1 UNIVERSITI TEKNOLOGI PETRONAS FOUNDATION PROGRAM FAP0025 PHYSICS II TUTORIAL QUESTIONS SOLUTIONS REFERENCE: NMM, BSMS & HM, FOUNDATION PHYSICS, Prentice Hall, 2006. By Dr. John Ojur Dennis CHAPTER 15 Q12 When two identical ions are separated by a distance of 6.2 × 10 −10 m, the electrostatic force each exerts on the other is 5.4 × 10 −9 N. How many electrons are missing from each ion? F =k q= q r = e e Q16 q2 r2 Fr 2 k F 6.2 × 10 −10 m 5.4 × 10 −9 N = = 3.0 .m k 1.60 × 10 −19 C 8.99 × 109 NC Suppose the charge q2 in Figure 19–30 can be moved left or right along the line connecting the charges q1 and q3 . Given that q = +12 μC, find the distance from q1 where q2 experiences a net electrostatic force of zero? (The charges q1 and q3 are separated by a fixed distance of 32 cm.) F21 = −k F23 = k q1 q 2 2. 0 q 2 k = − 2 r21 x2 q 2 q3 6.0q 2 k = 2 r23 (0.32m − x )2 F21 + F21 = 0 − 2.0q 2 6.0q 2 + =0 x2 (0.32m − x )2 2.0 6.0 = x 2 (0.32m − x )2 2.0 x 2 + 0.64 x − 0.10 = 0 − 0.64 ± 0.64 2 − 4(2.0)(− 0.10 ) = 0.12 m, − 0.44 m 2(2.0 ) x = 0.12 m = 12 cm x= 2 Q29. Two identical point charges in free space are connected by a string 6.6 cm long. The tension in the string is 0.21 N. (a) Find the magnitude of the charge on each of the point charges. (b) Using the information given in the problem statement, is it possible to determine the sign of the charges? Explain. (c) Find the tension in the string if +1.0 μC of charge is transferred from one point charge to the other. Compare with your result from part (a). (a) T= kq 2 r2 T q=r = (0.066 m ) k 0.21 N = 3.2 × 10 −7 C 2 N.m 8.99 × 10 9 C2 (b) (c) No, because the repulsive force which causes the tension in the string depends only upon the magnitude, not the sign, of the charges. Regardless of whether the charges are positive or negative, a transfer of + 1.0 μC will result in the charges having opposite signs. So, the force between the charges will be attractive and the tension will be zero. Q69. A square with sides of length L has a point charge at each of its four corners. Two corners that are diagonally opposite have charges equal to +2.25 μC; the other two diagonal corners have charges Q. Find the magnitude and sign of the charges Q such that each of the +2.25 μC charges experiences zero net force. Place the square in the first quadrant with the 2.25-µC charges at (0, 0) and (L, L), and the others with charge Q at (L, 0) and (0, L). Let the 2.25-µC charges be called q. Note that the sign Q must be negative or all of the charges would fly away from each other. Because of the symmetry of the problem, if Q is found such that one of the 2.25-µC charges experiences zero net force, the other one will also. Set the net force on the charge at the origin equal to zero. ∑F x =0= kqQ kq 2 + cos 225 o 2 2 L 2L ( ) ( 1 ⎛ 2⎞ 2 ⎟= q = − q⎜ − 2.25 × 10 −6 C = 7.95 × 10 −7 C ⎜ ⎟ 2 ⎝ 2 ⎠ 4 ) The sign of Q is negative. 45 × 10 6 m / s.0° ( ) (b) Q78. Find (a) the magnitude of the electric field and (b) the tension in the string. F = ma = qE q a= E m .81 m ⎛ mg ⎞ 2 mg −2 s ⎟ T = QE cos θ = Q cos θ ⎜ = = 6.3 × 10 N ⎟= ⎜ tan θ θ tan30.0 ° Q sin ⎠ ⎝ ( ) Figure 19–38 shows an electron entering a parallel-plate capacitor with a speed of 5. (a) Find the acceleration.7 g and charge Q = +44 μC is attached to a string and placed in a uniform electric field that is inclined at an angle of 30. The electric field of the capacitor has deflected the electron downward by a distance of 0. (0. Find (a) the magnitude of the electric field in the capacitor and (b) the speed of the electron when it exits the capacitor.3 Q76 An object of mass m = 3.0037 kg) 9.81 m mg s2 = = 1.0037kg) 9. The object is in static equilibrium when the string is horizontal.0° with the horizontal (Figure 19–37).618 cm at the point where the electron exits the capacitor. (a) ∑F ∑F E= x y = 0 = QE cos θ − T = 0 = QE sin θ − mg (0.6 × 10 3 N/C −6 Q sin θ (44 × 10 C) sin 30. 22 × 10 m s 2 CHAPTER 16 Q7. A particle with a mass of 3. What is the charge on the ion? q= ΔU −1.37 × 10−15 J = = −4.60 × 10 −19 C)(0.00618 m ⎞ m⎞ ⎛ = ⎜ 5.045 μC is released from rest at point A in Figure 20–21.0 cm.0225 m) 2 ( ) 2 (b) vx = v ⎛ 2v 2 Δy ⎞ ⎛ x ⎞ 2vΔy v y = at = ⎜ ⎟ ⎟= ⎜ x2 ⎟ ⎜ x ⎝ ⎠⎝ v ⎠ vf = vx 2 + v y 2 = v2 + 4 v 2 Δy 2 x2 2 ⎛ Δy ⎞ = v 1+ 4⎜ ⎟ ⎝ x ⎠ ⎛ 0.00e ΔV 2850 V Q19. x = vt Δy = a= ⎛ x2 ⎞ 1 2 1 ⎛ x⎞ 2 ⎟ at = a⎜ ⎟ = a⎜ ⎜ 2v 2 ⎟ 2 ⎝ v⎠ 2 ⎝ ⎠ 2v 2 Δy q = E x2 m Solve for E. (a) The particle has a positive charge.5 g and a charge of + 0. −31 6 m 2mv 2 Δy 2(9. .4 Solve the equations of motion for a.81× 10−19 C = −3.13 × 103 N/ C qx 2 (1. (a) In which direction will this charge move? (b) What speed will it have after moving through a distance of 5. which is the negative x-direction. less than or equal to its increase in speed in the first 5. When an ion accelerates through a potential difference of 2850 V its electric potential energy decreases by 1.0 cm? Explain.45 × 10 s (0.0225 m ⎠ 6 = 6.00618 m) E= = = 4. Will its increase in speed for the second 5.0 cm be greater than.37 × 10 −15 J.0 cm? (c) Suppose the particle continues moving for another 5. so it will move in the direction of the electric field.45 × 106 ⎟ 1 + 4 ⎜ ⎟ s ⎠ ⎝ ⎝ 0.11 × 10 kg) 5. .3 μ C charge more than it is attracted by the 2. How much energy is stored in the capacitor now? (c) How much work is required to increase the separation of the plates from 2.86 J (b) The −6.μC charge to infinity.7 μ C charge.3 × 10− 6 C + ⎟ (2.99 × 109 ⎜ C2 ⎝ = −0.25 m (0. Figure 20–23 shows three charges at the corners of a rectangle.1 μ C q2 = +2.1 μ C charge is repelled by the −3.050 m) C 0.16 m)2 ⎠ ⎣ ⎤ ⎥ ⎥ ⎦ ⎛ N ⋅ m2 = − ⎜ 8.25 mm to 4.μC charge to infinity? Explain.99 × 109 ⎜ C2 ⎝ = 0.50 mm? Explain your reasoning.25 m)2 + (0. (c) Calculate the work needed to move the − 6. (a) How much work must be done to move the + 2.7 × 10− 6 C) ⎢ ⎟ ⎢ 0.7 μ C q3 = −3.0 cm because v is proportional to the square root of the distance traveled. Its increase in speed will be less than its increase in speed in the first 5.045 × 10− 6 C) 1200 N (0.μC charge to infinity? (b) Suppose.1× 10− 6 C) ⎢ ⎥ ⎟ 0. W = −ΔU = − kq1q2 kq1q3 − r12 r13 ⎛ q2 q3 ⎞ = − kq1 ⎜ + ⎟ ⎝ r12 r13 ⎠ ⎛ N ⋅ m2 = − ⎜ 8. A parallel-plate capacitor has plates with an area of 405 cm2 and an air-filled gap between the plates that is 2.50 mm. Is the work required in this case greater than. less than. which is less than the work required in part (a).54 J (c) ⎞ ⎡ 2. The capacitor is charged by a battery to 575 V and then is disconnected from the battery.7 . (a) How much energy is stored in the capacitor? (b) The separation between the plates is now increased to 4.9 cm/s (c) Q31.3 μ C W = −ΔU = − kq2 q1 kq2 q3 − r21 r23 ⎛ q1 q3 ⎞ = − kq2 ⎜ + ⎟ ⎝ r21 r23 ⎠ ⎡ ⎞ − 6. (a) q1 = −6.25 mm thick. instead.7 × 10− 6 C −3. The work required will be negative.1× 10− 6 C −3.3 × 10− 6 C ⎤ + ⎟ ( − 6.5 (b) 1 2 mv = − qΔV = qEd 2 v= 2qEd = m 2(0.1 .16 m ⎦ ⎢ 0.0035 kg ( ) = 3.7 . that we move the − 6.1 .25 m ⎥ ⎠ ⎣ Q60.μC charge to infinity. or the same as when we moved the + 2. A second point charge.37 × 10− 6 C) 0.85 × 10−12 = ( C2 N⋅m 2 ) (405 ×10 −4 m 2 )(575 V) 2 2(2. A point charge Q = + 87. but Q stays the same.99 × 109 ( = (b) N ⋅m 2 C2 ) (87.37 μC.0526 kg 1 1 ⎛ ⎞ − ⎜ ⎟ 0.1 μC is held fixed at the origin.1×10 −6 C)(−2.27 × 10−5 J 2 2 (c) Q69. with mass m = 0. (b) If the second charge is released from rest. is placed at the location (0.1 m/s .6 (a) U= 1 CV 2 2 1 ⎛ κε A ⎞ = ⎜ 0 ⎟V 2 2⎝ d ⎠ 1. 0).0526 kg and charge q = − 2. C C0 1 1 QV = Q0 (2V0 ) = Q0V0 = 2 E0 = 5.1×10 −6 C)(−2.99 × 109 ( N ⋅m 2 C2 ) (87.63 ×10−5 J .323 m = −5.121 m ⎝ ⎠ = 19. C is reduced by one half. so V is doubled because V= U= Q Q = 2 0 = 2V0 .121 m. what is its speed when it reaches the point (0.75 J Ki + U i = K f + U f kQq 1 2 kQq 0+ = mv + ri rf 2 ⎛1 1⎞ 1 2 mv = kQq ⎜ − ⎟ 2 ⎝ ri rf ⎠ 2kQq ⎛ 1 1 ⎞ v= ⎜ − ⎟ m ⎝ ri rf ⎠ = 2 8. Since the energy stored in the capacitor increased.323 m.00059 8. (a) Find the electric potential energy of this system of charges.25 × 10−3 m) = 2. W = ΔU = 2.323 m 0.63 × 10−5 J (b) By doubling the separation.37 × 10− 6 C) 0. 0)? (a) U= kQq r 8. 7 ⎟ 1. or equal to the potential at point B? Explain.2 6.7 6.0 A. It costs 2.0 A I2 = A− 6. calculate the current in the circuit 5.7 13.2 6. (a) I = I1 + I 2 KK (I) 0 = 12 V − I (3.6 cents to charge a car battery at a voltage of 12 V and a current of 15 A for 120 minutes.0 V − I (9. (b) Is the potential at point A greater than.6 cents = = kWh power × time (15 A)(12 V)(120 min) 1 h 60 min ( ) ⎛ 103 ⎞ ⎜ ⎟ = 7.7 I= 1 + 13. 9.7 I KK (II) 1.7 1.7 3 13.8 Ω : 0.Ω resistor between A and B. 1. (c) Determine the potential difference between the points A and B.2 6.7 ⎠ ⎝ The currents through each resistor are as follows: 3.7 CHAPTER 17 Q23.7 3 A 10 A + 6.7 ⎠ ⎝ 13.0 ms after the switch is .7 A ⎞ ⎤ ⎜ ⎟ ⎥ (1.9 Ω . (b) The potential at point A is greater than that at point B because the potential has been decreased by the 1.72 A 3 ⎛ 13.7 A− I+ I 1. less than.7 10 A + 6. (a) Find the current in each resistor in Figure 21–37.8 Ω) I1 = 10 A − 13.7 ⎞ + ⎜1 + ⎟I 1.2 6.7 + 13.2 6. (a) What capacitance must be used in this circuit if the time constant is to be 2.9 Ω) − I 2 (6.2 6.7 ⎜ 1 + 13.7 6.8 Ω) I2 = 3 13.7 I = 0.2 Ω : 2 A.5 ms? (b) Using the capacitance determined in part (a).7 + 13.2 .7 + 13.2 ⎜ 1 + 13.7 ⎠ ⎝ 3 ⎛ 3 13.7 Substitute (II) and (III) into (I).7 A− I KK (III) 6.7 Ω) − 9. I = 10 A − ⎛ 13.2 ⎜ 1 + 13.2 cents/kWh ⎜ k ⎟ ⎝ ⎠ Q51.7 ⎛ 10 A + 6. 3 ⎡ 13.2 Ω) − I (9.2 0 = 12 V − I (3.7 A ⎞ ⎜ ⎟=2A I1 = 10 A − 1.7 10 A + 6. What is the cost of electrical energy per kilowatt-hour at this location? cost cost 2.7 ⎟ ⎥ ⎢ 1.7 3 A = 10 A + 6.7 Ω :1.72 A.7 6.2 Ω) = 2 V VA − VB = ⎢10 A − 1.7 ⎟ 1. The resistor in an RC circuit has a resistance of 145 Ω.7 A ⎞ ⎜ ⎟ = −1.9 Ω) − I1 (1.7 ⎝ ⎠⎦ ⎣ (c) Q64.7 + 13. 6. ..5 ms) I (5. Then.5 × 10−3 s = 17 μ F 145 Ω R 9.0 ms) /(2.. Assume that the capacitor is uncharged initially and that the emf of the battery is 9. connect this pair in parallel with the 413.8 closed.Ω resistor.. What force does this particle experience when it moves with a speed of 6.8 m ⎛ F ⎞ v s F2 = qv2 ⎜ 1 ⎟ sin θ 2 = 2 F1 sin θ 2 = 2...7 × 10−5 N v1 23 m ⎝ qv1 ⎠ s ( ) ..(II) Subtract (II) from (I)..65 A) ⎛ ⎜ ⎞ ⎟ = 44 V ⎝ 0.. A battery has an emf ε and an internal resistance r..4 mA I (t ) = ε e −t / τ Q70. 521 Ω...65 A)(25 Ω) − (0.65 A.0 ms) = e 145 Ω = 8... When the battery is connected to 55 .45 A)(55 Ω) − (0.65 A)(25 Ω) + (0. and 146 Ω.. You are given resistors of 413 Ω... 0 = ε − (0. Find the battery’s emf and internal resistance.(I) 0 = ε − (0.. (a) (b) C= τ R = 2..Ω resistor..Ω resistors in series... Connect the 521.Ω resistor.45 A. 0 = 8..0 V −(5. When the battery is connected to a 25 .Ω and 146 ...45 A) r.20 A)r r = 43 Ω 8. the current is 0.65 A) r... the current through the battery is 0. Describe how these resistors must be connected to produce an equivalent resistance of 255 Ω.5 V ε = (0.2 × 10−4 N sin 25° = 2...5 V − (0.0 V..2 × 10−4 N when it moves at right angles to a magnetic field with a speed of 23 m/s. 1 1 ⎞ ⎛ + Req = ⎜ ⎟ ⎝ 521 Ω + 146 Ω 413 Ω ⎠ −1 = 255 Ω Q91.20 A ⎠ CHAPTER 18 Q6 A particle with a charge of 14 μC experiences a force of 2.8 m/s at an angle of 25° relative to the magnetic field? F1 = qv1B sin 90° F B= 1 qv1 6... τ square L ⎛ L2 ⎞ ⎛ 4π IA B A = s = s = 4 = ⎜ ⎟⎜ ⎟ 2 τ circle IAc B Ac π L 2 ⎜ ⎝ 16 ⎠ ⎝ L 2π (b) ( ) ( ) 2 π ⎞ ⎟= 4 ⎠ Q60 A long. as indicated in Figure 22–38. perpendicular to each . the maximum torque of the square loop is less than the maximum torque of the circular loop. less than.180 T) = = 1. given that its charge has a magnitude of 160 −27 . u. × 10 kg. Next to the wire is a square loop with sides 1. as shown in Figure 22–44.7 m)(0. (b) Calculate the ratio of the maximum torques.9 Q19 When a charged particle enters a region of uniform magnetic field it follows a circular path. × 10 −19 C. (a) A circle has a larger area than a square when the perimeters of each are equal. (a) If these loops carry the same current and are placed in magnetic fields of equal magnitude.0 m in length. have one turn made from wires of the same length. Since the particle is experiencing a force to the right. I= (0. m= erB (1.0 × 106 m / s.5 A Q36 Two current loops.81 T) sin 90° = 2. is the maximum torque of the square loop greater than. and the radius of its path is 52. or the same as the maximum torque of the circular loop? Explain. where 1 u = 167 (a) According to the RHR. one square the other circular.67 × 10−27 u 6. (b) Calculate the magnitude of the net force acting on the loop.0 cm. (a) Is this particle positively or negatively charged? Explain. the particle’s speed is 6.60 ×10−19 C)(0. The loop carries a current of 2.180 T. straight wire carries a current of 14 A. atomic mass units. (a) The current-carrying wire generates a magnetic field that is directed into the page. The force on each side of the loop is directed away from the center of the loop. Find the mass of .7 m and a mass of 0. (a) What is the direction of the net force exerted on the loop? Explain. perpendicular to all four sides of the square loop.5 A in the direction indicated.0 × 106 m s (b) Q29 ( )( ) A wire with a length of 3. What is the minimum current needed to levitate the wire? F = mg = ILB sin θ mg I= LB sin θ I is minimized when θ = 90°. (b) Suppose that the magnetic field has a magnitude of 0. a positively charged particle would experience a force to the left. τ square / τ circle . So. it must be negatively charged .5 u kg v 1.84 T.75 kg is in a region of space with a magnetic field of 0. since the torque is directly proportional to the area of the loop.81 m 2 s ( ) (3. Give your result in the particle.75 kg) 9.520 m)(0. .22 A)(110 Ω) = 3. (b) If the current in the wire increases. the magnetic field through the circuit does not vary with time.2 m ⎠ = 3 × 10−5 N CHAPTER 19 Q14 A single conducting loop of wire has an area of 7. And.3 × 102 T/s Q22 Figure 23–30 shows a current-carrying wire and a circuit containing a resistor R.28 T. (a) If the current in the wire is constant. since the magnetic field is directed out of the page.2 m 1.10 side. Fnet = ILBclose − ILBfar ⎛ μ0 I ′ μ I′ ⎞ = IL ⎜ − 0 ⎟ ⎝ 2π rclose 2π rfar ⎠ μ0 LII ′ ⎛ 1 1 ⎞ = − ⎜ ⎟ 2π ⎝ rclose rfar ⎠ = 2π ⋅m (1. counterclockwise. so the induced current is zero .4 × 10 −2 m 2 and a resistance of 110 Ω.5 A)(14 A) ( 4π ×10−7 TA ) ⎛ 1 1 ⎞ − ⎜ ⎟ ⎝ 0. the forces due to the sides of the loop cancel. or zero? Explain. Because the force is stronger closer to the wire.32 A? ε = IR = N IR ΔB = = A Δt 7. the current in the circuit flows clockwise . should the resistance R be greater than or less than the value found in part (a)? Explain. Perpendicular to the plane of the loop is a magnetic field of strength 0.11 J.0 m)(2. Since the current in the wire is increasing. is the induced current in the circuit clockwise. the magnetic field through the circuit is increasing.4 × 10− 2 m 2 AΔB ΔΦ = (1) Δt Δt (0. the induced current in the circuit will induce a magnetic field into the page. or zero? Explain. the net force is towards the wire . (a) What is the value of the resistance R? (b) If it is desired that more energy be stored in the inductor. So. and in the plane of the loop. is the induced current in the circuit clockwise. At what rate (in T/s) must this field change if the induced current in the loop is to be 0. (b) Due to symmetry. Q52 After the switch in Figure 23–36 has been closed for a long time the energy stored in the inductor is 0. counterclockwise. (a) (b) Since the current in the wire is constant. So.5 Ω ⎟ ⎟ ⎝ ⎠ −1 ⎡ 2(0. the rod falls with constant speed. as shown in Figure 23–39. When the acceleration reaches zero. the value of R should be less than the value found in part (a). Q79 A conducting rod of mass m is in contact with two vertical conducting rails separated by a distance L. the magnetic flux increases. resulting in a clockwise flowing current. mg = ILB mg I= LB v= ε LB = IR ⎛ mg ⎞ R mgR =⎜ = ⎟ LB ⎝ LB ⎠ LB L2 B 2 . The induced emf opposes the increase.11 1 2 1 ⎛ ε LI = L ⎜ ⎜ Req 2 2 ⎝ 1 1 2U + = R 7. Assuming the rod slides without friction. the equivalent resistance is decreased.5 Ω ⎦ ⎢ (62 × 10 H)(12 V) ⎥ ⎣ = 40 Ω −1 (b) The energy stored in the inductor is inversely proportional to the square of the equivalent resistance. the energy stored is increased.5 Ω Lε 2 U= 2 ⎞ 1 1⎞ ⎛ 1 + ⎟ ⎟ = Lε 2 ⎜ ⎟ 2 ⎝ 7. the induced current begins to flow. and the rod is acted upon by an upward magnetic force. If the equivalent resistance is decreased.11 J) 1 ⎤ =⎢ − ⎥ 2 −3 7. (b) (c) As the rod falls. The entire system is immersed in a magnetic field of magnitude B pointing out of the page.5 Ω R ⎠ ⎠ 2 (a) ⎛ 2U 1 ⎞ R=⎜ ⎜ Lε 2 − 7. This causes the acceleration to decrease. If R is decreased. (b) What is the direction of the induced current (clockwise or counterclockwise) in the circuit? (c) Find the speed of the rod after it has fallen for a long time. the rod falls with the acceleration of gravity. As the rod falls. (a) Initially. (a) describe the motion of the rod after it is released from rest. 0 V ω C 2π (120 s−1 )(22 × 10− 6 F) (c) ⎛ ⎞ I 0.10 A ⎛ ⎞ − 90° ⎟ = (9.10 A I ⎛ ⎞ − 90° ⎟ = (9.72 A.72 A ⎠ ⎣ ⎢ 2π (150 s )(13 × 10 F) ⎦ ⎥ ⎝ I rms ⎠ ⎝ ω C ⎠ 2 . (a) What is the maximum voltage of the generator? (b) What is the voltage across the capacitor when the current in the circuit is 0. What is the resistance in this circuit? Z= Vrms ⎛ 1 ⎞ = R2 + ⎜ ⎟ I rms ⎝ω C ⎠ 2 2 2 ⎤ ⎛ Vrms ⎞ ⎛ 1 ⎞2 1 ⎛ 95 V ⎞ ⎡ R= ⎜ ⎥ = 100 Ω ⎟ −⎜ ⎟ = ⎜ ⎟ −⎢ −1 −6 ⎝ 0. so the current will increase if the frequency increases.12 CHAPTER 20 Q13 The maximum current in a 22 . (c) Find the rms current in this capacitor at a frequency of 410 Hz.04 V) sin ⎜180° − sin −1 − 90° ⎟ = 6. I rms = ω CVrms = 2π (410 s −1 )(5.15 A I ⎝ ⎠ max ⎝ ⎠ Vmax = X C I max = I max 0. or stay the same? Explain. (a) (b) (c) C= I 1 27 × 10−3 A = rms = = 6.μF capacitor connected to an ac generator with a frequency of 120 Hz is 0.15 A ⎝ ⎠ ⎝ ⎠ Q16 A capacitor has an rms current of 21 mA at a frequency of 60.7 V V = Vmax sin θ ′ = Vmax sin ⎜180° − sin −1 I max 0.04 V) sin ⎜ sin −1 − 90° ⎟ = −6.10 A and decreasing? (a) (b) ⎛ ⎞ 0.97 × 10− 6 F)(12 V) = 180 mA Q21 The rms current in an RC circuit is 0.15 A.10 A and increasing? (c) What is the voltage across the capacitor when the current in the circuit is 0. decrease. The capacitor in this circuit has a capacitance of 13 μF and the ac generator has a frequency of 150 Hz and an rms voltage of 95 V.0 s −1 )(12 V) The current in the capacitor is directly proportional to the frequency.15 A = = 9.7 V V = Vmax sin(θ − 90°) = Vmax sin ⎜ sin −1 0. (a) What is the capacitance of this capacitor? (b) If the frequency is increased. will the current in the capacitor increase.0 Hz when the rms voltage across it is 14 V.0 μ F ω X C ω Vrms 2π (60. 100 × 10−6 F) Since the rms voltages are not all in phase.0° (b) By inspection of the diagram. ( X L − X C )2 = [ω L − 1/(ω C )]2 will increase. less than.0 ×103 s −1 )(0. what is the impedance? (b) If the frequency in this circuit is increased. find I rms . L.100×10−6 F) ⎥ ⎦ 2 (2.00 V? Explain.84 V I 69. where we assume that the values of R.3 mA)(2.50 Ω) = 0. (a) cos φ = Z= R Z R 525 Ω = = 606 Ω cos φ cos 30. and C are the same as in the previous problem. or equal to 6.173 V Vrms. L.00 V ⎤ 1 s −1 )(0. is doubled to 60. C. Calculate the rms voltage across (a) the resistor.3 mA Vrms.L = I rmsω L = (69. their sum will be greater than the overall rms voltage of 6.) Q70 The phasor diagram for an RLC circuit is shown in Figure 24–33.50 Ω)2 + ⎡ 2π (60. (b). (b) the inductor. If the frequency is increased. we see that XL > XC. R. Again.0×103 ⎣ = 69. and (c) the capacitor. and (c) to be greater than. . or stay the same? Explain. and that the rms voltage of the generator is still 6.00 V.13 Q49 Consider the ac circuit shown in Figure 24–32.00 V. however. decrease.300 × 10−3 H) = 7.R = I rms R = (69.3 mA)2π (60.3 mA (a) (b) (c) (d) Vrms.0 × 103 s −1 )(0. Since Z = R 2 + ( X L − X C )2 . (The sum of the magnitudes of out-of-phase phasors is greater than the magnitude of their vector sum. (d) Do you expect the sum of the rms voltages in parts (a).0 × 103 s−1 )(0.0 kHz. will the impedance increase.84 V ωC 2π (60. I rms = Vrms = Z = Vrms R 2 + ω L − ω1 C ( ) 2 6. the impedance will increase if the frequency is increased.300 × 10−3 H) − ⎢ 2π (60.C = rms = = 1. The frequency of the generator. (a) If the resistance in this circuit is 525 Ω. 5pF.8 μ H. 2 • We find the time from t= (235 J ) U = = 1.00 × 10 8 m s 6.75 × 10 −9 T. ⎝ 2π ⎠ ⎝ LC ⎠ ⎛ 1⎞ 96. the B field oscillates vertically and has a frequency of 80. The rms strength of the electric field is E rms = cBrms = 3.14 CHAPTER 21 Q4 In an EM wave traveling west.1-MHz FM station has an inductance of 1.194 W m ⎝ ⎠ ( ) Q32 The oscillator of a 96. ( )( ) The electric field is perpendicular to both the direction of travel and the magnetic field.1 × 106 Hz = ⎜ ⎟ ⎝ 2π ⎠ ⎡ ⎤2 1 ⎢ ⎥ . What are the frequency and rms strength of the electric field.21 × 10 7 s = 140 days. 2⎞ − 4 2 AS 1.75 × 10 −9 T = 2. Q19 The magnetic field in a traveling EM wave has an rms strength of 28.00 × 10 m s )(28. so the electric field oscillates along the horizontal north-south line. How long does it take to deliver 235 J of energy to 1.00 × 10 m ⎛ 0.0 kHz and an rms strength of 6.5 nT.03 V m .194 W m . ⎢ 1.00 cm 2 of a wall that it hits perpendicularly? The energy per unit area per unit time is S = cB rms 2 μ0 (3.5 × 10 −12 F = 1. What value must the capacitance be? We find the capacitance from the resonant frequency: ⎛ 1 ⎞⎛ 1 ⎞2 f0 = ⎜ ⎟ ⎜ ⎟ .8 × 10 −6 H C ⎥ ⎣ ⎦ 1 1 ( ) . which gives C = 1.0 kHz.5 × 10 T ) = (4π × 10 T m A ) 8 −9 −7 2 = 0. and what is its direction? The frequency of the two fields must be the same: 80. The image appears to be 18.0 cm behind the mirror. Q29 Rays of the Sun are seen to make a 31. which gives f = −18.33)sin 31. When the beam strikes Earth.0° − θ 1 = 90. which gives θ = 43. The rms electric field strength of the beam is given by S= (10 ×10 W) = 3. Find the rms electric field strength of the beam. What kind of mirror is it. and find the focal length from ⎛ 1⎞ ⎛ 1⎞ 1 ⎜ ⎟ +⎜ ⎟ = .2°.2° = 46.0 × 10 ( ⎛ ⎞ ⎜ ⎟ 3 P 2 = cε 0 E rms .0 cm. Thus the angle above the horizon is 90.0°.15 Q35 A satellite beams microwave radiation with a power of 10 kW toward the Earth’s surface. 1 1 n1 sin θ1 = n2 sin θ 2 . ⎥ f ⎦ Because the focal length is negative.0° − 43. ⎝ do ⎠ ⎝ d i ⎠ f ⎛ 1⎞ ⎡ 1 ⎢ ⎜ ⎟ +⎢ ⎝ ∞⎠ −18.00)sin θ = (1.8°. its circular diameter is about 1500 m. At what angle above the horizon is the Sun? We find the incident angle in the air from (1.5V m.0° angle to the vertical beneath the water. the mirror is convex. and E rms = 1. . A 2 8 ⎝ π (750 m) m s 8. )( ) 2 ⎠ which gives E rms 2 = 2.0 cm ⎣ ( ) ⎤ ⎥ = 1 .85 × 10 −12 C2 N • m2 E rms . 550 km away. CHAPTER 22 Q16 The image of a distant tree is virtual and very small when viewed in a curved mirror.1314 V2 m2 . and what is its radius of curvature? We take the object distance to be ∞. 52)sin θ .2°. A beam of light from outside the aquarium strikes the glass at a 43.16 Q30 An aquarium filled with water has flat glass sides whose index of refraction is 1. . 23–49). which gives θ 3 n2 sin θ 2 = n2 sin θ 3 .00)sin 43.00)sin 43. θ 3 is independent of the presence of the glass.9° = (1. 3 = 31. we would have (1.9°. Note that. the refraction angle from the first surface is the incident angle at the second surface.5° angle to the perpendicular (Fig.2°. (b) Because the surfaces are parallel. 31. 2 = 26. (c) If there were no glass. which gives θ 2 n1 sin θ1 = n2 sin θ 2 .52)sin 26.5° = (1. which gives θ ′ = 3 3 n1 sin θ1 = n3 sin θ 3 ′. because the sides are parallel. and then (b) the water? (c) What would be the refracted angle if the ray entered the water directly? (a) We find the angle in the glass from the refraction at the air–glass surface: (1. We find the angle in the water from the refraction at the glass–water surface: (1. What is the angle of this light ray when it enters (a) the glass.5° = (1.52.33)sin θ .33)sin θ ′ . We find the angle inside the glass from nair sin θ = n sin φ . If the angles are small. the exit beam will be traveling in the same direction as the original beam. [Hint: for small θ . Show that if the incident angle θ is small. . (1. where t is the thickness of the glass and θ is in radians. d = L sin θ − φ ≈ t θ − φ = t ⎢θ − ⎜ ⎟ ⎥ = n ⎢ ⎣ ⎝ n⎠⎥ ⎦ ( ) ( ) ( ) . or φ = θ n We find the distance along the ray in the glass from L= t cos φ ≈ t. where φ is in radians. and sin φ ≈ φ . we use cos φ ≈ 1. 23–22.17 Q35 A light ray is incident on a flat piece of glass with index of refraction n as in Fig.] Because the glass surfaces are parallel. We find the perpendicular displacement from the original direction from ⎡ ⎛ θ ⎞ ⎤ tθ n − 1 . sin θ ≈ tan θ ≈ θ in radians. the emerging ray is displaced a distance d = tθ (n − 1) n from the incident ray.00)θ = nφ. 920 nm = m ′ + 1 2 ( For the first three values of m ′. 3. 3.K . m ′ = 0.0. ⎝ do ⎠ ⎝ di ⎠ f ⎛ 1 ⎞ ⎛ 1⎞ 1 . What wavelength of visible light would have a minimum at the same location? For constructive interference of the second order for the blue light. d sin θ = m ′ + 1 2 ( When the two angles are equal. ( )( ) ) ) For destructive interference of the other light.1. . which gives λ = 1. CHAPTER 23 Q7 In a double-slit experiment. 920 nm = 0 + 1 2 920 nm = 1 + 1 λ . 2. (a) Where is the image located? (b) What is the magnification? (a) We locate the image from ⎛ 1 ⎞ ⎛ 1⎞ 1 ⎜ ⎟ +⎜ ⎟ = . 2 1 2 ( ) ( ) 920 nm = ( 2+ ) λ .K . we have λ . we have d sin θ = m λb = 2 460 nm = 920 nm. we get λ. The only one of these that is visible light is 613nm. which gives λ = 613nm.18 Q47 A stamp collector uses a converging lens with focal length 24 cm to view a stamp 18 cm in front of the lens.84 × 103 nm. it is found that blue light of wavelength 460 nm gives a second-order maximum at a certain location on the screen. ⎜ ⎟ +⎜ ⎟ = ⎝ 18 cm ⎠ ⎝ di ⎠ 24 cm The negative sign means the image is 72 cm behind the lens (virtual). which gives di = −72 cm. (b) We find the magnification from m= − di do =− (−72 cm )= (18 cm ) + 4.1. we get λ . m ′ = 0. 2. which gives λ = 368 nm. 1 1 ⎢ 9700 lines cm ⎥ ⎣ ⎦ ⎡ ⎤ 1 ⎢ ⎥ 10−2 m cm sin 36.4° = 0. ( ) The distances from the central white line on the screen are y 410 = L tan θ 410 = 2.2°. Thus the width of the spectrum is y 750 − y 410 = 1. The number of lines per meter is 1 d = 1 1. 36.12 × 10−7 m = 612 nm.79 × 105 lines m .73 × 10−6 m.6°. and 47. Q37 A He-Ne gas laser which produces monochromatic light of a known wavelength λ = 6.05 m. y 750 = L tan θ 750 ( ) =( 2.855 m. which gives λ = 7. ⎡ ⎤ 1 ⎢ ⎥ 10−2 m cm sin 31.73 × 10−6 m = 5.5° = 6. What wavelengths are these? We find the wavelengths from d sin θ = m λ .6° = 1. How wide is the first-order spectrum on a screen 2.4°.4°.855 m = 1.328 × 10 −7 m is used to calibrate a reflection grating in a spectroscope.90 m − 0.3485. which gives 410 ⎢ 8500 lines cm ⎥ ⎣ ⎦ sin θ 410 = 0.30 m away? We find the angles for the first order from d sin θ = m λ = λ.90 m .60 × 10−7 m = 760 nm. 2 2 ⎢ 9700 lines cm ⎥ ⎣ ⎦ ⎡ ⎤ 1 ⎢ ⎥ 10−2 m cm sin 47. d sin 21. so θ 750 = 39.2° = 1 λ . 3 3 ⎢ 9700 lines cm ⎥ ⎣ ⎦ ( ( ( )( )( )( ) ) ) () () () Q36 White light containing wavelengths from 410 nm to 750 nm falls on a grating with 8500 lines cm .328 × 10−7 m. so θ 410 = 20.5° to the incident beam. which gives λ = 6. ( ( )( )( ) ) ( ) ⎡ ⎤ 1 ⎢ ⎥ 10−2 m cm sin θ = 750 × 10−9 m . How many lines per meter are there on the grating? We find the angle for the first order from d sin θ = m λ = λ. which gives 750 ⎢ 8500 lines cm ⎥ ⎣ ⎦ sin θ 750 = 0.6375. which gives d = 1. ⎡ ⎤ 1 ⎢ ⎥ 10−2 m cm sin θ = 410 × 10−9 m . .34 × 10−7 m = 534 nm.4° = 1 λ .19 Q33 Light falling normally on a 9700-line cm grating is revealed to contain three lines in the first-order spectrum at angles of 31.30 m tan 20.5° = 1 λ . which gives λ = 5. The firstorder diffraction line is found at an angle of 21.30 m ) tan 39.5°. 0)( 5273 K)4 = 2. (a) λmax T = 2.280 nm is scattered by a metal foil.12 × 10−13 m.63 eV. How much energy does one of its photons have expressed in (a) joules and (b) electron-volts? (a) `(b) E = hf = hc λ = 1.67 × 10 W/(m2·K4)](4π)(7.27 eV .12 × 10−13 m = 2.24 × 103 eV·nm − 3.80 × 10−10 m + 7.32 × 10−18 J . .90 × 10−3 m·K = = 550 × 10−9 m (b) P = σAeT 4 = [5.81 × 10 −10 m = 0.0 × 108 m)2(1. If the material is illuminated with monochromatic light ( λ = 300 nm). −3 T= 2.5 eV.24 × 103 eV·nm = 8.6 × 10−19 J/eV) = 8. −8 Q25 A source of UV light has a wavelength of 150 nm.27 × 103 K .5 eV)(1.5 eV = 0.63 × 10−34 J·s Q47 A monochromatic beam of X-rays with a wavelength of 0.20 CHAPTER 25 Q14 The wavelength at which the Sun emits its maximum energy light is about 550 nm. 6. hc λ − φo = 1. estimate (a) its surface temperature and (b) its total emitted power. What is the wavelength of the scattered X-rays observed at an angle of 45° from the direction of the incident beam? Δλ = λC (1 − cos θ) = (2.90 × 10−3 m·K λmax 2. 5. what are (a) the stopping potential and (b) the cutoff frequency? (a) eVo = Kmax = hf − φo = So (b) Vo = 0.7 × 1026 W . E = 8. 300 nm fo = h = φo (3.27 eV = 1. 150 nm As in (a).4 × 1014 Hz . So λ = λo + Δλ = 2. Q34 The work function of a material is 3.281 nm .43 × 10−12 m)(1 − cos 45°) = 7.63 V . Assuming the Sun radiates as a blackbody.90 × 10 m·K. (a) Will its de Broglie wavelength (1) increase. 2 ni f 3 1. will the electron have (1) a shorter. or (3) decrease. what are their de Broglie wavelengths? (a) The electron will have (3) a longer de Broglie wavelength due to its smaller mass. 6. λ = mv . ΔE21 = (13.6 eV) 22 − 52 = 2.2 eV. (a) What are the wavelengths of the emitted photons? (b) Would any of the emitted light be in the visible region? (a) 1 1 1 .856 eV. (2) an equal. or (3) a longer de Broglie wavelength? Why? (b) If the speed of the electron and proton is 100 m s .2 Yes. the transition from n = 5 to n = 2 is in the visible region (violet). and the de h Broglie wavelength is inversely proportional to speed.6 eV) ΔE (in eV) ΔE = (−13. The proton gains speed from the potential difference.63 × 10 J·s −9 = 3. CHAPTER 26 Q8 An electron and a proton are moving with the same speed. 10. (b) The initial kinetic energy of the proton is .11 × 10 kg)(100 m/s) −34 λproton = Q13 6. −27 (1. It then makes a transition directly to the n = 2 level before returning to the ground state. due to the potential difference? Why? (b) By what percentage does the de Broglie wavelength of the proton change? (a) Its de Broglie wavelength will (3) decrease due to the potential difference.24 × 103 nm = 122 nm .28 × 10 m .63 × 10 J·s h −6 (b) λelectron = mv = = 7. (a) Compared with the proton.67 × 10 kg)(100 m/s) −34 A proton traveling at a speed of 4.24 × 10 1. ΔE52 = (13.24 × 103 λ52 = nm = 2.5 × 104 m/s is accelerated through a potential difference of 37 V.6 eV) (n1 2 − ) ( ) (11 (b) 2 − 1 22 )= 10. λ = mv .856 nm = 434 nm . The de Broglie wavelength is inversely proportional to the h mass. (2) remain the same. −31 (9. λ21 = 1.21 Q66 A hydrogen atom in its ground state is excited to the n = 5 level.97 × 10 m . 5 × 10 m/s. 2 mvo = 2 (1.32 × 10−10 m.69 × 10−18 J = 10.50 V nm.6 V + 37 V = 47.50 V nm. 28. it is = 3.67 × 10 4 Since the initial speed of the proton is 4. the speed of the electron in the first Bohr orbit is 2. = 0.15 nm? d sin θ = mλ.3: λ= d sin θ (0.6 V and V2 = 10. From Eq.6 V 1 = 0. Q16 According to the Bohr theory of the hydrogen atom. 28. Q14 V1 − V2 1 = 10.15 × 10−9 m) sin 25° = = 6. So V1 = 10. so Therefore.50 λ 2 1.529 × 10−10 m) = 3.6 V − − 47.32 × 10−10 m .11 × 10−31 kg)(2.7 × 102 V. it had taken 10.3: λ= λ∝ 1 . . the percentage difference is From Eq. (a) What is the wavelength of the matter wave associated with the electron? (b) How does this wavelength compare with the circumference of the first Bohr orbit? (a) (b) 6. 1.6 V.6 V to accelerate the proton from rest to this speed.34 × 10−11 m m 1 λ= 1. (9. Therefore.22 −27 1 2 Ko = 1 kg)(4.5 × 104 m/s)2 = 1. So they are the same.7 × 102 eV .63 × 10−34 J·s h λ = mv = = 3.0634 nm.6 eV. V= 1. λ = 2πr1.50 = (0. because it is a standing wave .0634)2 = 3.19 × 106 m/s) λ = 2πr1 = 2π(0. V λ2 − λ1 λ2 = −1= λ1 λ1 −53% (a decrease) .53 = What is the energy of a beam of electrons that exhibits a first-order maximum at an angle of 25° when diffracted by a crystal grating with a lattice plane spacing of 0.19 × 106 m s . and 11e .) α–β: β–α: Q41 209 82 0 Pb + –1 e → 209 81 Tl. For 24Mg: 12p . (c) With a +1 charge. . 24 − 12 = 12n . ΔN ΔNo − λ t = λ N = λ No e − λ t = e . (b) With a −2 charge. neutrons. (a) In an α decay. A lead-209 nucleus results from both alpha–beta sequential decays and beta–alpha sequential decays. For 25Mg: 12p . or (4) four? Why? (b) Write the nuclear equation for this decay. 25 − 12 = 13n . So the decay product has (3) two fewer protons.23 CHAPTER 27 Q10 and 25 Mg are two isotopes of magnesium.3 years. the ion has one fewer electron. 25 − 12 = 13n . (a) The product of its decay has how many fewer protons than polonium-214: (1) zero. Δt Δt e−λ t = 0. For 24Mg: 12p . the proton number decreases by two. and 11e . (b) the ion has a −2 charge. and (c) the ion has a +1 charge? 24 Mg (a) A neutral Mg atom has the same number of protons as electrons (12). and 12e . What are the numbers of protons. the ion has two more electrons than the neutral atom. For 25Mg: 12p . 213 Pb + 4 2He → 84 Po. and 14e . 24 − 12 = 12n . 24 − 12 = 12n . What period of time is required for a sample of radioactive tritium ( 3 H) to lose 80. (2) one. (3) two. What was the grandparent nucleus? (Show this result for both decay routes by writing the nuclear equations for both decay processes. and electrons in each if (a) the atom is electrically neutral. For 25Mg: 12p . (b) Q25 α–decay: 214 84 4 Po → 210 82 Pb + 2He .0% of its activity? Tritium has a half-life of 12. For 24Mg: 12p . 209 81 Tl + 4 2He → 0 Po + –1 e→ 213 83 Bi . 25 − 12 = 13n . 209 82 213 84 213 83 Bi . Q24 Polonium-214 can decay by alpha decay. and 14e .20. and 12e . So yes .75 × 1019 nuclei .3 years) ln 0. Calculate the amounts of energy released in the following fusion reactions: (a) 2 2 2 3 H+1 H→ 3 He + 1 n.008 665 u](931.75 × 1019 nuclei) = 2. o Therefore. N = e − λ t = e–3.6 MeV .994435 u − me) c < 0.0313. (a) (a) (b) (c) Q30 22 10 Ne → 22 11 Na + 0 −1 e. (a) How many nuclei are initially 87 present in a 25. (223 u)(1. (b) CHAPTER 28 Q11 What would the daughter nuclei in the following decay equations be? Can these decays actually occur spontaneously? Explain your reasoning in each case. 1/2 N Thus. N = No e − λ t . (b) 1 H+1 H→ 4 He + 1 n.693 So Q49 λ Francium-223 ( 223 Fr) has a half-life of 21.66 × 10−27 kg/u).465 = 0. there are (0.014 102 u + 3.8 min. Q = (226.016 029 u − 4. So no .24 t=− ln 0. So no .025406 u − 222.5 MeV/u) = 17.5 MeV/u) = 3.000000 − 4.994915 u − 12.002603 u) c2 > 0. (b) Q = (mH-2 + mH-3 − mHe − mn)c = (2.008 665 u)(931. 25.016 029 u − 1.002 603 u − 1.27 MeV . Q = (21.002603 u) c2 < 0. 2 2 .693 0.465.11 × 1018 nuclei . 1 2 0 2 0 (a) Q = (2mH-2 − mH-3 − mn)c = [2(2. the Q value must be greater than zero.0 × 10−6 kg So there were = 6.66 × 10−27 kg/u) 0.0313)(6. C . 2 22 11Na .991384 u − 21.20 (12. (b) 226 88 Ra → 222 86 Rn + 4 He . Q = (15. 2 (c) 16 8 O→ 12 6 C +4 He 2 For a reaction to occur spontaneously.8 min (60 min + 49 min) = 3.6 years . 0.693 λt = t t = 21. 222 86 12 6 Rn .20 =− t1/2 ln 0.017574 u − 4.20 − = 28.014 102 u) − 3.0-mg sample of this isotope? (b) How many nuclei will be present 1 h and 49 min later? (a) The mass of one nuclei is (223 u)(1.693 = 0. 00 MeV. (d) Be(α . END OF TUTORIAL SOLUTIONS . (d) n or 1 0n . p)16 N 7 Solution (a) 1 1H . E = pc ).) From Exercise 30.6 × 10−19 J/eV) λ . Determine the de Broglie wavelength of the neutrino. (See Exercise 39. The kinetic energy of an electron emitted from a 32 P nucleus that beta decays into a 32 S nucleus is observed to be 1.65 × 106 eV)(1. Assuming the neutrino has no mass.710 MeV. What is the energy of the accompanying neutrino of the decay process? Neglect the recoil energy of the daughter nucleus. (c) 93 38 (b) 59 28 Ni . 6 (e) 16 8 O (n.972 072 u)(931.69 × 10−13 m . O. (2. (e) 16 8 Sr . n )12 C.65 MeV. So the energy of the neutrino is 1.39. 1 2 2 U+1 n → 138 Xe + 51 n+ 0 54 0 (b) 23 38 58 28 2 Ni + 1 H→ 9 4 59 28 Ni + 1 H 1 235 92 Sr . Assume that it is massless. and thus has the same relationship between its energy and momentum as a photon (that is.63 × 10−34 J·)(3.710 MeV − 1 MeV = 0.5 MeV/u) = 1.25 Q37 A neutrino created in a beta-decay process has an energy of 2. E = pc = Therefore Q41 hc (6.973 908 u − 31.00 × 108 m/s) hc λ= E = = 4. the maximum energy released in the decay is Q = (Mp − MD)c2 = (31. Q64 Complete the following nuclear reactions: (a) (c) 6 3 Li + 1 H →3 He + 4 He .71 MeV .