This preview has intentionally blurred parts. Sign up to view the full document

View Full Document

Unformatted Document Excerpt

(dea457) sarceno Ch17-h3-extra yao (57465) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Consider a parallel plate capacitor system with plate charge Q (Q > 0) and cross section A of each plate. Top plate +Q Egap d Bottom plate 1. F = Q2 , 2 0A 2. F = Q2 , 0A Etop = Q2 3. F = , 0A Etop = 4. F = 5. F = 2 2 0A Q2 0A , , Etop Q , 0A Etop = Etop Q 2 0A , Etop Etop Q , 0A Etop Etop Q = , 0A Q Etop = 2 0A Q2 , 2 0A Q Etop = 2 0A Etop Etop , correct 6. F = Explanation: Recall from chapter 16 that the magnitude of the electric eld between two oppositely charged parallel conducting plates of area A with charge magnitude |Q| is given by |Ecap | = Q/ 0 A, and that this is a superposition of the (equal) contributions of both plates. Therefore, the magnitude of the eld due to the top plate must be given by |Etop | = Q/2 0 A. By denition, this eld must point from the positive to the negative plate, so the direction is downward. Finally, the force is simply obtained from F = qE , so F = Q 2 /2 0 A . Plate area is A Q Denote the magnitude of the attractive force by which the charge of the top plate pulls on the charge of the bottom plate by F , and denote the electric eld vector due to the charges on the top plate by Etop . Find i) the magnitude F of the force; ii) the magnitude of Etop at the bottom plate; and iii) the direction of Etop . indicates downward and indicates upward. Q2 , 4. 1 Etop Etop 002 10.0 points Given two parallel plate capacitors 1 and 2. They both have a plate-charge Q and platearea A. The gap of 1 is lled with a dielectric medium with dielectric constant K1 = 2 and the gap of 2 is lled with a dielectric constant K2 = 4. Determine the ratio of the polarized q1 charge . q2 4 1. 9 4 2. 3 1 3. 3 5. 6. Etop Etop Etop 7. 8. 9. 7. F = 8. F = Q2 0A Q2 0A 1 , Etop = , Etop = Etop Q , 0A Q 2 0A , Etop Etop Etop 10. 8 9 7 9 2 9 5 9 2 correct 3 1 9 Explanation: E Q = E Qq q = Q 1 2. 1 K V K= 2 kQ 6R sarceno (dea457) Ch17-h3-extra yao (57465) r R 2R 3R correct q1 = Q 1 1 2 Q 2 = 3. q2 = Q 1 1 4 = V For capacitor 2, we know that K2 = 4 kQ 6R For capacitor 1, we know that K1 = 2 R 2R 3R 3Q 4 kQ 6R q1 (Q/2) 2 = = q2 (3Q/4) 3 r V 4. 003 10.0 points Consider a conducting sphere with radius R and charge +Q , surrounded by a conducting spherical shell with inner radius 2 R, outer radius 3 R and net charge +Q . R 2R 3R r Explanation: The charge on the inner sphere is +Q , concentrated on its surface. The induced charge on the inner surface of the spherical shell is Q , so the charge on the outer surface of the spherical shell is +Q Qnet Qinner = +Q (Q) = +2 Q . +Q What potential vs radial distance diagram describes this situation? +Q kQ 6R +Q V 1. R 2R 3R r Q on surface Q on surface 2 Q on surface The potential within a conductor is constant and the electric eld within a conductor is zero. The potential for 3 R < R < (outside the conductors) is sarceno (dea457) Ch17-h3-extra yao (57465) +Q + (+Q) Vr = k =k r 2Q r . For 2 R r 3 R (inside the conducting shell), V3R = Vr = V2R = k 2Q 3R . For R < r < 2 R (between the conductors), +Q Q 2 Q + + r 2R 3R 1 1 = kQ . + r 6R Vr = k 3 k (q + Q) Ri kQ k (q + Q) 8. Ro r 7. Explanation: There are 3 sources of charge a) q+Q distributed at the outer shell b) -Q at the inner shell c) the point charge Q at O. In the region Ri < r < Ro, the contributions from the three sources are Va (r ) = k (q + Q) Ro For 0 < r R (inside the conducting sphere), Vb (r ) = kQ 1 1 =7 + . VR = Vr = V0 = k Q R 6R 6R Vc (r ) = 004 (part 1 of 2) 10.0 points A thick conducting shell with inner radius Ri and outer radius Ro is centered at O, the origin. There is a point charge +Q located at O. The equilibrium charge distributions on the shell is as follows. There is -Q uniformly distributed on the inner surface of the shell, and Q+q uniformly distributed on the outer surface of the shell. Notice here we assume that the net charge on the thick shell is q. The potential at r in the region Ri < r < Ro , is given by 1. 2. 3. 4. 5. 6. k (q + Q) Ro k (q + Q) Ro k (q + Q) r k (q + Q) Ro k (q + Q) Ro k (q + Q) Ro kQ kQ + Ri r kQ Ri V (r ) = Va (r ) + Vb (r ) + Vc (r ) V (r ) = k (q + Q) Ro 005 (part 2 of 2) 10.0 points The potential at r in the region r < Ri , is given by 1. 2. 3. kQ r 5. correct 6. kQ Ri 7. + kQ r So, the resultant potential at r is 4. + kQ r k (q + Q) Ri k (q + Q) Ro k (q + Q) r k (q + Q) Ro k (q + Q) Ro k (q + Q) Ro k (q + Q) Ro kQ kQ + correct Ri r kQ r kQ + r kQ + Ri sarceno (dea457) Ch17-h3-extra yao (57465) 8. k (q + Q) kQ Ro Ri Explanation: The potentials contributed by three sources for r < Ri are Va (r ) = k (q + Q) Ro Vb (r ) = 4 Let : e = 1.602 1019 C , q = +2 e = 3.204 1019 C , qnucl = +79 e = 1.26558 1017 C , m = 6.64 1027 kg , vi = 1.44 107 m/s , and ke = 8.98755 109 N m2 /C2 . By conservation of energy kQ Ri (KE + U )i = (KE + U )f . 1 Since the potential varies as and distance the alpha particle is initially very far from the nucleus, we may approximate Ui = 0. At the turning point, the speed of the alpha particle is zero which implies that Kf = 0, so kQ Vc (r ) = r So, the resultant potential at r is V (r ) = Va (r ) + Vb (r ) + Vc (r ) V (r ) = 1 q qnucl 2 m vi = ke 2 rturn 2 ke q qnucl rturn = 2 m vi k (q + Q) kQ kQ + Ro Ri r = 2 8.98755 109 N m2 /C2 006 10.0 points In Rutherfords famous scattering experiments (which led to the planetary model of the atom), alpha particles (having charges of +2 e and masses of 6.64 1027 kg) were red toward a gold nucleus with charge +79 e. An alpha particle, initially very far from the gold nucleus, is red at 1.44 107 m/s directly toward the gold nucleus. v=0 + 2e ++ + + + 79e ++ + + + d How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary. The fundamental charge is 19 1.602 10 C and the Coulomb constant is 8.98755 109 N m2 /C2 . Correct answer: 5.29371 1014 m. Explanation: 3.204 1019 C 6.64 1027 kg 1.26558 1017 C (1.44 107 m/s)2 = 5.29371 1014 m . 007 10.0 points Three charges are situated at three corners of a rectangle, as shown. 8.5 C + 7.0 cm 3.1 cm + + 1.9 C 4.9 C How much electrical potential energy would be expended in moving the 8.5 C charge to innity? The value of the Coulomb constant is 8.98755 109 N m2 /C2 and the acceleration due to gravity is 9.8 m/s2 . sarceno (dea457) Ch17-h3-extra yao (57465) Correct answer: 9.57179 109 J. q1 = 8.5 C , q2 = 1.9 C , q3 = 4.9 C , a = 3.1 cm , b = 7.0 cm , and ke = 8.98755 109 N m2 . r1,2 /C2 = a and a 2 + b2 , r1,3 = U1,tot = U1,2 + U1,3 = ke so q1 q3 q1 q2 + ke r1,2 r1,3 q2 q3 = ke q1 + a a 2 + b2 = (8.98755 109 N m2 /C2 ) (8.5 106 C) 1.9 106 C 0.031 m + 1 kZe 2R kZe 4. R 1 kZe 5. 2R (kZe)2 6. R kZe 7. R (kZe)2 8. R 3. Explanation: Let : 4.9 106 C (0.031 m)2 + (0.07 m)2 = 9.57179 109 J . 008 10.0 points A nucleus contains Z protons that on average are uniformly distributed throughout a tiny sphere of radius R. Assume that there are no electrons or other charged particles in the vicinity of this bare nucleus. Using Gausss Law, it can be shown that the electric eld inside uniformly charged sphere of radius R and charge Q is Explanation: 0 Vr=0 = Calculate the potential (relative to innity) at the center of the nucleus. E.dr R Vr=0 = 0 E.dr + Vr=0 = 0 kZe R R kZer dr R3 kZe kZe 3 = R R Vr=0 E.dr R Vr=0 = 0 r2 2 R 3 kZe 2R 009 (part 1 of 3) 10.0 points Consider a solid conducting sphere with an inner radius R1 and total charge Q surrounded by a concentric thick conducting spherical shell of inner radius R2 and outer radius R3 and no net charge. q2 = 0 kQr E= R3 3 kZe correct 1. 2R 3 kZe 2. 2R 5 R1 R2 R3 Q O B A sarceno (dea457) Ch17-h3-extra yao (57465) Find the potential at A. OA = a and OB = b. Consider the potential at zero to be innity. 1. VA = 2. VA = 3. VA = 4. VA = 7. VA = 8. VA = 9. VA = 10. VA = kQ kQ + R3 R2 kQ kQ + R2 R1 2 2kQ a kQ correct a kQ kQ + R2 a 6. VB = 0 kQ kQ + R3 b kQ 8. VB = b 2kQ 9. VB = R1 2kQ 10. VB = b Explanation: B is between the shell and the sphere. Consider a Gaussian surface through B concentric to the system. Let us start from the inside and use superposition to add contributions as we go outward. Outside of the sphere, we can treat its charge Q as a point charge, and its potential is 7. VB = 2kQ R1 2kQ a kQ kQ + R3 a 2kQ a 5. VA = 0 6. VA = 6 kQ a kQ a Q . b The inner surface of the shell carries an induced charge of q2 = Q so its potential is V1 = k Explanation: A is outside of the entire charge distribution a distance a from the center, so the enclosed charge Qencl = Q + q2 = Q can be treated as a point charge, and VC = k V2 = k The outer surface of the shell carries a charge of q2 = q2 q2 = Q , so its potential is Q . a 010 (part 2 of 3) 10.0 points Determine the potential at B . 2kQ 1. VB = b kQ kQ kQ 2. VB = + R2 R1 b kQ kQ + 3. VB = R2 b 2 2kQ 4. VB = b kQ kQ kQ + correct 5. VB = R3 R2 b Q q2 = k . R2 R2 V3 = k Q q2 =k . R3 R3 Thus the total potential is VB = V1 + V2 + V3 Q Q Q =k +k . k b R2 R3 011 (part 3 of 3) 10.0 points Find the potential at O , the center of the system. 1. VO = 2. VO = kQ kQ + R3 R1 sarceno (dea457) Ch17-h3-extra yao (57465) 3. VO = 4. VO = 5. VO = 6. VO = 7. VO = 8. VO = 7 2kQ R 1 2 2kQ R1 2kQ R1 + R2 2kQ R1 kQ kQ + R2 R1 kQ kQ kQ + correct R3 R2 R1 9. VO = 0 kQ 10. VO = R1 Explanation: O is inside the sphere, so we expect all contributions to have a constant radius R1 , R2 or R3 in the denominator. (A conducting body is an equipotential body.) The last two contributions are the same, with the spheres contribution now V1 = Q , so the total potential is k R1 The dielectric will 1. remain in place. 2. be pushed out of the capacitor. 3. be pulled back into the capacitor. correct Explanation: The capacitance of a capacitor with a dielectric slab is Cin = Cout , where > 1 . NOTE When the battery remains connected, the electric potential dierence across the plates of the capacitor will remain constant. The dierence in the potential energy stored in the capacitor is 1 1 Qout V Qin V 2 2 VA = V1 + V2 + V3 1 1 = Qout V Qout V Q Q Q 2 2 =k k +k . R1 R2 R3 1 = (1 ) Qout V , 2 and the energy drained from the battery is Ubat = (Qout Qin ) V 012 10.0 points = (1 ) Qout V , Consider the following steps: so the total energy dierence is a) An isolated capacitor has a dielectric slab 1 Uin Uout = (1 ) Qout V between its plates; 2 b) The capacitor is charged by a battery; (1 ) Qout V c) After the capacitor is charged, the battery 1 1 remains connected; Qout Qout V = 2 2 d) The dielectric slab is then moved half way 1 out of the capacitor; = ( 1) Qout V = ( 1) Uout , e) Finally, the dielectric is released and is set 2 free to move on its own. Uin < Uout , Ucap = so sarceno (dea457) Ch17-h3-extra yao (57465) where Uout is with an air-lled gap and Uin is with a dielectric-lled gap. A system will move to a position of lower potential energy. After the dielectric is moved half way out of the capacitor, the potential energy stored in the capacitor will be larger than it would have been with the dielectric left in place. Therefore, the dielectric will be pulled back into the capacitor. 013 10.0 points A capacitor with air between its plates is charged to 100 V and then disconnected from the battery. When a piece of glass is placed between the plates, the voltage across the capacitor drops to 43 V. What is the dielectric constant of this glass? (Assume the glass completely lls the space between the plates) Correct answer: 2.32558. Explanation: Given : V0 = 100 V Vd = 43 V . and 8 2 mm + + + + + + + + 1 2 3 4 0.5 mm 1 mm 0.5 mm Calculate the potential dierence V1 V2 . Correct answer: 200 V. Explanation: Lets use the notation V12 V1 V2 for this problem. The E-eld in the regions 1-2 and 3-4 is solely due to the plates. Since the charge on the plates stays the same, then the voltage across the vacuum gaps remains the same. Since the E-eld is constant in these regions, the potential is linear and you can use the ratio of the width of the region to the width of the plates. Thus 0.5 mm 2 mm 0.5 mm = (800 V) 2 mm V12 = V0 The capacitance is C= A Q =, d V so the dielectric constant is = Vf Cf 100 V = = = 2.32558 . Ci Vi 43 V 014 (part 1 of 4) 10.0 points An isolated large-plate capacitor (not connected to anything) originally has a potential dierence of 800 V with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 4, is inserted into the middle of the air gap as shown in the gure below. = 200 V . 015 (part 2 of 4) 10.0 points Now nd the potential dierence V2 V3 . Correct answer: 100 V. Explanation: This process is similar to part 1, although we have to be sure and include the dielectric constant, since V = We have Vvacuum . K sarceno (dea457) Ch17-h3-extra yao (57465) V23 = V0 1 mm 2 mm K 1 mm 800 V 2 mm = 4 = 100 V . 016 (part 3 of 4) 10.0 points Find V3 V4 . Correct answer: 200 V. Explanation: Since this situation is exactly the same as the situation in part 1, the answers are the same. 017 (part 4 of 4) 10.0 points Find V1 V4 . Correct answer: 500 V. Explanation: Here we simply add the potential dierences weve already found: V14 = V12 + V23 + V34 = (200 V) + (100 V) + (200 V) = 500 V . 9 ... View Full Document

End of Preview

Sign up now to access the rest of the document