Physics 6B. Practice Final Solutions

Transcription

1 Physics 6B Practice Final Solutions

2 . Two speakers placed 4m apart produce sound waves with frequency 45Hz. A listener is standing m in front of the left speaker. Describe the sound that he hears. Assume the speed of sound is 40 m/s. First find the distance from the right speaker (use the pythagorean theorem or recognize the -4-5 right triangle). 4m Now find the difference in the distances to each speaker: (5m m) = m m 5m We need to compare this to the wavelength of the sound. v f 40 m s 45Hz 0.8m m 0.8m.5 This means the difference is.5 wavelengths, yielding destructive interference (a quieter sound)

3 . Charge q = -5.4µC is placed at the origin, and charge q = -. µc is on the x-axis at x=m. Where should a charge q be placed between q and q so that the net force acting on it is zero? Here is a diagram, with q in between. We just need to find the forces on q due to the other charges, and make them equal. q F(q on q) q F(q on q) q We are not given the sign of charge q, but it shouldn t matter. The forces due to q and q will be in opposite directions in either case. F F F elec q q q kqq' r kqq q x kqq x x -x Setting these equal, and plugging in the given values: kq q x kq q q q 5.4. x x x x x x.48 x. x.8x. x 0. m. x 5.4 6

4 . An object with a charge of.6µc and a mass of g experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. Find the direction and magnitude of the electric field. Electric force is simply charge times electric field. F elec E qe mg E 0.0kg9.8 m C s mg q,000 N C F elec This is the magnitude of the field. The direction must be downward to give an upward force on a negative charge. One way to remember this is that the direction of the field is always the direction of the force on a positive charge. Since our charge is negative, it must be the opposite direction. mg

5 4. Two conducting spheres have net charges q=+8 µc and q=- µc. The spheres touch and some charge is transferred. How many electrons are transferred, and to which sphere? BEFORE AFTER q +8 µc q - µc q + µc q + µc When the spheres touch, charges are transferred until both spheres have equal charge. The total charge before they touch is +6µC, so the charge on each sphere after they touch must be +µc. So there must be a total of 5µC of charge transferred. We need to divide by the charge on the electron to find the # of electrons transferred: C C electron.0 electrons Since sphere gets less positive, it must be gaining the negative electrons. Conversely, sphere is getting more positive, so it is losing electrons. Thus the electrons must be transferred to sphere.

6 5. During a lightning strike, electrons are transferred from the bottom of a thundercloud to the ground. During this process, the electrons: a) gain potential energy as they move toward a higher potential b) lose potential energy as they move toward a lower potential c) gain potential energy as they move toward a lower potential d) lose potential energy as they move toward a higher potential Since the electron is moving downward, the ground must be positively charged, and the bottom of the cloud must be negative. So the electron is moving toward a higher (positive) potential. Because its own charge is negative, however, the potential energy of the electron is getting more negative as it moves toward the ground. So it is losing potential energy e

7 6. During a lightning strike, electrons are transferred from the bottom of a thundercloud to the ground. This occurs due to dielectric breakdown of the air, when the electric field is greater than x0 6 V/m. The distance from the ground to the cloud is 000m. Find the magnitude of the potential difference between the cloud and the ground. Assuming a constant electric field, we can use the formula V=Ed V 6 0 V 000m 0 9 Volts m m

8 7. An electric dipole consists of two equal charges, +q and q, a distance d apart. Find the total electric potential at a point that is a distance of d/ to the right of the positive charge, as shown. kq The electric potential due to a point charge is given by: r In this formula we include the sign of the charge to tell us the sign of the potential. All we need to do is find the potential due to each charge, and then add them together. left right total k k k q d q d q kq d d kq d kq d 4 kq d -q +q d d/

9 8. How much charge is on each plate of the capacitors in the circuit shown? The battery has voltage V, and the capacitances are: C =µf, C =µf, C =4µF. First we need to find the equivalent capacitance for the circuit. C and C are in parallel, so we just add them together. Next we see that C is in series with the (C +C ) combination, so we need to use the reciprocal formula. C eq C C C C 4 eq F V C C The total charge for the system is given by Q=CV. Q total =4µC Now we need to work backward to find the charge on each capacitor. First look at capacitor C. There are no parallel paths connected to it, so all the charge must land on its plates. Q =4µC V C Next we notice that since the other capacitors are in parallel, the charge must split up so that the total adds up to 4µC. C +C C Option Option Since C has half the capacitance of C, it will only get half the charge. So we need Q =8µF and Q =6µF. This adds up to 4, and Q is half as much as Q. The voltage across C and C must be equal because they are in parallel. Using V=Q/C we get V =8volts. This leaves 4volts for the others. Now using Q=CV again we get Q =8 and Q =6 V C eq =µf

10 9. A parallel-plate capacitor is initially charged by a battery with voltage V. The battery is disconnected, and a dielectric with constant k is inserted between the plates. What happens to the energy stored in the capacitor? When the battery is disconnected, the charges that have built up on the plates have nowhere to go, so the charge on the capacitor will remain constant when the dielectric material is inserted. To calculate the amount of charge we could use the formula Q=CV. Now what happens when the dielectric is inserted? The capacitance is increased by a factor of k (the dielectric constant). So what does that do to the energy? We have versions of the potential energy formula for capacitors: Let s use version (). We already know that the charge (Q) is not going to change. Since the capacitance is increased by a factor of k, the energy must decrease by that factor (the C is in the denominator of the formula). () U () U () U cap cap cap CV QV Q C So the answer is c) The stored energy decreases by a factor of k Note: if the battery had remained connected, the energy would have increased by a factor of k instead. Think about why this happens.

11 0. How much power is dissipated in each of the resistors in the circuit shown? The battery has voltage V, and the resistances are: R=4Ω, R=Ω, R=6Ω. First we need to find the equivalent resistance for the circuit. R and R are in parallel, so we use the reciprocal trick. V R R eq Req R R 6 Next we see that R is in series with the (R and R ) combination, so we just add them together, For a total of 6Ω. V V Now we can use Ohm s law to find the current: I A R 6 All of this current must go through R, since there is no other path parallel to it. We can use the formula for power dissipated by a resistor: P=I R = IV = V /R. In this case the first one works fine: P A 4 6 W V R R The current will split when it gets to the R /R parallel combination: Option Option Ω R Since R has half the resistance of R, it will get twice the current. So we get. I 4 A and I A This adds up to, and I is half as much as I. Now use P=I R again: 4 6 P W 5.W P 6 8 W.7W The voltage across R and R must be equal because they are in parallel. Using V=IR we get V =8volts. This leaves 4volts for the others. Now using P=V /R we get: P P 4V 4V W 5.W W.7W V 6Ω

12 . In the circuit below all the bulbs have the same resistance. When the switch is closed, what happens to bulb A? First realize that since we assume that the resistance stays constant, we know that bulbs get brighter when the current increases. A B C Next, see from the diagram that all the current must go through bulb A. So now think about what happens when the switch is closed. The path through bulb B is closed, creating a parallel path to bulb C. This has the effect of reducing the equivalent resistance (adding a parallel path will always reduce the overall resistance). Since the resistance is reduced, the current must increase. (V=I/R) So the current through bulb A increases, and bulb A gets brighter. Bonus question what happens to bulb C?

13 . Resistor is a solid cylinder with length L and diameter D. Resistor is made of the same material, but it has length L and diameter D. Compared to the resistance of resistor, resistance is: Use the formula for resistance: R L A The resistivity is the same for each resistor. The length is twice as large. The area is 4 times as big, because area depends on the diameter squared Now compare the resistances: R R L A L 4A L A So the answer is b) half as large R

14 . A coil has 000 turns of wire and a cross-sectional area of 0.5m. A uniform magnetic field of magnitude Teslas is directed along the axis of the coil. Find the voltage induced in the coil as the strength of the magnetic field decreases steadily to zero in a time of 4 seconds. There will be an induced EMF (voltage) whenever there is a changing magnetic flux through the coil. The B-field is decreasing in strength, and the total change is Teslas in a time of 4 seconds. Now we can use our formula for induced voltage (N=000 turns of wire): B A T 0.5m V induced N N 50 t t 4s 000 V Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

15 4. Two singly ionized isotopes of uranium (one electron has been removed) are projected at v=.5x0 5 m/s into a region with a uniform magnetic field of strength 0.75T directed into the page, as shown. The particles land at distances d =68.cm and d =69.cm from the entry point. Find the difference between the masses of the particles. Each charge feels a magnetic force directed toward the center of the circular path. Setting centripetal force equal to magnetic force: B mv r qvb m qbr v The charge, magnetic field, and velocity are the same for both charges. Only the radius is different. Notice that the distance given are diameters, so we must divide them by. m m m m qb v C 0.75T r r 7 kg.50 5 m s 0.69m 0.68m v

16 5. Consider the three electric charges shown below. Charge B is equidistant from charges A and C. List the charges in order of the magnitude of the force they experience, from smallest to largest. Each charge will feel forces from the other charges. Remember same sign repels, opposites attract. F F F BonA AonB AonC kqq ; F r ConA kqq ; F r kqq ; F ConB kqq r F kqq F r kqq F net, A net, B 5 4 kqq BonC net, C r r 4 r r kqq r kqq -q A r +q +q B r C

17 6. A beam of electrons is passing through a region with a uniform magnetic field directed downward. If the electrons are initially moving East, which direction are they deflected when they enter the field? Use a right-hand-rule for this. Start with fingers along the velocity, then bend them into the page (toward the magnetic field). The thumb gives the direction of the force on a positive charge. B In this case we get the thumb pointing Northward. However, our charge is an electron, so the force is in the opposite direction. You can flip your hand over, or use your left hand if you want. e - F mag v Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18 7. A circuit is arranged with a sliding wire that can move vertically, as shown. There is a uniform magnetic field directed out of the page. The sliding wire is released from rest in the position shown. As the wire falls: a) induced current is clockwise and the bulb gets brighter until it reaches a steady intensity. b) induced current is clockwise and the bulb glows with a constant brightness. c) induced current is counter-clockwise and the bulb gets brighter until it reaches a steady intensity. d) induced current is counter-clockwise and the bulb glows with a constant brightness. First note that there is magnetic flux through the loop, directed out of the page. As the wire slides down, the loop gets smaller, so the amount of flux decreases. v The reaction to this decreasing flux is to induce current in the wire in such a way that magnetic flux is created to oppose the change out of the page. A right-hand rule will tell us that the induced current flows counter-clockwise. I induced B The wire is accelerating downward, so the flux is decreasing faster and faster as the wire falls. This means the induced current must get stronger and stronger, making the bulb glow brighter and brighter. Eventually the induced current is strong enough to produce enough magnetic force to balance the weight of the wire, and the wire falls at a steady speed. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

19 8. A metal ring is falling toward a wire with a steady current flowing to the right. What direction is the induced current in the ring? The magnetic field makes concentric circles around the wire, so there is some magnetic flux through the ring, out of the page. I induced v As the ring falls, the flux increases because the field is stronger closer to the wire. The induced current will try to oppose this change. I To oppose the change, the induced current must create magnetic flux through the ring, into the page. Using a right-hand-rule, this means the induced current must be clockwise, as shown.

20 9. A metal ring is falling along a wire with steady current flowing upward. What direction is the induced current in the ring, as viewed from above? The magnetic field makes concentric circles around the wire, so there is no magnetic flux through the ring (as long as the ring is flat, as shown). With no flux, there is no change in flux, thus no induced current. I

21 0. In Europe, the standard voltage available from a wall socket is 40V rms. What is the maximum voltage in this case? The formula for RMS voltage is V rms V max So the maximum voltage is 40V V V 40 max