General Physics – PH 213 Midterm II (Ch 28 - 32) August 15, 2013 • • • • Name:____________ Exam is closed book, closed notes. Use only your note card with 10 formulae on it. Write all work and answers in the space provided. Use color papers for additional space or scratch. Show all your work and explain your reasoning (except for true/false or multiple choices ones) Partial credit will be given. No credit will be given if no work is shown (except for true/false or multiple choices ones) Useful Constants: N ⋅m2 C2 − 31 m e = 9.11× 10 kg C2 N ⋅m2 − 27 m p = 1.67 × 10 kg mili (m) = 10 −3 micro ( µ ) = 10 − 6 k = 9.0 × 10 9 ε 0 = 8.85 × 10 −12 q e = 1.6 × 10 −19 C µ 0 = 4π × 10 − 7 T ⋅m A 1eV = 1.6 × 10 −19 J nano (n) = 10 −9 Part I – (3 pts each)\True or False: For questions 1 – 8, state whether each statement is true or false. You do not need to show any work for T/F questions 1. __ True __ Volts per meter, V/m, are equivalent to newtons per coulomb, N/C. 2. __True___ To produce higher voltages, the cells of a battery are connected in series. 3. __True___ The current in a circuit flows in the opposite direction to the direction in which the electrons move. 4. __True___ If R is resistance and C is capacitance, then the units associated with RC are seconds. 5. __False___ The magnetic field near a current carrying wire is directly proportional to the distance from the wire. (B-field is inversely proportional to distance from the wire) 6. __True___ For any displacement of a charged particle, the work done by the magnetic force on the charged particle moving in a magnetic field is zero J.. 7. __False___ Cutting a bar magnet near its north end results in a smaller mostly north pole magnet and a larger mostly south pole magnet. 8. __True___ Capacitors in series share the same charge and capacitors in parallel share the same voltage. Part II – (3 pts each) Multiple choice questions: For questions 9 – 18 select the one alternative that best completes the statement or answers the question. 9. If a current of 2.0 A is flowing from point a to point b, the potential difference between the points is A) 6 V D) 8 V B) 14 V E) 20 V C) 22 V 10. The electric potential inside a capacitor A) is constant. B) increases linearly from the negative to the positive plate. C) decreases linearly from the negative to the positive plate. D) decreases inversely with distance from the negative plate. E) decreases inversely with the square of the distance from the negative plate. 11. If the voltage on a capacitor is doubled, then the energy stored by the capacitor A) stays the same B) doubles C) quadruples D) halves 12. For the capacitor shown, the potential difference V between points a and b is A) 3 V B) 3 tan30o V C) 3 sin30o V D) 3 cos30o V E) 3/sin30o V F) 3/cos30o V 13. The switch in the circuit is initially in the open position (as shown). After the switch is closed, what happens to the current through the resistor? A) It starts out at zero and gradually builds up to a maximum. B) It suddenly rises to a maximum and gradually decreases to zero. C) It starts out at zero and gradually builds up to a maximum and drops sharply down to zero. V D) It remains at a constant nonzero value. E) It is zero at all times. R C 14. A proton is released from rest at point A where the potential is 0 After release, the proton A) moves toward C with a constant speed. B) moves toward C with an increasing speed. C) moves toward B with a constant speed. D) moves toward B with an increasing speed. E) remains at rest. 15. Three charged metal spheres of different radii are connected by a thin copper wire. Which of the following is true for the electric field and potential at the surface of each sphere A) V1 < V2 < V3 and E1 = E2 = E3 B) V1 > V2 > V3 and E1 = E2 = E3 1 C) V1 = V2 = V3 and E1 = E2 = E3 V1 , E1 D) V1 = V2 = V3 and E1 > E2 > E3 V2 , E2 E) V1 > V2 > V3 and E1 > E2 > E3 V3 , E3 F) V1 < V2 < V3 and E1 > E2 > E3 G) Both potential and electric field are zero at each surface. 16. A moving electron can be accelerated without changing its speed by A) an electric field . D) This situation can not be possible. B) a magnetic field. E) none of these. C) magnetic and electric field. V. 17. When a bar magnet is broken in two pieces as shown, then the magnetization of each piece A) is twice as strong as the original piece. B) is half as strong as the original piece. N S C) is as strong as the original piece. D) no longer exists. E) is a random value and have no relation to one another. 18. A positively charged particle is moving northward in a magnetic field. The magnetic force on the particle is northeast. What is the direction of the magnetic field? A) Up D) South B) Down E) This situation can not exist. C) West 19. A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its south pole pointing down. What is the direction of the induced current in the coil? A) counterclockwise C) There is no current in the coil. B) clockwise D) None of the other answers is correct. 20. In the plane of the paper, a group of equipotential lines are shown at right. At which location will an electron experience a force directed toward the top of the page (along y-axis)? At B. Electric field is pointing in direction of decreasing potential and perpendicular to equipotential surfaces (lines). At location B, electric field is pointing straight down (–y direction). An electron will accelerate in opposite direction of the field. 21. In the circuit shown at right, resistors R1 = R2 = R3. How do power dissipated in each resistor compare? A) P1 = (½)P2 =(½)P3 D) P1 = P2 + P3 B) P1 = P2 = P3 E) P1 = 2P2 + 2P3 C) 4P1 = P2 = P3 22. In the circuit shown at right, the current in the 6.0β¦ resistor is 3A. The currents in other two resistors are: A) I1 = 1A and I2 = 1.5A B) I1 = 6A and I2 = 9A C) I1 = 9A and I2 = 6A D) The answer cannot be obtained without knowing the emf ε of the battery. 23. The phenomenon of magnetism is best understood in terms of A) the existence of magnetic poles. B) the magnetic fields associated with the movement of charged particles. C) gravitational forces between nuclei and orbital electrons. D) electrical fluids. E) None of these is correct. 24. What is the direction of the magnetic field around a wire carrying a current directly into this page? A) The field is parallel to and in the same direction as the current flow. B) It is parallel to but directed opposite to the current flow. C) It is counterclockwise around the wire in the plane of the page. D) It is clockwise around the wire in the plane of the page. E) None of these is correct. 25. An electron traveling horizontally enters a region where a uniform magnetic field is directed into the plane of the paper as shown. Which one of the following phrases most accurately describes the motion of the electron once it has entered the field? A) upward and parabolic B) upward and circular C) downward and circular D) upward, along a straight line E) downward and parabolic Part III - Questions and Problems Please explain or show your work clearly and completelyfor each of the following questions or problem. 26. (10 pts) The graph of electric potential versus position in the x-direction for a region is shown at right. Six positions in this region are labeled. A) How do you calculate the electric field if potential function is given? οΏ½βπ, in one dimension this equation Recall πΈοΏ½β = −∇ dV will be: πΈοΏ½β = − dx π€Μ. From the graph the slope of dV the curve provides the dx and appling the slope dV values to πΈοΏ½β = − dx π€Μ will show E-field. B) Rank the electric field at the labeled positions from largest (greatest positive) to smallest (greatest negative), using successive > sign or = sign where applies Again First calculate slopes at points A – F, and then multiply the values by –1 and the rank the result. ED > EC = EE > EF > EB = EA 27. (5 pts)Rank in order, from largest to smallest, the energies U1 to U4 stored in each capacitor shown 1 below. Recall: ππΆ = 2 πΆπ 2 < 1 π1 = 2 πΆπ 2 Ranking: 1 1 π2 = 2 (4πΆ)(3 π)2 2 π2 = 9 πΆπ 2 U4 > U1 > U3 > U2 1 1 π3 = 2 (2πΆ)(2 π)2 1 π3 = 4 πΆπ 2 1 1 π4 = 2 οΏ½2 πΆοΏ½ (2π)2 π4 = πΆπ 2 28. (15 pts)Two bulb and a capacitor are connected to a battery as shown. Immediately after the switch is closed: A) Rank the currents IA , IB , IC , and Ibat. Explain your reasoning. At t = 0 (immediately after the switched is closed) capacitor is empty and acts like a short (meaning all current goes through capacitor and almost none in bulb B. Ibat = IA = IC > IB ≈ 0 B) A long time after the switch has been closed: Rank the currents IA , IB , IC , and Ibat. Explain your reasoning. Long time after switch is closed (t = = ∞), capacitor is fully charged and acts like open switch (no current will pass throw it). All current goes throw IB. Ibat = IA = IB > IC ≈ 0 C) A long time after the switch has been closed: Rank the potential difference across bulb A (VA), bulb B (VB), across the capacitor (VC), and the battery (Vbat)? Explain. Voltage of battery is divided on each balb equally and since capacitor is in parallel to bulb B, hence will have same voltage as bulb B. Vbat > VA = VB = VB 29. (10 pts) Consider two bar magnets placed at right angles to each other, as shown. A) If a small compass is placed at point P, what direction does the painted end of the compass needle point? Show the vector diagram. Outside of a magnet, magnetic field lines are leaving north pole and enter south pole. The compass needle will be diffelected in the direction of the net B-field. Assuming both Magnets are identical, the deflection of compass will be 45o clockwise. B) If the compass needle instead pointed 15 degrees clockwise of where you predicted in A), what could you qualitatively conclude about the relative strengths of the two magnets? From the vector addition of the fields of magnet a and b we can write π©π tan 30π = = 0.58 π©π Therefore magnet b is weaker than magnet a. Magnet b is about 58% as strong as magnet a. 30. (10 pts) An electron is initially traveling at 3.0 x 106 m/s to the left at point A when it enters a uniform electric field. The equipotential surfaces associated with the field is shown at right. The equipotential surfaces are 25.0 cm apart from each other A) Is the speed of electron increased, decreased or remains constant when enters the electric field? Electric field lines are perpendicular to equipotential surfaces and in direction of decreasing potentials as shown. An electron entering field is accelerated in opposite of field direction. Since electron had an initial velocity to the left (opposite of field direction), it will be accelerated by the field (speed increases). B) What is the velocity of the electron after it has traveled 10.0 cm in the field? βπΎπΈ = βπ = πβπ Let’s assume the equipotential surfaces are separated by distance of 25 cm. therefore: 0 + 50π π0 − π−50 οΏ½ β 10ππ = = +20π βπ = οΏ½ 25ππ 2.5 1 1 ππ£22 − ππ£12 = πβπ 2 2 1 π 2 1 (9.11 × 10−31 ππ)π£22 − (9.11 × 10−31 ππ) οΏ½3.0 × 106 οΏ½ = 1.6 × 10−19 πΆ(+20π) 2 π 2 v2 =4.0×106 m/s 31. (10 pts) Use the circuit at right to answer the following: A) What is the resistance between points a and b for the resistor combination shown? Show the new equivalent circuit after every resistors reduction. The three resistors at the end are in series and will add up to 12β¦. This 12β¦ and other 12β¦ already in the circuit are in parallel and combine to make and equivalent of 6β¦. The pattern repeats in similar way. The net resistance is 11β¦. n(see the sequence of circuit reduction) B) If a 6.0V battery is placed between point a and b, what will be current throw resistor R1 =2 β¦ The current through the 2β¦ is the total current that is drawn from the battery which is determined by the net resistance (11β¦) in circuit. πΌ = π π ππ = 6π 11 πβπ = 0.55π΄ 32. (10 pts) Consider the circuit shown at right. Let I1 = 2.0A, I2 = 4.0A, and I3 = 5.0A. A) Calculate the current values of I4 and I5. Apply the junction rule at point b: I1 + I2 = I3 + I4 ο I4 = 2.0A + 4.0A – 5.0A ο I4 = 1.0A Now apply the junction rule at point e: I1 + I5 = I3 ο I5 = 5.0A – 2.0A ο I3 = 3.0A B) If ε1 = 20V and Let R1= 5.0β¦, find the value of R3 . Applying the loop rule for the loop abe while going CW: ε1 – I1R1 – I3R3 = 0 20 + 20V – 2.0A . 5.0 β¦ = 5.0A R3 ο R3 = 2.0 β¦ 33. (20 pts)Suppose in the circuit shown the switch has been closed long enough for the capacitor to be fully charged. A) Find the current (steady) in each resistor. A fully charged capacitor acts like an open switch, hence no current flows in capacitor branch. All currents goes through 6β¦ and 12β¦ resistors, Since they are in series, the current through each is the same. 9.0π πΌ6πΎ = πΌ12πΎ = = 0.50 ππ΄ 6kΩ + 12kΩ B) C) D) Find he charge on the capacitor. Q = CV where the voltage across capacitor is the same as that of across 12.0β¦ resistor. ππΆ = π12πΎ = 0.50ππ΄ β 12000Ω = 6.0V π = 12.0ππΉ β 6.0π = 72ππΆ Now switch is opened at t = 0. What is the time constant? When the switch is open the battery and the 6.0kβ¦ resistor are out of the circuit. The capacitor will discharge into the 12.0kβ¦ resistor and 4.0kβ¦ resistor that are in series (i.e. only the right loop will be energized.) Time constant τ = ReqC =(4.0kβ¦ +12.0kβ¦)(12.0µF) ο τ = 0.192 s = 192 ms How long will it take for the charge on capacitor to fall to one-fourth of its original charge? π‘ π(π‘) = π0 π −π ο 1 π‘ π = π0 π −0.192 π 4 0 ο 1 π‘ ln οΏ½4οΏ½ = − 0.192π ο t = 0.27 sec 34. (15 pts)A long rigid conductor lying along the x-axis carries a current of 5.0A in the –x direction. A magnetic field B = 2i + 7x2j is present, where x is in meters and B in mT. Calculate the force on the 1.6-m segment of the conductor that lies between x = 1.4 m and x = 3.0 m. οΏ½β where dl is along the negative x-axis (same as direction of current) and can be πΉβ = ∫ πΌπποΏ½β × π΅ replaced by dx π€Μ οΏ½π₯ οΏ½β = πΌ οΏ½−ππ₯ 0 πΌπποΏ½β × π΅ 2 7π₯ 2 ποΏ½ 2 0 οΏ½ = −7πΌπ₯ ππ₯ ποΏ½ 0 3.0 πΉβ = οΏ½ οΏ½ −7πΌ π₯ 2 ππ₯οΏ½ ποΏ½ = (−7(5.0π΄) 1.4 3 π₯3 οΏ½ )ποΏ½ = −283 π ππ ποΏ½ ππππππ‘πππ 3 1.4 35. (20 pts)A thin rod of length 2L is centered on the x axis as shown. The rod carries a uniformly distributed charge Q distributed along its length 2L with linear charge density λ = Q/2L. A) Determine the electric potential for any point x > L along the x axis. The electric potential at point P due small charge dq a distance x away is given by: ππ ππ = π (π₯−π) where dq = λ dX Total potential due to entire rod is: +πΏ π= οΏ½ π −πΏ Recall λ = Q/2L. ο πππ π₯−π π = −ππ ln(π₯ − π)|+πΏ −πΏ π (π₯ ) = ππ 2πΏ ln π₯+πΏ π₯−πΏ οΏ½βπ to determine the electric field due to the charged rod at a point x > L B) Use the fact that πΈοΏ½β = −∇ along x axis. οΏ½β = − ππ π€Μ οΏ½βπ reduces to πΈ In one dimension πΈοΏ½β = −∇ ππ₯ πΈοΏ½β = − π ππ π₯ + πΏ οΏ½ π€Μ πΈοΏ½β = − οΏ½ ln ππ₯ 2πΏ π₯ − πΏ 1 1(π₯ − πΏ ) − 1(π₯ + πΏ ) ππ οΏ½ οΏ½οΏ½ οΏ½ π€Μ (π₯ − πΏ ) 2 2πΏ π₯ + πΏ π₯−πΏ ππ πΈοΏ½β = 2 π€Μ π₯ − πΏ2 C) How much work is done to move a particle of mass m and charge +q from the point x = 3L to the point x =2L? The work done to move a particle of charge q from one point to another is just the potential difference between the points multiplied q by definition of potential. Thus we have π = −πΔπ = −(+π)(ππ₯=3πΏ − ππ₯=2πΏ ) π = −π( πππ ππ 3πΏ + πΏ ππ 2πΏ + πΏ ln − ln )= (ln 3 − ln 2) 2πΏ 2πΏ 3πΏ − πΏ 2πΏ 2πΏ − πΏ π= πππ 3 ln 2πΏ 2