Answer, Key Homework 10 David McIntyre 1

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1 Answer, Key Homework 10 Dvid McIntyre 1 This print-out should hve 22 questions, check tht it is complete. Multiple-choice questions my continue on the next column or pge: find ll choices efore mking your selection. The due time is Centrl time. Chpter 26 prolems. 001 (prt 1 of 1) 0 points A prllel-plte cpcitor is chrged y connecting it to ttery. If the ttery is disconnected nd the seprtion etween the pltes is incresed, wht will hppen to the chrge on the cpcitor nd the electric potentil cross it? 1. The chrge increses nd the electric potentil decreses. 2. The chrge decreses nd the electric potentil increses. 3. The chrge nd the electric potentil increse. 4. The chrge decreses nd the electric potentil remins fixed. 5. The chrge nd the electric potentil decrese. 6. The chrge remins fixed nd the electric potentil increses. correct 7. The chrge nd the electric potentil remin fixed. 8. The chrge increses nd the electric potentil remins fixed. 9. The chrge remins fixed nd the electric potentil decreses. Chrge is conserved, so it must remin constnt since it is stuck on the pltes. With the ttery disconnected, Q is fixed. C ɛ A d A lrger d mkes the frction smller, so C is smller. Thus the new potentil V Q C is lrger. 002 (prt 1 of 2) 5 points Given: A cpcitor network is shown elow. 14 µf 20 µf 7 µf 116 V Find the equivlent cpcitnce C etween points nd for the group of cpcitors. Correct nswer: µf. Given : C 1 14 µf, C 2 20 µf, C 3 7 µf, E B 116 V. C 1 C 2 C 3 E B nd The cpcitors C 1 nd C 2 re in series, so ( 1 C s + 1 ) 1 C 1 C 2 C 1 C 2. C s is prllel with C 3, so C C s + C 3 C 1 C 2 + C 3 (14 µf) (20 µf) 14 µf + 20 µf + 7 µf µf + 7 µf µf.

2 Answer, Key Homework 10 Dvid McIntyre (prt 2 of 2) 5 points Wht chrge is stored on the 7 µf cpcitor on the lower portin of the prllel circuit? Correct nswer: 812 µc. Since C s nd C 3 re prllel, the sme potentil E B V is cross oth, so q 3 C 3 V (7 µf) (116 V) 812 µc. 004 (prt 1 of 2) 5 points Four cpcitors re connected s shown in the figure. c 97 V 21 µf 68 µf d 35 µf 84 µf Find the cpcitnce etween points nd. Correct nswer: µf. A good rule of thum is to eliminte junctions connected y zero cpcitnce. C 1 C 4 C 2 C 3 The definition of cpcitnce is C Q V. The series connection of C 2 nd C 3 gives the equivlent cpcitnce 1 C C 2 C 3 C 2 C 3 C 2 + C 3 (35 µf) (68 µf) 35 µf + 68 µf µf. The totl cpcitnce C etween nd cn e otined y clculting the cpcitnce in the prllel comintion of the cpcitors C 1, C 4, nd C 23 ; i.e., C C 1 + C 4 + C µf + 84 µf µf µf. Given : C 1 21 µf, C 2 35 µf, C 3 68 µf, C 4 84 µf, E 97 V. C 1 c nd 005 (prt 2 of 2) 5 points Wht is the chrge on the 35 µf uppercentered cpcitor? Correct nswer: µc. The voltges cross C 2 nd C 3, respectively, (the voltge etween nd ) re V V V, nd we hve E B C 3 d C 2 C 4 Q 23 Q 3 Q 2 V C 23 (97 V) ( µf) µc. 006 (prt 1 of 2) 0 points

3 Answer, Key Homework 10 Dvid McIntyre 3 Two cpcitors of 21 µf nd 4.1 µf re connected prllel nd chrged with 131 V power supply. Clculte the totl energy stored in the two cpcitors. Correct nswer: J V 4.78 µf 5.6 µf 13.8 µf U C V 2 2 C prllel. Ech cpcitor hs voltge V, so U 1 2 ( ) V 2 Wht is effective cpcitnce C entire cpcitor network? Correct nswer: µf. Given: of the 1 (21 µf µf) (131 V) J. 007 (prt 2 of 2) 0 points Wht potentil difference would e required cross the sme two cpcitors connected in series in order for the comintion to store the sme energy s in the first prt? Correct nswer: V. When in series the equivlent cpcitnce is E B C µf, C µf, C µf, E B 17.1 V. nd C 1 C 2 C 3 C series C 1 C 2 (21 µf) (4.1 µf) 21 µf µf µf. Since U C eq V 2, we hve 2 V 2 U C eq 2 ( J) µf 106 µf 1 F V. 008 (prt 1 of 7) 2 points Given: A cpcitor network is shown in the following figure. C 1 nd C 2 re in series with ech other, nd they re together re prllel with C 3. So C C 1 C 2 + C 3 (4.78 µf) (5.6 µf) µf 4.78 µf µf µf. 009 (prt 2 of 7) 2 points Wht is the voltge cross the 5.6 µf upper right-hnd cpcitor? Correct nswer: V. Since C 1 nd C 2 re in series they crry the sme chrge C 1 V 1 C 2 V 2,

4 Answer, Key Homework 10 Dvid McIntyre 4 nd their voltges dd up to V, voltge of the ttery V 1 + V 2 V C 2 V 2 + V 2 V C 1 C 2 V 2 + C 1 V 2 V C 1 V 2 V C 1 (17.1 V)(4.78 µf) 4.78 µf µf V. 010 (prt 3 of 7) 2 points If dielectric of constnt 4.18 is inserted in the 5.6 µf top right-hnd cpcitor (when the ttery is connected), wht is the electric potentil cross the 4.78 µf top left-hnd cpcitor? Correct nswer: V. Given : κ When the dielectric is inserted, the cpcitnce formerly C 2 ecomes C 2 κ C 2, nd the new voltge cross C 1 is V 1 V C 2 C 1 + C 2 κ V C 2 C 1 + κ C 2 (4.18)(17.1 V)(5.6 µf) 4.78 µf + (4.18)(5.6 µf) V. 011 (prt 4 of 7) 1 points If the ttery is disconnected nd then the dielectric is removed, wht is the chrge on 4.78 µf top left-hnd cpcitor? Correct nswer: µc. Immeditely efore the ttery ws disconnected the chrges on the cpcitors hd een Q 3 C 3 V (13.8 µf)(17.1 V) µc Q 1 Q 2 C 12V ( µf)(17.1 V) µc. When we remove the dielectric, the sum of the chrges stys the sme, nd the voltges on C 3 nd on C 12 (where C 12 is the equivlent cpcitnce of C 1 nd C 2 in series) re equl to ech other Therefore Q 1 + Q 3 Q 1 + Q 3 Q 1 Q 3. C 12 C 3 Q 1 Q 1 + Q C 3 C µc µc 13.8 µf µf µc. 012 (prt 5 of 7) 1 points Wht is now the voltge drop cross the 5.6 µf top right-hnd cpcitor? Correct nswer: V. V 2 Q 2 C 2 Q 1 C µc 5.6 µf V. 013 (prt 6 of 7) 1 points Wht ws the energy stored in the system

5 Answer, Key Homework 10 Dvid McIntyre 5 efore the dielectric ws removed? Correct nswer: J. The totl cpcitnce of the system efore the dielectric ws removed hd een C C 12 + C µf µf µf so the energy stored in the system ws U 1 2 C V µj J. 014 (prt 7 of 7) 1 points Wht is the energy stored in the system fter the dielectric ws removed? Correct nswer: J. Since the ttery is disconnected, the totl chrge is conserved: Q Q C V ( µf)(17.1 V) µc. Therefore, the energy stored is U Q 2 2 C µj J. 015 (prt 1 of 1) 0 points A sheet of mic is inserted etween the pltes of n isolted chrged prllel-plte cpcitor. Which of the following sttements is true? 1. The potentil difference cross the cpcitor decreses. correct 2. The energy of the cpcitor does not chnge. 3. The cpcitnce decreses. 4. The chrge on the cpcitor pltes decreses. 5. The electric field etween the cpcitor pltes increses. Since the cpcitor is isolted, the chrge on the cpcitor pltes remins the sme. On the other hnd, the cpcitnce is incresed y introducing dielectric. Therefore, from V Q, the potentil difference cross the C cpcitor is decresed. 016 (prt 1 of 1) 0 points A prllel plte cpcitor is ttched to ttery which mintins constnt potentil difference of V etween the pltes. While the ttery is still connected, glss sl is inserted so s to just fill the spce etween the cpcitor pltes. V The stored energy will 1. remin the sme. 2. increse. correct 3. decrese. The energy stored in the cpcitor is given y U V Q2 2 κc κ C 2 V 2 Without the glss, the stored energy is U C V 2 After inserting the glss, it ecomes U 1 2 κ C V 2

6 Answer, Key Homework 10 Dvid McIntyre 6 Since κ > 1, the stored energy will increse. 017 (prt 1 of 6) 2 points A prllel-plte cpcitor hs plte re of 105 cm 2 nd plte seprtion of 2.25 mm. A potentil difference of 5.87 V is pplied cross the pltes with only ir etween the pltes. The ttery is then disconnected, nd piece of glss (κ 8.53) is inserted to completely fill the spce etween the pltes. Wht is the cpcitnce efore the dielectric is inserted? Correct nswer: F. Given : ɛ C 2 /N m 2, A 105 cm 2, nd d 2.25 mm. The cpcitnce efore the dielectric is inserted is C 1 ɛ 0 A ( d C 2 /N m 2) (105 cm 2 ) (2.25 mm) F. 018 (prt 2 of 6) 2 points Wht is the cpcitnce fter the dielectric is inserted? Correct nswer: F. Given : κ The cpcitnce fter the dielectric is inserted is C 2 κ ɛ 0 A d (8.53) (ɛ 0) (105 cm 2 ) (2.25 mm) F. Wht is the chrge on the pltes efore the dielectric is inserted? Correct nswer: C. Given : V 5.87 V. The chrge on the pltes efore the dielectric is inserted is given y Q 1 C 1 V ( F) (5.87 V) C. 020 (prt 4 of 6) 2 points Wht is the chrge on the pltes fter the dielectric is inserted? Correct nswer: C. The chrge on the pltes doesn t chnge fter the dielectric is inserted, so it is given y Q 2 Q C. 021 (prt 5 of 6) 1 points Wht is the potentil difference cross the pltes efore the dielectric is inserted? Correct nswer: 5.87 V. The potentil difference cross the pltes efore the dielectric is inserted is given y V 1 V 5.87 V. 022 (prt 6 of 6) 1 points Wht is the potentil difference cross the pltes fter the dielectric is inserted? Correct nswer: V. The potentil difference cross the pltes fter the dielectric is inserted is given y V 2 V κ 5.87 V V. 019 (prt 3 of 6) 2 points