1 of 11 Concept Tests -- Final Review CTF-1. Calculator question: 105 = ? A) 1E5 B) 10E5 C) something else. Answer: 1E5 means 1×105 = 105 CTF-2. A cube is made larger and larger... As the cube grows, the ratio areavolume.. A) increases B) decreases C) remains constant. Answer: 23area 6L 6volume L L== decreases with increasing L Phys1120 Dubson ©University of Colorado at Boulder2 of 11 CTF-3. A box with mass M and charge +Q is pushed along a rough surface at constant velocity by a uniform horizontal electric field of magnitude E. The coefficient of kinetic friction between the box and the floor is µ. What is the net force on the box? A) QE B) µ Μ g C) QE – Mg D) QE – Mg + FN – µ Mg . E) None of these M, Q v = constant E µ Answer: the magnitude of the net force must be zero. This is actually answer D! FN = Mg and QE = µ FN = µ Mg, so the sum is zero. But that was a dirty trick so I'll accept answer E as correct, if you thought that the answer is zero. CTF-4. A row of positive charges is stationary on the ground. A person with a gauss-meter (which measures the magnetic field) is running to the right along the row of charges, at the same height as the charges and in front of them (in the diagram below). Does the person measure a non-zero B-field? A) Yes B) No v What is the direction of the B-field which the moving observer measures? A) Up↑ B) Down↓ C) Forward direction → D) backward direction ← E) no direction, B-field zero Answers: In the frame of reference of the person, the charges are moving to the left, which constitutes a current to the left. There is a B-field in the person's frame of reference and its direction is up (at the location of the person). Phys1120 Dubson ©University of Colorado at Boulder3 of 11 CTF-5. A charge q is located at the center of an gaussian cubical box as shown. A student is asked: What is the electric flux ΦE though the top face of the cube? Is the answer hard to calculate or easy to calculate? A: Easy B: Hard. q Answer: It's easy if you remember Gauss’s LawinsideoQEdAε⋅=∫GGv. The total flux through the entire box is q/εo. By symmetry, the flux through one face is 1/6 th the flux through the whole box. Flux = (1/6) q/εo. CTF-6. A capacitor with capacitance C is attached to a battery with voltage V. What is the flux GGEda⋅z through the cubical volume shown? (The end faces of the cube are within the metal plates of the capacitor. V C A) CVoε B) 2CVoε C) zero D) None of these. Answer: Zero. In electrostatic situations, the E-field inside a conductor is zero. Since E=0 on the right and left ends of the cubical (imaginary) box, the flux through those ends is zero. On the sides of the box, GGEda⋅=0, since GGEda⊥. So the total flux is zero. Phys1120 Dubson ©University of Colorado at Boulder4 of 11 Since GGEda⋅=z0, then by Gauss's Law, the total charge must be zero within the box must be zero, so we have proven that the two plates must have equal and opposite charge densities. CTF-7. Points A and B are distances r and 3r respectively from a point charge q. What is the voltage difference between points A and B? A)23kqr B) 3kqr C) 89kqr D) 34kqr E) None q r 2r A B Answer: Voltage due to a point charge is kq/r. The voltage difference between points A and B is (kq/r – kq/(3r)) = 2kq/3r. CTF-8. Consider a point in empty space near several charges, which might be positive, negative, or both. Consider the following statements. I. The E-field can be zero while the voltage is non-zero. II. The voltage can be zero while the E-field is non-zero. Which of these statements can be true? A) both can be true B) neither can be true C) only I can be true D) only II can by true Answer: Both can be true. Phys1120 Dubson ©University of Colorado at Boulder5 of 11 CTF-9. A capacitor has a voltage V across its plates. An electron, initially at rest, is released from at a point very close to the negative plate of a capacitor and it accelerates toward the positive plate. The electron has charge –e and mass m. There is no gravity in this problem. What is the final kinetic energy of the electron just before it collides with the positive plate? VA) B) 2eV m e V⋅⋅ C) (1/2)mV2 D) eV E) None of these Answer: |∆KE| = |∆PE| = |q∆V| = eV CTF-10. A resistor with resistance R is plugged into a 120VAC wall socket. The graph below is either resistance R, voltage V across, current I thru, or power P dissipated in the resistor vs. time. time ? What could the graph be? A) V only B) I only C) V or I only D) V, I, or P E) Some other combination Answer: The graph could be V or I only. Power is always positive. Phys1120 Dubson ©University of Colorado at Boulder time ? P I or V6 of 11 CTF-11. Consider the circuit shown, with the switch initially open. When the switch is closed, the current I1 through resistor R1 VR1R2I1IbatA) increases, B) decreases, C) stays the same. When the switch is closed, the current from the battery Ibat A) increases, B) decreases, C) stays the same. Answers: I1 stays the same. I1 = V/R1 regardless of whether or not the switch is closed. The battery current increases when the switch is closed. CTF-12. When a resistor is plugged into a standard AC outlet, the fuse blows and all the lights go out. If we want to repeat this dangerous experiment, and not have the fuse blow, we need a resistor that is A) larger B) smaller C) Same R, but larger power rating Answer: larger. At constant voltage (here V = 120 VAC), the power is P = V2/R. Larger resistance draws less power. CTF-13. A charge q is released from rest at point in empty space were there may be a static E- and/or a static B-fields. There are no forces on the charge except for the forces due to the E and/or B-fields (no gravity, etc.). The charge is observed for a short while and is seen to move along a curved path. q Consider the following possibilities I. There is only an E-field present and no B-field. II. There is only a B-field present and no E-field. III. There is both an E-field and a B-field present. Which possibilities could account for the observed motion? A) all three B) I and III only C) II and III D) III only Phys1120 Dubson ©University of Colorado at Boulder7 of 11 The correct answer is: There might be only a
View Full Document
Unlocking...