Ch17-h3-extra-solutions
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Ch17-h3-extra-solutions

Course Number: PHY Phy 303l, Spring 2013

College/University: University of Texas

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guajardo (jmg4256) Ch17-h3-extra chiu (57425) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points Consider a parallel plate capacitor system with plate charge Q (Q > 0) and cross section A of each plate. Top plate +Q d Bottom plate -Q Egap Plate area is A 1 Explanation: Recall from chapter 16 that the...

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(jmg4256) guajardo Ch17-h3-extra chiu (57425) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points Consider a parallel plate capacitor system with plate charge Q (Q > 0) and cross section A of each plate. Top plate +Q d Bottom plate -Q Egap Plate area is A 1 Explanation: Recall from chapter 16 that the magnitude of the electric field 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 field due to the top plate must be given by |Etop | = Q/2 0 A. By definition, this field 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 = Q2 /2 0 A. 002 0.0 points Given two parallel plate capacitors 1 and 2. They both have a plate-charge Q and platearea A. The gap of 1 is filled with a dielectric medium with dielectric constant K1 = 2 and the gap of 2 is filled with a dielectric constant K2 = 4. Determine the ratio of the polarized q1 charge . q2 1. 1 2. 3. 5 9 8 9 2 9 4 9 1 9 4 3 2 correct 3 1 3 7 9 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 field 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. 1. F = 2. F = 3. F = correct 4. F = 5. F = 6. F = Q2 , 0A Q2 , 0A Q2 2 0A Q2 , 2 0A Q2 0A Q2 Etop = Etop = Q , 0A Q , 0A Q 2 0A Etop Etop Etop , , 2 0A Etop = , Etop Etop 4. 5. 6. Etop Etop Etop Etop Etop = Etop = , Q , 0A Q 2 0A , , 7. 8. 9. Etop = Q 2 0A Etop Etop Etop Etop Etop Etop Etop 7. F = 8. F = Q2 2 0A Q2 0A , Etop = Etop = Q , 0A Q 0A 10. , 2 , Explanation: guajardo (jmg4256) Ch17-h3-extra chiu (57425) E Q = E Q-q 1 K 2 2. q = Q 1- V kQ 6R K = correct R 2R 3R r For capacitor 1, we know that K1 = 2 q1 = Q 1 - 1 2 = Q 2 3. V kQ 6R For capacitor 2, we know that K2 = 4 q2 = Q 1 - 1 4 = 3Q 4 003 0.0 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 . 4. V kQ 6R q1 (Q/2) 2 = = q2 (3Q/4) 3 R 2R 3R r 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 Qnet - Qinner = +Q - (-Q) = +2 Q . +Q +Q What potential vs radial distance diagram describes this situation? +Q Q on surface -Q on surface 2 Q on surface +Q kQ 6R 1. V R 2R 3R r The potential within a conductor is constant and the electric field within a conductor is zero. The potential for 3 R < R < (outside the conductors) is guajardo (jmg4256) Ch17-h3-extra chiu (57425) k(q + Q) correct Ro kQ k(q + Q) + 8. Ro r 7. 3 +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), Vr = k +Q -Q 2 Q + + r 2R 3R 1 1 = kQ . + r 6R 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 kQ r For 0 < r R (inside the conducting sphere), kQ 1 1 =7 + . VR = Vr = V0 = k Q R 6R 6R 004 (part 1 of 2) 0.0 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) r k(q + Q) Ro k(q + Q) Ri k(q + Q) Ro k(q + Q) Ro - kQ Ri Vb (r) = - Vc (r) = kQ r So, the resultant potential at r is V (r) = Va (r) + Vb (r) + Vc (r) k(q + Q) Ro V (r) = 005 (part 2 of 2) 0.0 points The potential at r in the region r < Ri , is given by 1. 2. 3. k(q + Q) Ro k(q + Q) Ro k(q + Q) Ro k(q + Q) r k(q + Q) Ri k(q + Q) Ro k(q + Q) Ro kQ Ri kQ kQ - + correct Ri r kQ + r + + kQ Ri 4. 5. kQ r kQ kQ - + Ri r - 6. 7. kQ Ri kQ - r - guajardo (jmg4256) Ch17-h3-extra chiu (57425) 8. k(q + Q) Ro 4 Explanation: The potentials contributed by three sources for r < Ri are Va (r) = k(q + Q) Ro kQ Ri Let : e = 1.602 10-19 C , q = +2 e = 3.204 10-19 C , qnucl = +79 e = 1.26558 10-17 C , m = 6.64 10-27 kg , vi = 1.25 107 m/s , and ke = 8.98755 109 N m2 /C2 . By conservation of energy (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 1 q qnucl 2 m vi = ke 2 rturn 2 ke q qnucl rturn = 2 m vi = 2 8.98755 109 N m2 /C2 3.204 10-19 C 6.64 10-27 kg 1.26558 10-17 C (1.25 107 m/s)2 Vb (r) = - Vc (r) = kQ r So, the resultant potential at r is V (r) = Va (r) + Vb (r) + Vc (r) k(q + Q) kQ kQ - + Ro Ri r V (r) = 006 0.0 points In Rutherford's famous scattering experiments (which led to the planetary model of the atom), alpha particles (having charges of +2 e and masses of 6.64 10-27 kg) were fired toward a gold nucleus with charge +79 e. An alpha particle, initially very far from the gold nucleus, is fired at 1.25 107 m/s directly toward the gold nucleus. 2e ++ v=0 = 7.02531 10-14 m . 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: 7.02531 10-14 m. Explanation: + + + 79e + + 007 0.0 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 Gauss's Law, it can be shown that the electric field inside uniformly charged sphere of radius R and charge Q is E = kQr R3 + + + + Calculate the potential (relative to infinity) the at center of the nucleus. 1. 3 kZe correct 2 R guajardo (jmg4256) Ch17-h3-extra chiu (57425) 1 kZe 2. - 2 R 1 kZe 3. 2 R (kZe)2 4. - R kZe 5. R (kZe)2 6. R 3 kZe 7. - 2 R kZe 8. - R Explanation: 0 5 Find the potential at A. OA = a and OB = b. Consider the potential at zero to be infinity. 1. VA = 2. VA = 3. VA = 4. VA = 5. VA = 6. VA = 7. VA = E.dr 0 Vr=0 = - R 8. VA = E.dr R 0 Vr=0 = - Vr=0 = Vr=0 E.dr + 9. VA = kQ kQ + R3 a 2kQ R1 kQ kQ + R2 a 2 2kQ a kQ kQ - + R2 R1 2kQ a kQ kQ - + R3 R2 2kQ a kQ correct a kQ a kQ a kZe - R R kZer dr R3 r2 2 0 R 10. VA = 0 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 Q . a kZe kZe = - 3 R R Vr=0 = 3 kZe 2 R 008 (part 1 of 3) 0.0 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 = 0 009 (part 2 of 3) 0.0 points Determine the potential at B. kQ kQ kQ - + correct R3 R2 b 2kQ 2. VB = b 1. VB = 3. VB = 0 R1 R2 Q O B A R3 kQ kQ kQ - + R2 R1 b 2kQ 5. VB = b 4. VB = guajardo (jmg4256) Ch17-h3-extra chiu (57425) 2 2kQ 6. VB = b kQ kQ 7. VB = + R2 b kQ kQ 8. VB = + R3 b kQ 9. VB = b 2kQ 10. VB = R1 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 Q . b The inner surface of the shell carries an induced charge of q2 = -Q so its potential is V1 = k V2 = k q2 Q = -k . R2 R2 3. VO = 0 4. VO = 5. VO = 6. VO = 7. VO = 8. VO = 9. VO = 10. VO = kQ kQ kQ - + correct R3 R2 R1 2kQ R 1 2 2kQ R1 kQ kQ + R2 R1 kQ kQ + R3 R1 2kQ R1 6 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 sphere's contribution now V1 = Q k , so the total potential is R1 The outer surface of the shell carries a charge of q2 = q2 - q2 = Q , so its potential is q2 Q =k . R3 R3 Thus the total potential is V3 = k VB = V1 + V2 + V3 Q Q Q -k +k . =k b R2 R3 010 (part 3 of 3) 0.0 points Find the potential at O, the center of the system. kQ R1 2kQ 2. VO = R1 + R2 1. VO = VA = V1 + V2 + V3 Q Q Q =k -k +k . R1 R2 R3 011 0.0 points Consider the following steps: a) An isolated capacitor has a dielectric slab between its plates; b) The capacitor is charged by a battery; c) After the capacitor is charged, the battery is removed; d) The dielectric slab is then moved half way out of the capacitor; e) Finally, the dielectric is released and is set free to move on its own. guajardo (jmg4256) Ch17-h3-extra chiu (57425) 7 between the plates, the voltage across the capacitor drops to 36 V. What is the dielectric constant of this glass? (Assume the glass completely fills the space between the plates) Correct answer: 1.66667. The dielectric will 1. be pulled back into the capacitor. correct 2. be pushed out of the capacitor. 3. remain in place. Explanation: The capacitance of a capacitor with a dielectric slab is Cin = Cout , where > 1 . Explanation: Given : V0 = 60 V and Vd = 36 V . The capacitance is C= A Q = , d V so the dielectric constant is = Cf Vf 60 V = = = 1.66667 . Ci Vi 36 V NOTE When the battery is removed, the charge on the plates of the capacitor will remain constant. Charge is neither created nor destroyed. 1 Q2 , and 2 Cout 1 Q2 1 Q2 1 = = = Uout 2 Cin 2 Cout < Uout , 013 (part 1 of 4) 0.0 points An isolated large-plate capacitor (not connected to anything) originally has a potential difference of 900 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 figure below. 2 mm + + + + + + + + 1 - - - - - - - - 4 Uout = Uin where Uout is with an air-filled gap and Uin is with a dielectric-filled 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. 012 0.0 points A capacitor with air between its plates is charged to 60 V and then disconnected from the battery. When a piece of glass is placed 2 3 0.5 mm 1 mm 0.5 mm Calculate the potential difference V1 - V2 . Correct answer: 225 V. Explanation: Let's use the notation V12 V1 - V2 for this problem. The E-field 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 guajardo (jmg4256) Ch17-h3-extra chiu (57425) voltage across the vacuum gaps remains the same. Since the E-field 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 V12 = V0 0.5 mm 2 mm 0.5 mm = (900 V) 2 mm = 225 V . 014 (part 2 of 4) 0.0 points Now find the potential difference V2 - V3 . Correct answer: 112.5 V. Explanation: This process is similar to part 1, although we have to be sure and include the dielectric constant, since V = We have V23 = V0 1 mm 2 mm 8 V14 = V12 + V23 + V34 = (225 V) + (112.5 V) + (225 V) = 562.5 V . 017 0.0 points Three charges are situated at three corners of a rectangle, as shown. 7.9 C + 5.7 cm 3.7 cm + 1.9 C + 4.7 C Vvacuum . K How much electrical potential energy would be expended in moving the 7.9 C charge to infinity? The value of the Coulomb constant is 8.98755 109 N m2 /C2 and the acceleration due to gravity is 9.8 m/s2 . Correct answer: 8.55668 J. Explanation: K 1 mm 900 V 2 mm = 4 = 112.5 V . Let : 015 (part 3 of 4) 0.0 points Find V3 - V4 . Correct answer: 225 V. Explanation: Since this situation is exactly the same as the situation in part 1, the answers are the same. 016 (part 4 of 4) 0.0 points Find V1 - V4 . Correct answer: 562.5 V. Explanation: Here we simply add the potential differences we've already found: r1,2 = a q1 = 7.9 C , q2 = 1.9 C , q3 = 4.7 C , a = 3.7 cm , b = 5.7 cm , and ke = 8.98755 109 N m2 /C2 . and r1,3 = a 2 + b2 , so U1,tot = U1,2 + U1,3 = ke = ke q1 q1 q2 q1 q3 + ke r1,2 r1,3 q2 q3 + 2 + b2 a a = (8.98755 109 N m2 /C2 ) (7.9 10-6 C) 1.9 10-6 C 0.037 m + (0.037 m)2 + (0.057 m)2 4.7 10-6 C guajardo (jmg4256) Ch17-h3-extra chiu (57425) = 8.55668 J . 9

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