PHYS 202 (DDE) Solved problems in electricity 1 1. Common static electricity involves charges ranging from nanocoulombs to microcoulombs. How many electrons are needed to form a charge of -2.00 nc? How many electrons must be removed from a neutral object to leave a net charge of 0.500 µc? 2.00 10 9 C 1.6 10 19 C/e = 1.25 1010 electrons 0.500 10 6 C 1.6 10 19 C/e = 3.13 1012 electrons 2. An amoeba has 1.00 10 16 protons and a net charge of 0.300 pc. How many fewer electrons are there than protons? If you paired them up, what fraction of the protons would have no electrons? 0.300 10 12 C 1.6 10 19 C/e = 1.88 106 fewer electrons 1.88 10 6 16 = 1.88 10 10 1.00 10 3. What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of -30.0 nc? F = k q 1q 2 r 2 = (9.0 10 9 N-m 2 /C 2 ) (-30 10 9 C) 2 (0.080 m) 2 = 1.27 10 3 N 4. How far apart must two point charges of 75.0 nc (typical of static electricity) be to have a force of 1.00 N between them? so F = k q 1q 2 r 2 r = kq1 q 2 F = 9.0 10 9 N-m 2 /C 2 (7.5 10 8 C) 1.0 N = 7.1 mm
PHYS 202 (DDE) Solved problems in electricity 2 5. By what factor must you change the distance between two point charges to change the force between them by a factor of 10? Force is proportional to 1/r 2. Therefore to change the force by a factor of 10, you must change the quantity r 2 by a factor of 10. So you must change r by a factor of 10 or 3.16. 6. A simple and common technique for accelerating electrons is shown in the figure, where a uniform electric field is created between two plates. Electrons are released, usually from a hot filament, near the negative plate, and there is a small hole in the positive plate that allows the electrons to continue moving. Calculate the acceleration of the electron if the field strength is 2.50 10 4 N/C. Explain why the electron will not be pulled back to the positive plate once it moves through the hole. F = ma so a = F/m, and F = Eq. a = Eq m = (2.50 104 N/C)(1.6 10 19 C) 9.1 10 31 kg = 4.4 10 15 m/s 2 e The electric field outside the plates is small. Since each plate is charged with equal numbers of charges (but opposite signs), all of the field lines starting at the positive plate will terminate at the negative charges on the other plate. They do not pass through the plates. 7. The earth has a net charge that produces an electric field of approximately 150 N/C downward at its surface. What is the magnitude and sign of the excess charge, noting the electric field of a conducting sphere is equivalent to a point charge at its center? What acceleration will the field produce on a free electron near the earth s surface? (c) What mass object with a single extra electron will have its weight supported by this field? E = kq/r 2 so q = Er 2 /k. q = (150 N/C)(6.4 106 m) 2 9.0 10 9 N-m 2 /C 2 = 6.83 10 5 C, negative charge Since F = ma, a = F m = Eq m
PHYS 202 (DDE) Solved problems in electricity 3 = (150 N/C)(1.6 10 19 C) 9.1 10 31 kg = 2.6 10 13 m/s 2 (c) Drawing a free-body diagram, the upward electrical force, Eq is equal to the downward gravitational force, mg: Eq = mg m = Eq g = (150 N/C)(1.6 10 19 C) 9.8 m/s 2 = 2.44 10 18 kg This is a mass about equal to 1.5 10 9 protons, or a volume of water 1.3 µm on a side: about as big as a bacterium. 8. Two identical 0.500 g insulating balls with equal charges hang on 20.0 cm long string as shown in the figure. Find their charge. Draw a free-body diagram: 10 T F e T = tension = electric force F e L = 20 cm We know mg 10 10 ΣF y = 0 so T cos 10 = mg and q q ΣF x = 0 so T sin 10 = F e = k q2 r 2 Divide the bottom equation by the top one, and remember that sine/cosine = tangent: r
PHYS 202 (DDE) Solved problems in electricity 4 T sin 10 T cos 10 = kq2 /r 2 mg q 2 = mgr2 tan10 k mg q = r tan 10 k But from the diagram (above right), r = 2(20 cm)sin 10 = 6.95 cm. Then q = (0.0695 m) (5.0 10 4 kg)(9.8 m/s 2 ) 9.0 10 9 N m 2 /C 2 tan 10 = 2.15 10 8 C 9. If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain. Yes, a test charge can be moved between them with zero net work being done. However, this cannot in general be done without exerting a force. For example, when you pick a book up from a table and transfer it to another location on the table, the net work done on the book is zero. But certainly a force was exerted; the potential energy of the book went up when it was raised and back down when it was lowered onto the table. In the same way, forces might have to be exerted to overcome electric fields present between the two points and to increase the electrical potential energy of the charge temporarily between the initial and final location. 10. What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy? The potential difference is the change in electrical potential energy per unit charge. The unit for potential energy is J/C (= volt) while that for electric potential energy is just J. 11. Car batteries are rated in ampere-hours (A h). To what physical quantity do amperehours correspond, and what relationship do ampere-hours have to energy content? Amp-hours correspond to charge, since A = C/s; when you multipy C/s by time you get some number of coulombs. The energy content of a battery is its voltage multiplied
PHYS 202 (DDE) Solved problems in electricity 5 by its charge capacity. (To get joules, however, you have to use a charge capacity in coulombs.) 12. Why do incandescent light bulbs grow dim late in their lives, particularly just before their filaments break? As a light bulb ages, the metal in the filament evaporates. That is why the glass of the bulb ends up with a dark coating on it. As the filament evaporates, it gets thinner and its resistance increases. This reduces the current through the bulb and it gets dimmer. If the filament is not perfectly uniform, some part of the filament will be thinner than the rest and will have a locally greater resistance. Since P = IR, this thin spot will dissipate more power per unit length than the rest of the filament, and will therefore be hotter. This hot spot will evaporate more quickly than the rest of the filament, making it still thinner and hotter, until it gets so hot the filament melts and breaks in two. 13. The 200 A current through a spark plug moves 0.300 mc of charge. How long does the spark last? We know that Q = It so t = Q I = 0.30 10 3 C 200 A = 1.5 10 6 s or 1.5 microseconds. 14. A clock battery wears out after moving 10,000 C of charge through the clock at a rate of 0.500 ma. How long did the clock run? How many electrons per second flowed? One ampere is a coulomb per second. The charge is then found from Q = It. So t = Q I = 10,000 C 0.00050 A = 2.0 10 7 s, or 231 days. There is 1.6 10 19 C per electron. Thus the flow in e /s is 5.0 10 4 A I = 1.6 10 19 C/electron = 3.13 10 15 electrons/s
PHYS 202 (DDE) Solved problems in electricity 6 15. The diameter of 0 gauge copper wire is 8.252 mm. Find the resistance of a 1.00 km length of such wire used for power transmission. We know R = ρ L A and we can find the value of ρ for copper in a table of resistivities. Then R = (1.72 10 8 1000 m Ω-m) π(4.126 10 3 m) 2 = 0.322 Ω 16. Find the voltage drop in an extension cord having a 0.0600 Ω resistance and through which 5.00 A of current is flowing. A cheaper cord utilizes thinner wire and has a resistance of 0.300 Ω. What is the voltage drop in it when 5.00 A flows? (The voltage to whatever appliance is supplied by the cords is reduced by these amounts.) V = IR = (5.0 A)(0.060 Ω) = 0.30 V V = IR = (5.0 A)(0.30 Ω) = 1.5 V 17. Find the power dissipated in each of the extension cords in the previous problem P = I 2 R = (5.0 A) 2 (0.060 Ω) = 1.5 W P = I 2 R = (5.0 A) 2 (0.30 Ω) = 7.5 W
PHYS 202 (DDE) Solved problems in electricity 7 18. If two household light bulbs rated 60 W and 100 W are connected in series to household power, which will be brighter? The 60 W bulb will be brighter. The power ratings are valid only for bulbs connected to the usual household voltage of 115 V. Since P = V 2 /R, the 60 W bulb has a higher resistance. When the bulbs are hooked in series, they both pass the same current, and since P = I 2 R, the higher resistance bulb (the 60 W one) will dissipate the most power and be brightest. 19. What is the resistance of a 100 Ω, a 2.50 kω, and a 4.00 kω resistor connected in series? In parallel? R eq = 100Ω 2500Ω 4000Ω = 6600 Ω or 6.6 kω 1 R eq = R eq 1 100Ω 1 2500Ω 1 4000Ω = 0.01065Ω 1 = 94 Ω 20. What resistance do you need to connect to a 137 Ω resistor to get a total of 53.0 Ω? We must use a parallel connection to get a lower resistance. 1 = 1 1 R eq R 1 R 2 1 53.0Ω = 1 137Ω 1 R 2 1 1 = R 2 53.0Ω 1 137Ω = 0.01157Ω 1 = 86.4 Ω R 2
PHYS 202 (DDE) Solved problems in electricity 8 21. Standard automobile batteries have six lead-acid cells in series, creating a total emf of 12.0 V. What is the emf of an individual lead-acid cell? Each cell supplies 2.0 V. When in series, the voltages add. 22. Given a 48.0 V battery, and 24.0Ω and 96.0Ω resistors, find the current and power for each when connected in series. Repeat when the resistances are in parallel. I = V R = 48 V 24 Ω 96 Ω = 48 V 120Ω = 0.40 A for both resistors. Now, P = I 2 R, so the power dissipated is P 24 P 96 = (0.4 A) 2 (24 Ω) = 3.84 W for the 24 Ω resistor. = (0.4 A) 2 (96 Ω) = 15.4 W for the 96 Ω resistor. P = IV = (0.4 A)(48 V) = 19.2 W supplied by the battery. I 24 = 48 V = 2.0 A for the 24 Ω resistor. 24Ω I 96 = 48 V = 0.50 A for the 96 Ω resistor. 96Ω I batt = 2.0 A 0.5 A = 2.5 A supplied by the battery. P 24 = V 2 (48 V)2 = = 96 W for the 24 Ω resistor. R 24 Ω P 96 = V 2 (48 V)2 = = 24 W for the 96 Ω resistor. R 96 Ω P = IV = (2.5 A)(48 V) = 120 W supplied by the battery. 23. What is the internal resistance of an automobile battery that has an emf of 12.0 V and a terminal voltage of 15.0 V while a current of 8.00 A is charging it? The voltage drop across the internal resistance is 15.0 V - 12.0 V = 3.0 V. Therefore R = V I = 3.0 V 8.0 A = 0.38 Ω terminal 12.0 V r i terminal 15.0 V
PHYS 202 (DDE) Solved problems in electricity 9 24. A 1.58 V alkaline cell with a 0.200Ω internal resistance is supplying 8.50 A to a load. What is its terminal voltage? What is the value of the load resistance? (c) What is unreasonable about these results? (d) Which assumptions are unreasonable or inconsistent? The voltage drop across r i is V = Ir i = (8.5 A)(0.20 Ω); V = 1.70 V across r i. This would make the terminal voltage a negative 0.12 volts! The load resistance would have to be negative also, which does not make sense. You would have 1.58 V R = V I = -0.12 V 8.50 A = -0.014 Ω (c) The terminal voltage cannot be negative if the load is just a resistor. (d) The current is far too large and/or the internal resistance is unreasonable high. r i R L
PHYS 202 (DDE) Solved problems in electricity 10 25. In the diagram below, find the current through the 20-ohm resistor. 6 Ω 20 Ω 4 Ω 8 Ω 12.0 V Solution: 6 Ω 20 Ω 4 Ω 6 Ω 24 Ω 6 Ω 6 Ω 8 Ω 8 Ω 12.0 V The parallel resistance is found from 12.0 V 12.0 V 1 R p = 1 24 Ω 1 8 Ω = 4 24 Ω so R p = 6 Ω The total current is I = V/R = (12.0 V)/(12.0 Ω) = 1.0 A. Therefore, the voltage drop across the parallel combination is V = IR = (1.0 A)(6 Ω) = 6.0 V. The current through the 20 Ω plus 4 Ω series combination is I = V R = 6.0 V 24 Ω = 0.25 A