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Version 001 – Exam 1 – shih – (57480) µN 5 50 16 1. 9m 5 + 5 nC µN 9m ◦ 14.0362 17 + 7 nC 9m + 7 nC 4 6 µN q1 q2 r2 ~ net = F ~1 + F ~3 F q 2 2 Fnet = Fnet,x + Fnet,y Felectric = kC 9m + 5 nC Find the magnitude of the net electric force on the 7 nC charge. The Coulomb constant is 8.98755 × 109 N · m2 /C2 . 1. 1.33978e-08 2. 2.01278e-08 3. 5.23443e-09 4. 1.74007e-09 5. 3.88861e-09 6. 2.53508e-09 7. 2.13158e-08 8. 2.26446e-09 9. 2.83544e-08 10. 1.48865e-09 Correct answer: 2.26446 × 10−9 N. Explanation: Let : 2. 264 94 3 nC + 9m 1. Holt SF 17Rev 21 001 (part 1 of 2) 10.0 points Three positive point charges are arranged in a triangular pattern in a plane, as shown below. 3 nC + 9m This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 1 q1 = +3 nC = +3 × 10−9 C , q2 = +7 nC = +7 × 10−9 C , q3 = +5 nC = +5 × 10−9 C , (x1 , y1 ) = (0 m, 9 m) , (x2 , y2 ) = (9 m, 0 m) , (x3 , y3 ) = (0 m, −9 m) , and kC = 8.98755 × 109 N · m2 /C2 . The distances between the 7 nC charge and q1 and q2 are r12 = (x2 − x1 )2 + (y2 − y1 )2 = (x2 − 0)2 + (0 − y1 )2 = x22 + y12 = (9 m)2 + (−9 m)2 = 162 m2 , and r32 = (x3 − x2 )2 + (y3 − y2 )2 = (0 − x2 )2 + (y3 − 0)2 = x22 + y32 = (−9 m)2 + (−9 m)2 = 162 m2 . Consider the magnitudes of the forces. The repulsive force q2 q1 F1 = kC 2 r1 = 8.98755 × 109 N · m2 /C2 7 × 10−9 C 3 × 10−9 C × 162 m2 = 1.16505 × 10−9 N acts downward and to the right along the line connecting q2 and q1 , and the repulsive force q2 q3 F3 = kC 2 r3 = 8.98755 × 109 N · m2 /C2 7 × 10−9 C 5 × 10−9 C × 162 m2 = 1.94175 × 10−9 N Version 001 – Exam 1 – shih – (57480) acts upward and to the right along the line connecting q2 and q3 . Now, lets consider the directions of the forces. Since |x2 | = |y1 | = |y3 | , θ21 = −θ23 = 45◦ , and Fx = F1 cos 45◦ + F3 cos 45◦ = (1.16505 × 10−9 N) cos 45◦ + (1.94175 × 10−9 N) cos 45◦ = 2.19684 × 10−9 N Fy = −F1 sin 45◦ + F3 sin 45◦ = −(1.16505 × 10−9 N) sin 45◦ + (1.94175 × 10−9 N) sin 45◦ =q 5.49211 × 10−10 N Fx2 + Fy2 Fnet = = h 2.19684 × 10−9 N 2 + 5.49211 × 10−10 N = 2.26446 × 10−9 N . 2 i 21 002 (part 2 of 2) 10.0 points What is the direction of this force (measured from the positive x-axis as an angle between −180◦ and 180◦ , with counterclockwise positive)? 1. 18.4349 2. 30.9638 3. 26.5651 4. -23.1986 5. 11.3099 6. 14.0362 7. -15.9454 8. 12.9946 9. 21.8014 10. -32.4712 Correct answer: 14.0362 . ◦ Explanation: tan θ = Fy Fx θ = arctan Fy Fx 2 = tan −1 5.49211 × 10−10 N 2.19684 × 10−9 N = 14.0362◦ . The net force on the 7 nC charge is 2.26446 × 10−9 N acting 14.0362 ◦ above the positive x-axis. Magnitude of Force 003 10.0 points There are two identical small metal spheres with charges 64.3 µC and −41.622 µC. The distance between them is 6 cm. The spheres are placed in contact then set at their original distance. Calculate the magnitude of the force between the two spheres at the final position. The Coulomb constant is 8.98755 × 109 N · m2 /C2 . 1. 2952.86 2. 67.3856 3. 320.989 4. 9.64534 5. 8.28741 6. 57.464 7. 133.051 8. 45.6884 9. 3.22108 10. 39.5515 Correct answer: 320.989 N. Explanation: Let : qA = 64.3 µC = 6.43 × 10−5 C , qB = −41.622 µC = −4.1622 × 10−5 C , d = 6 cm = 0.06 m . and When the spheres are in contact, the charges will rearrange themselves until equilibrium is reached. Each sphere will then have half of the original total charge: q= qA + qB 2 Version 001 – Exam 1 – shih – (57480) 6.43 × 10−5 C−4.1622 × 10−5 C 2 −5 = 1.1339 × 10 C . 3 = The force between the two spheres is q2 d2 = 8.98755 × 109 N · m2 /C2 F =k (1.1339 × 10−5 C)2 × (0.06 m)2 Let : L = 18.8 cm = 0.188 m and q = −8.63 µC = −8.63 × 10−6 C . Call the length of the rod L and its charge q. Due to symmetry Z Ey = dEy = 0 and Ex = = 320.989 N . Z dE sin θ = ke Z dq sin θ , r2 where dq = λ dx = λ r dθ, so that 18.8 cm Uniformly Charged Bent Wire 01 004 10.0 points A uniformly charged insulating rod of length 18.8 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8.98755 × 109 N m2 /C2 . −8.63 µC O y θ b x ke λ Ex = − r =− If the rod has a total charge of −8.63 µC, find the horizontal component of the electric field at O, the center of the semicircle. Define right as positive. 1. -46219500.0 2. -8843070.0 3. -13788500.0 4. -15471100.0 5. -65540400.0 6. -11306900.0 7. -50340000.0 8. -31098000.0 9. -18157300.0 10. -58994900.0 Correct answer: −1.37885 × 107 N/C. Explanation: =2 where λ = Z 3π/2 cos θ dθ π/2 ke λ (sin θ) r 3π/2 π/2 ke λ , r q L and r = . Therefore, L π 2 ke q π L2 2 (8.98755 × 109 N m2 /C2 ) = (0.188 m)2 × (−8.63 × 10−6 C) π Ex = = −1.37885 × 107 N/C . Since the rod has negative charge, the field is pointing to the left (towards the charge distribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. Version 001 – Exam 1 – shih – (57480) 4 T cos θ = m g − q Ey T sin θ q Ex tan θ = = T cos θ m g − q Ey An Oblique Field 005 (part 1 of 2) 10.0 points A charged cork ball is suspended on a light string in the presence of a uniform electric field as in the figure. The ball is in equilibrium. (m g − q Ey ) tan θ = q Ex q Ex m g − q Ey = tan θ ~ = (2.5 × 105 N/C)ı̂ + (3.4 × 105 N/C) ̂ . E q Ex tan θ mg q Ex q= − Ey Ey tan θ m g tan θ q= Ey tan θ + Ex The acceleration of gravity is 9.8 m/s2 . q Ey = m g − ̂ 0. 5 ı̂ E m q= 34◦ 1.3 g (0.0013 kg) (9.8 m/s2 ) tan 34◦ (3.4 × 105 N/C) tan 34◦ + (2.5 × 105 N/C) = 17.9275 nC . Find the charge on the ball. 1. 11.1438 2. 6.80226 3. 11.5353 4. 12.0853 5. 15.6364 6. 17.9275 7. 19.3324 8. 9.39524 9. 4.79127 10. 10.6053 006 (part 2 of 2) 10.0 points Find the tension in the string. 1. 0.00976323 2. 0.00685318 3. 0.00384903 4. 0.00303988 5. 0.00432643 6. 0.00962097 7. 0.0080149 8. 0.00511639 9. 0.0086204 10. 0.00491469 Correct answer: 17.9275 nC. Explanation: Correct answer: 0.0080149 N. Let : Ex Ey θ mb 5 = 2.5 × 10 N/C , = 3.4 × 105 N/C , = 34◦ , and = 1.3 g = 0.0013 kg . Explanation: The first equation for force equilibrium gives In the ı̂ and ̂ directions, force equilibrium tells us q Ex − T sin θ = 0 q Ey + T cos θ − m g = 0 , T sin θ = q Ex (1) (2) q Ex sin θ (17.9275 nC) (2.5 × 105 N/C) = sin 34◦ = 0.0080149 N . T = Charge Inside a Box 02 Version 001 – Exam 1 – shih – (57480) 007 10.0 points A cubic box of side a, oriented as shown, contains an unknown charge. The vertically directed electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface. E 5 (inward is negative) and the bottom Φbottom = 2 E a2 . The total electric flux is ΦE = −E a2 + 2 E a2 = E a2 . Using Gauss’s Law, the charge inside the box is Qencl = ǫ0 ΦE = ǫ0 E a2 . a 2E How much charge Q is inside the box? 1. Qencl = 0 2. Qencl = ǫ0 E a2 correct 3. Qencl = 2 E ǫ0 a2 4. Qencl = 2 ǫ0 E a2 5. insufficient information E ǫ0 a2 E =3 ǫ0 a2 6. Qencl = 7. Qencl 8. Qencl = 3 ǫ0 E a2 9. Qencl = 6 ǫ0 E a2 1 ǫ0 E a2 2 Explanation: Electric flux through a surface S is, by convention, positive for electric field lines going out of the surface S and negative for lines going in. Here the surface is a cube and no flux goes through the vertical sides. The top receives 10. Qencl = Φtop = −E a2 Energy gained from A to B 008 10.0 points A proton is released from rest in a uniform electric field of magnitude 1.9 × 105 V/m directed along the positive x axis. The proton undergoes a displacement of 0.2 m in the direction of the electric field as shown in the figure. 1.9 × 105 V/m + − + − + − + − + − + − + vA = 0 v0 − + + − + − 0.2 m + − B − + A + − + − Apply the principle of energy conservation to find the amount of the kinetic energy gained after it has moved 0.2 m. 1. 8.65176e-15 2. 9.13241e-15 3. 9.61306e-15 4. 8.33132e-15 5. 1.92261e-15 6. 6.08827e-15 7. 1.53809e-14 8. 6.72914e-15 9. 4.80653e-15 10. 5.28719e-15 Correct answer: 6.08827 × 10−15 J. Explanation: Version 001 – Exam 1 – shih – (57480) The change in the potential energy of the proton is ∆U = qp ∆V = (1.60218 × 10−19 C) (−38000 V) = −6.08827 × 10−15 J . Conservation of energy in this case is ∆K + ∆U = Kf − Ki + ∆U = 0 . So Kf − Ki = −∆U = 6.08827 × 10 −15 J . Q r 2 9. Φ = ǫ0 R Q R 10. Φ = ǫ0 r Explanation: Gauss’ law states that I ~ · dA ~ = qin = Φ . E ǫ0 The net charge inside the sphere of radius r is qin = ρ Uniformly Charged Sphere 05 p 009 (part 1 of 2) 10.0 points Consider the uniformly charged sphere with radius R, shown in the figure below. Q is the total charge inside the sphere Find the total flux passing through the Gaussian surface (a spherical shell) with radius r. 2. Φ = 3. Φ = 4. Φ = 5. Φ = 6. Φ = 7. Φ = 8. Φ = Φ= Q r3 . ǫ0 R3 010 (part 2 of 2) 10.0 points Find the electric field at radius r . r Q ǫ0 Q r2 ǫ0 R3 3 Q R ǫ0 r Qr ǫ0 R Q R2 ǫ0 r 3 Q r 3 correct ǫ0 R 2 Q R ǫ0 r Q ǫ0 r 4 Q 4 r3 π r3 = π r3 = Q 3 , 4 3 3 R π R3 3 so the total flux through the surface is R p 1. Φ = 6 QR r3 Q R2 2. E = k r3 Q 3. E = k r Q 4. E = k 2 r Q r2 5. E = k R3 Q 6. E = k R Q 7. E = k 2 R Q R2 8. E = k r4 Qr 9. E = k 3 correct R Q r2 10. E = k R4 Explanation: 1. E = k Version 001 – Exam 1 – shih – (57480) 7 From Gauss’ law, 2. 6.77 eV 3. 101.3 eV 4. 709.1 eV 5. 4.37 eV 6. 5.14 eV correct q = 1.6 × 10−19 C , k = 8.99 × 109 N · m2 /C2 , d = 2.8 × 10−10 m . W = ∆K + ∆U = 0 − Ui k (−e)e k q2 =− = d d 9 (8.99 × 10 N · m2 /C2 ) = 2.8 × 10−10 m = 5.14 eV . − 5.8 × 10−9 C 1.2 cm Calculate the electric potential at the midpoint of the base if the magnitude of the positive charge is 4.4 × 10−9 C and the magnitude of the negative charges are 5.8 × 10−9 C. 1. -10456.8 2. -4738.42 3. -6892.22 4. -4802.96 5. -7145.57 6. -8588.8 7. -16657.1 8. -7645.36 9. -5124.04 10. -3688.97 Explanation: and Let : q1 = 4.4 × 10−9 C , q2 = q3 = −5.8 × 10−9 C , a = 5.5 cm , b = 1.2 cm , and ke = 8.98755 × 109 N · m2 /C2 . The energy is × (1.6 × 10 5.8 × 10−9 C − Correct answer: −16657.1 V. Explanation: Let : cm 1. 8.91 eV 5. 5 Tipler PSE5 23 24 011 10.0 points The distance between the K+ and Cl− ions in KCl is 2.8 × 10−10 m. Find the energy required to separate the two ions to an infinite distance apart, assuming them to be point charges initially at rest. The elemental charge is 1.6 × 10−19 C and the Coulomb constant is 8.99 × 109 N · m2 /C2 . 5. 5 so the electric field is Φ Q r3 1 E= = 4 π r2 ǫ0 R3 4 π r 2 Qr = 4 π ǫ0 R3 kQr = . R3 cm Holt SF 18Rev 39 012 10.0 points The three charges shown in the figure are located at the vertices of an isosceles triangle. The Coulomb constant is 8.98755 × 109 N · m2 /C2 and the acceleration of gravity is 9.8 m/s2 . 4.4 × 10−9 C + E · 4 π r2 = Φ , −19 C) 2 1 eV 1.6 × 10−19 J Let r1 , r2 and r3 be the distances of the respective charges from the midpoint of the base of the triangle. Version 001 – Exam 1 – shih – (57480) + a − r1 r2 r3 − b r2 = r3 = 2 s 2 b 1p 2 2 4a − b2 r1 = a − = 2 2 q r Then the total electric potential is V = ke Vtot = V1 + V2 + V3 q1 q2 q3 = ke + ke + ke r r2 r3 1 2 q2 2 q3 2 q1 + + = ke √ b b 4a2 − b2 q1 q2 + q3 = 2 ke √ + b 4a2 − b2 q1 2 q2 = 2 ke √ + b 4a2 − b2 9 = 2 (8.98755 × 10 N · m2 /C2 ) " 4.4 × 10−9 C × p 4 (0.055 m)2 − (0.012 m)2 2 (−5.8 × 10−9 C) + 0.012 m = −16657.1 V . Connected Spheres 02 013 (part 1 of 2) 10.0 points Two charged spherical conductors are connected by a long conducting wire, and a charge of 15 µC is placed on the combination. One sphere has a radius of 4.05 cm and the other a radius of 6.05 cm. What is the electric field near the surface of the sphere with the smaller radius? 1. 31520800.0 2. 40954200.0 8 3. 85309100.0 4. 36123200.0 5. 41620400.0 6. 44828500.0 7. 35028000.0 8. 29420600.0 9. 32957700.0 10. 19693800.0 Correct answer: 3.29577 × 107 V/m. Explanation: Let : r1 = 4.05 cm = 0.0405 m , r2 = 6.05 cm = 0.0605 m , Q = 15 µC = 1.5 × 10−5 C , and ke = 8.98755 × 109 N · m2 /C2 . Since the two spheres are connected by a long conducting wire, they are at the same potential, and q1 q2 = (1) r1 r2 We are also given the total charge of the combination Q = q1 + q2 (2) With two equations and two unknowns, we can solve for the charge q1 on the smaller sphere. From (1), r1 (Q − q1 ) r1 q2 = r2 r2 q1 r2 = r1 Q − r1 q1 q1 = q1 (r1 + r2 ) = r1 Q r1 Q q1 = . r1 + r2 Hence the electric field near the surface of the sphere with radius r1 is q1 E = ke 2 r1 Q = ke (r1 + r2 ) r1 = 8.98755 × 109 N · m2 /C2 × 1.5 × 10−5 C (0.0405 m + 0.0605 m) (0.0405 m) = 3.29577 × 107 V/m . Version 001 – Exam 1 – shih – (57480) 014 (part 2 of 2) 10.0 points What is the potential of the larger sphere? 1. 1051180.0 2. 1334780.0 3. 1158920.0 4. 2662000.0 5. 1517900.0 6. 3300280.0 7. 2015180.0 8. 1948630.0 9. 1781320.0 10. 2105990.0 Correct answer: 1.33478 × 106 V. Explanation: Since the two spheres are at the same potential, it suffices to consider the potential of either sphere. Since q1 was obtained in the first part of the question, we may simply determine the potential of the smaller sphere. V = ke q1 r1 Q r1 + r2 = 8.98755 × 109 N · m2 /C2 = ke × 1.5 × 10−5 C 0.0405 m + 0.0605 m = 1.33478 × 106 V . Tipler PSE5 24 19 015 10.0 points What is the electrostatic potential energy of an isolated spherical conductor of 10 cm radius that is charged to 1000 V? The Coulomb constant is 8.99 × 109 N · m2 /C2 . 1. 208.565 2. 16.6852 3. 11.1235 4. 44.4939 5. 13.9043 6. 266.963 7. 75.0834 8. 278.087 9. 5.56174 10. 100.111 9 Correct answer: 5.56174 µJ. Explanation: Let : r = 10 cm = 0.1 m , V = 1 kV = 1000 V , and k = 8.99 × 109 N · m2 /C2 . The potential on a spherical conductor is kQ r rV Q= , k so the electrostatic potential energy is V = 1 QV 2 1 rV = V 2 k rV2 = 2k 106 µJ (0.1 m) (1000 V)2 · = 2 (8.99 × 109 N · m2 /C2 ) 1J U= = 5.56174 µJ . Sphere Within a Shell A01 016 10.0 points Consider a solid conducting sphere with an inner radius R1 surrounded by a concentric thick conducting spherical shell which has an inner radius R2 and outer radius R3 . There is a charge Q on the sphere and no net charge on the shell. For this problem, we adopt the standard convention of setting the electric potential at infinity to zero. q2 = 0 R1 R2 Ob R3 b q1 = Q B b A Version 001 – Exam 1 – shih – (57480) Find the potential at O, where RO < R1 . 1. VO = 0 2. VO = kQ kQ kQ − + correct R3 R2 R1 3. VO = ∞ 10 for a thin shell of radius a. Thus the potential due to the inner shell (radius R2 ) is V2 = k qinner Q = −k R2 R2 and that due to the outer shell (radius R3 ) is kQ kQ + R3 R1 2kQ 5. VO = R1 kQ kQ 6. VO = + R2 R1 √ 2kQ 7. VO = R1 kQ 8. VO = R1 2kQ 9. VO = R1 + R2 √ 2 2kQ 10. VO = R1 Explanation: We are still inside the spherical shell. The potential due to the shell requires knowing something about where charge exists on the shell. To this end, consider a Gaussian spherical surface inside the shell. This surface can have no flux through it since no electric field can be upheld inside a conductor. By Gauss’s Law, there can therefore be no charge enclosed. But we are already enclosing Q on the sphere, so a charge −Q must reside on the inner surface of the shell: 4. VO = qinner = −Q . Since the net charge on the shell is zero, the charge on the outer surface must be +Q for the inner and outer charges to add up to zero: V3 = k Q qouter =k , R3 R3 so the contribution from the two surfaces of the shell is V1 = k Q Q −k . R3 R2 However, we are now looking at a point inside the sphere. The charge is all on the surface of the sphere, so similarly to the situation for the shell we have V =k Q R1 everywhere inside the sphere. Thus the total potential at O is VO = k Q Q Q −k +k . R3 R2 R1 This potential is the same at O as it is everywhere inside the sphere. A conductor is an equipotential object. Add a Charge to Four 3 017 (part 1 of 2) 10.0 points Four charges are placed at the corners of a square of side a, with q1 = q2 = −q, q3 = q4 = +q, where q is positive. Initially there is no charge at the center of the square. q2 = −q q3 = +q qouter = +Q . Now that we know the exact distribution of charge, we can utilize the expression for potential inside of a spherical charge distribution: q V =k , a q q1 = −q q4 = +q Version 001 – Exam 1 – shih – (57480) Find the work required to bring the charge q from infinity and place it at the center of the square. 1. W = 2. W = 3. W = 4. W = 5. W = 6. W = −2 k q 2 a2 −2 k q 2 a 4 k q2 a 2 k q2 a2 8 k q2 a2 −4 k q 2 a2 7. W = 0 correct −4 k q 2 a 4 k q2 9. W = a2 2 k q2 10. W = a Explanation: Based on the superposition principle, the potential at the center due to the charges at the corners is 8. W = V = V1 + V2 + V3 + V4 kq = (−1 − 1 + 1 + 1) = 0 , r where r is the common distance from the center to the corners. The work required to bring the charge q from infinity to the center is then W = q V = 0. 018 (part 2 of 2) 10.0 points What is the magnitude of the total electrostatic energy of the final 5 charge system? It may be useful to consider the symmetry property of the charge distribution which leads to cancellations among several terms. k q2 . 1. U = 4 a 11 √ k q2 2 . correct a √ k q2 . 3. U = 4 2 a k q2 4. U = 2 . a √ k q2 5. U = 2 2 . a √ k q2 6. U = 4 2 2 . a √ k q2 7. U = 2 2 2 . a 2 kq 8. U = 4 2 . a k q2 9. U = 2 2 . a k q2 10. U = 8 2 . a Explanation: No work is done in bringing in the fifth charge and placing it at the center, so the total electrostatic energy is contributed only by 4 corner charges: 2. U = U = U12 + U13 + U14 + U23 + U24 + U34 1 1 k q2 +1 − √ − 1 − 1 − √ + 1 = a 2 2 2 √ kq =− 2 . a Cylindrical Capacitor 03 v2 019 10.0 points A 75 m length of coaxial cable has a solid cylindrical wire inner conductor with a diameter of 1.168 mm and carries a charge of 5.21 µC. The surrounding conductor is a cylindrical shell and has an inner diameter of 6.552 mm and a charge of −5.21 µC. Assume the region between the conductors is air. The Coulomb constant is 8.98755 × 109 N · m2 /C2 . What is the capacitance of this cable? 1. 3.5117 2. 2.62668 3. 1.73136 Version 001 – Exam 1 – shih – (57480) 4. 1.01978 5. 2.41954 6. 2.89136 7. 1.15634 8. 1.9711 9. 2.26507 10. 2.05129 12 4.43 µF 8.41 µF a b 8.66 µF 1.84 µF c d 11.8 V Correct answer: 2.41954 nF. Explanation: Let : ke = 8.98755 × 109 N · m2 /C2 , Q = 5.21 µC , ℓ = 75 m , a = 1.168 mm , and b = 6.552 mm . The charge per unit length is λ ≡ Z Q . ℓ b ~ · d~s E a Z b dr = −2 ke λ a r b Q . = −2 ke ln ℓ a V =− Find the equivalent capacitance between points a and d. 1. 2.54995 2. 2.21028 3. 1.29119 4. 1.55152 5. 2.3927 6. 1.32972 7. 1.52131 8. 1.82451 9. 1.35359 10. 1.77146 Correct answer: 1.35359 µF. Explanation: Let : The capacitance of a cylindrical capacitor is given by Q V ℓ = 2 ke C≡ 1 b ln a 75 m = 2 (8.98755 × 109 N · m2 /C2 ) 1 1 × 109 nF · × 6.552 mm 1F ln 1.168 mm C1 C2 C3 C4 EB = 8.41 µF , = 4.43 µF , = 8.66 µF , = 1.84 µF , = 11.8 V . and C2 a C1 b C3 c C4 d EB For capacitors in series, = 2.41954 nF 1 Capacitor Circuit 04 shortest 020 (part 1 of 2) 10.0 points Consider the group of capacitors shown in the figure. Cseries Vseries X 1 C X i = Vi , = and the individual charges are the same. Version 001 – Exam 1 – shih – (57480) For parallel capacitors, X Cparallel = Ci X Qparallel = Qi , and the individual voltages are the same. In the given circuit C2 and C3 are connected parallel with equivalent capacitance C23 = C2 + C3 = 4.43 µF + 8.66 µF = 13.09 µF . C1 , C23 , and C4 are connected in series with equivalent capacitance −1 1 1 1 Cba = + + C1 C23 C4 −1 1 1 1 + + = 8.41 µF 13.09 µF 1.84 µF = 1.35359 µF . 021 (part 2 of 2) 10.0 points Determine the charge on the 4.43 µF capacitor at the top center part of the circuit. 1. 34.3037 2. 3.95722 3. 16.2366 4. 16.4891 5. 10.9564 6. 11.6865 7. 5.40545 8. 14.9974 9. 5.05784 10. 10.1086 Correct answer: 5.40545 µC. Explanation: The voltage drops across C1 and C4 are then V1 = Q1 15.9723 µC = = 1.8992 V C1 8.41 µF The remaining voltage is Vremain = Vtotal − V1 − V4 = 11.8 V − 1.8992 V − 8.6806 V = 1.22019 V . This remaining voltage is the same across the parallel capacitors, so Q2 = C2 Vremain = (4.43 µF) (1.22019 V) = 5.40545 µC . Two Dielectric Slabs v1 022 (part 1 of 3) 10.0 points A parallel plate capacitor has capacitance C0 , plate separation d and plate area A. Two dielectric slabs of dielectric constants κ1 and d κ2 , each of thickness , are inserted between 2 the plates. Charges Q and −Q are put on the left and right plates, respectively. dielectric κ2 constant left dielectric constant κ1 right V4 = Q4 15.9723 µC = = 8.6806 V . C4 1.84 µF ℓ A = ℓ2 d Without the dielectrics, what is the field E0 between the plates? 1. E0 = 2. E0 = 3. E0 = 4. E0 = 5. E0 = and 13 6. E0 = Q d 2 ǫ0 A Q 4 ǫ0 A 2Q d ǫ0 A Q correct ǫ0 A Q d 4 ǫ0 A Q 2ǫ0 A Version 001 – Exam 1 – shih – (57480) 2Q 7. E0 = ǫ0 A Explanation: Solution: Qencl . From Gauss’ Law, we know that Φ = ǫ0 For a flat surface, like a plate, the flux is the normal component of the electric field times the area. Also, for a parallel plate capacitor, the field is only in the gap (neglecting edge effects), so Φ = E0 A Q Φ = . E0 = A ǫ0 A 023 (part 2 of 3) 10.0 points With the dielectrics inserted, the fields in κ1 and κ2 are respectively What is the resultant capacitance? 1. 2.06894e-07 2. 1.44507e-08 3. 8.05613e-09 4. 1.06711e-08 5. 3.21764e-08 6. 4.75003e-08 7. 9.74561e-09 8. 1.68571e-08 9. 2.02877e-08 10. 2.86385e-08 Correct answer: 2.02877 × 10−8 F. Explanation: The potential difference between the two plates is the sum of the potential differences across the two dielectrics. The potential difference across a region with a constant electric field is the product of the field and the distance, so V = V1 + V2 E0 d E0 d = + κ1 2 κ 2 2 Q d 1 1 = . + ǫ0 A 2 κ1 κ2 1. E1 = E0 κ2 and E2 = E0 κ1 E0 E0 and E2 = κ2 κ1 E0 E0 3. E1 = and E2 = correct κ1 κ2 E0 κ2 E0 κ1 4. E1 = and E2 = κ1 + κ2 κ1 + κ2 2. E1 = 5. E1 = E0 κ1 and E2 = E0 κ2 E0 κ1 κ2 E0 κ1 κ2 and E2 = κ1 + κ2 κ1 + κ2 E0 κ1 E0 κ2 7. E1 = and E2 = κ1 + κ2 κ1 + κ2 Explanation: In any region with dielectric constant κ, the permittivity ǫ0 becomes κǫ0 , so the fields in κ1 and κ2 are 6. E1 = E1 = Q E0 = κ1 ǫ0 A κ1 E2 = Q E0 = . κ2 ǫ0 A κ2 024 (part 3 of 3) 10.0 points Let κ1 = 3, κ2 = 4.41, A = 2.74 m2 , and d = 4.27 mm. 14 The capacitance is given by Q V 2 ǫ0 A = d C= ! 1 1 1 κ1 + κ2 κ1 κ2 ǫ0 A =2 · κ + κ2 d 1 (3) (4.41) =2 3 + 4.41 8.854 × 10−12 F/m (2.74 m2 ) × 0.00427 m = 2.02877 × 10−8 F . Charge in a Wire 025 10.0 points The current in a wire decreases with time according to the relationship I = (2.32 mA) × e−a t Version 001 – Exam 1 – shih – (57480) where a = 0.13328 s−1 . Determine the total charge that passes through the wire from t = 0 to the time the current has diminished to zero. 1. 0.0114046 2. 0.017407 3. 0.0105792 4. 0.0216086 5. 0.0281363 6. 0.0112545 7. 0.0197329 8. 0.0277611 9. 0.0129802 10. 0.0296369 15 Correct answer: 5.83517 A. Explanation: Let : V = 0.59 V , ℓ = 1.3 m , A = 0.72 mm2 = 7.2 × 10−7 m2 , ρ = 5.6 × 10 −8 The resistance is R= Correct answer: 0.017407 C. VA ρℓ (0.59 V) (7.2 × 10−7 m2 ) = (5.6 × 10−8 Ω · m) (1.3 m) dq I= dt q= I= t I dt Zt=0 ∞ = 5.83517 A . (0.00232 A)e−0.13328 s t e−0.13328 s t = (0.00232 A ) −0.13328 s−1 ∞ = V ρℓ = , I A so the current is Explanation: Z Ω · m. −1 dt t=0 −1 0 = 0.017407 C . Current in Tungsten Wire 026 10.0 points A 0.59 V potential difference is maintained across a 1.3 m length of tungsten wire that has a cross-sectional area of 0.72 mm2 and the resistivity of the tungsten is 5.6 × 10−8 Ω · m. What is the current in the wire? 1. 5.83517 2. 1.72556 3. 11.873 4. 8.71763 5. 13.7143 6. 16.7545 7. 8.0 8. 1.62175 9. 3.69643 10. 14.5357 and