1 of 7 Exam II review. RII-1. How many ohms in a coulomb? A) one gazillion B) one coullion C) 17 D) None of these Answer: None of these. Ohm is unit of resistance, coulomb is unit of charge. RII-2. Consider a positive point charge +Q just outside an imaginary sphere, as shown. How does the magnitude of the electric flux thru the left half of the sphere LeftEda⋅∫KKcompare to the magnitude of the electric flux thru the right halfRightEda⋅∫KK: A) |Flux Left| > |Flux Right| B) |Flux Left| < |Flux Right| C) |Flux Left| = |Flux Right| Answer: This is a tough one! |Flux Left| = |Flux Right|. Some of the flux lines that enter the right half, also exit the right half, so those lines do not contribute to the flux through the right half. All of the lines which exit the left half also enter the right half, and those are the only lines that contribute to the flux on the right half. +Q +Q does not contribute to flux thru right half 10/12/2009, Phys1120 M.Dubson ©University of Colorado at Boulder2 of 7 RII-3. Consider a point charge +Q off-center within a spherical metal shell. Does Gauss's Law allow you to compute the total charge on the inside surface of the shell? A: Yes B: No Answer: Yes, you can compute the total charge on the inside surface. Consider the flux over a spherical surface within the metal. Since the E-field must be zero everywhere within the conductor, then Since Qenc = 0, there must be a charge -Q spread over the inside surface of the conductor. Since the point charge +Q is off-center, the surface -Q will not be spread uniformly over the inside surface. SEda⋅∫GGvenc oSEda Q / 0.⋅= ε=∫GGv+Q RII-4. A point charge +q is brought from infinity to a point b near 3 other charges +Q,–Q, and +2Q. The charge q is brought along 3 different paths in turn, path 1, path 2, and path 3, as shown. Along which path is the most work done by the external agent carrying the charge +q? A) 1 B) 2 C) 3 D) Same work on all three paths +2Q +Q Answer: Same work on all paths. W = ∆PE = q ∆V=q(Vb – V∞) = qVb. RII-5. TRUE (A) or FALSE (B) ? If the E-field throughout a region of space is zero, the voltage throughout that region must be zero. –Q b path 1 path 2 path 3 10/12/2009, Phys1120 M.Dubson ©University of Colorado at Boulder3 of 7 Answer: False. If E = 0, that means that the voltage is not changing with position. The voltage can have any non-zero constant value. RII-6. TRUE (A) or FALSE (B) ? If the voltage throughout a region of space is zero, the E-field throughout that region must be zero. Answer: True. E-field is rate of change of voltage with position. |E| = |∆V/∆x|. If V = constant (zero or any constant), then |E| = |∆V/∆x| = 0. RII-7. Does the capacitance of a capacitor depend on the charge Q on the capacitor? A) Yes B) No C) Depends on the type of capacitor. Answer: No. The capacitance of a capacitor is fixed by its size and shape. C = Q / V is a constant ratio for a given capacitor: if Q increases, V increases so that the ration Q / V remains constant. RII-8. A capacitor is attached to a battery which maintains a constant voltage V across the capacitor plates. While the battery is attached, the plates are moved further apart. The energy stored in the capacitor.. V A) increased. B) decreased. C) remained constant. Hint: U = (1/2)QV = (1/2)Q2/C = (1/2) CV2 Answer: V is fixed by the battery. Q is not fixed, since charge can go on or off the plates through the attached wires. Use U = (1/2) CV2. C decreases, V constant, so U decreases. (Both the battery and the external agent do work, so work arguments are not easy to apply in this case.) 10/12/2009, Phys1120 M.Dubson ©University of Colorado at Boulder4 of 7 RII-9. A charged capacitor is isolated (so no charge can get on or off). The plates of the capacitor are slowly pulled apart. After the plates are pulled apart a bit, the electric field between the plates A) increased B) decreased C) remained constant The voltage difference between the plates.. A) increased B) decreased C) remained constant The capacitance .. A) increased B) decreased C) remained constant Answers: The E-field stays constant, because E due to a large plane of charge is constant, independent of distance from plane. The voltage difference increases since ∆V = E d (E is constant, d increases) The capacitance decreased. We can see this either from C = Q / V (Q stayed constant, V increases) or from C = εο A / d (d increased) RII-10. A copper wire and an aluminum wire both have the same dimensions (same length, same cross-sectional area). Which one statement is true? A) Both wires have the same resistance R and the same resistivity ρ . B) The wires have different R and different ρ. C) The wires have the same R but different ρ. D) The wires have different ρ but the same R. Answer: The wires have different resistances and different resistivities. R = ρ L / A Different materials have different resistivity ρ . Same L, same A, different ρ's, so different R's RII-11. Which has higher resistance ? A) 100 W bulb B) 60 W bulb C) Same R in both. Hint: P = IV = I2R = V2/R Answer: 60 W bulb has higher R. All light bulbs are meant to be used at constant voltage V = 120 VAC. So use P = V2 / R. 10/12/2009, Phys1120 M.Dubson ©University of Colorado at Boulder5 of 7 RII-12. The voltage between points a and b in the circuit shown is measured with an ideal voltmeter. What does the voltmeter read? V R a b R R Hint: Recall that an ideal voltmeter has infinite internal resistance. A) V/2 B) V/3 C) V/4 D) zero E) V Answer: A tough one! The answer is V/2. The current thru the voltmeter is (nearly) zero [since it has (nearly) infinite resistance]. So no current flows through the R in the upper right. K's voltage law says the voltage V is split evenly between the two equal resistances R. RII-13. Consider the circuit below. What is the total resistance which the battery sees? A) 1 Ω R2 =2Ω 1Ω R1 =2Ω 2Ω 1V I=? B) 2 Ω C) 0.5 Ω D) 0.25 Ω In the circuit above, the resistor R2 is increased to 5Ω. What happens to the current through from the battery? A) increases B) decreases C) remains constant Answers: The total resistance is 1 ohm. When any of the R's is increased, the battery current decreases, since the total resistance seen by the battery
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