C.0 C F C,A F C,B 0 N F C,A F C,B Ê ˆ Ê ˆ q C q C k (r C,A ) C (r C,B ) d (. m d) Ê d ˆÊ ˆ (. m d) Ê ˆ d (. m d) Ê ˆ d + q A (. m) d d 0.91 m (. m) C(. m).0 C C 0.91 m PTS: 5 DIF: IIIC OBJ:

9 34. ANS: 0.35 C r A,B r B,D r D,C r C,A q D q C C F B 0 N θ ABC θ DBC 45 Because charge B is in equilibrium and the charges are along the same line, the force exerted on charge B by charge A in the y-direction is zero, and only the force exerted on charge B by charge A in the x-direction is needed. In the x-direction, F A,B F C,B, which gives F B 0. F B F A,B + (cos θ ABC ) F C,B Because the distance between the charges is not known, any convenient value for r C,B can be substituted: r C,B 1.0 m Then, determine the relative distance between charge A and charge B. r A,B (cos θ ABC ) r C,B r A,B (0.707) (1.0 m) m Use the equation for the electric force between charges and set the two forces equal to each other. Ê ˆ F A,B (r A,B ) ; and F C,B (cos 45 ) k q C C (r C,B ) Ê ˆ q C (r A,B ) (cos 45 ) k C (r C,B ) Rearrange to solve for : Ê ˆ q C (r A,B ) (cos 45 ) (r C,B ) Substitute values and solve for. Ê ˆ Ê (0.707 m) 1.0 C (0.707) (1.0 m) (0.50 m )( 0.707) 1.0 C ˆ m 0.35 C PTS: 5 DIF: IIIC OBJ:

10 35. ANS: C q r total 1.0 m r 1 r 0.50 m E total N/C Nm /C E 1 E E total N/C N/C E 1 r 1 Rearrange to solve for. E 1 r Ê ˆ N/C 1 ( 0.50 m) C Nm /C q C PTS: 5 DIF: IIIB OBJ: