Samples of solutions to conceptual problems from chapter 19 Cutnell & Johnson 7E

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1 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E Samls of solutions to conctual oblms fom chat 9 Cutnll & Johnson 7E. A ositiv oint chag an a ngativ oint chag hav qual magnitus. On chag is fix to on con of a squa, an th oth is fix to anoth con. On which cons shoul th chags b lac, so that th sam otntial xists at th mty cons? Giv you asoning.. REASONING AND SOLUTION Th otntial at a oint in sac that is a istanc fom a oint chag q is givn by Equation 9.6: V kq/. Whn mo than on oint chag is snt, th total otntial at any location is th algbaic sum of th iniviual otntials cat by ach chag at that location. A ositiv oint chag an a ngativ oint chag hav qual magnitus. On of th chags is fix to on con of a squa. If th oth chag is lac oosit to th fist chag along th q +Q L iagonal of th squa, thn ach chag will b th sam istanc L fom th mty cons. Th otntial at ach of th mty cons will b L L q Q L k( + Q) k( Q) V + 0 L L Thfo, if th otntial at ach mty con is to b th sam, thn th chags must b lac at iagonally oosit cons as shown in th figu. 4. What oint chags, all having th sam magnitu, woul you lac at th cons of a squa (on chag con), so that both th lctic fil an th lctic otntial (assuming a zo fnc valu at infinity) a zo at th cnt of th squa? Account fo th fact that th chag istibution givs is to both a zo fil an a zo otntial. 4. REASONING AND SOLUTION Fou oint chags of qual magnitu a lac at th cons of a squa as shown in th figu at th ight. Th lctic fil at th cnt of th squa is th vcto sum of th lctic fil at th cnt u to ach of th chags iniviually. Th otntial at th q q cnt of th squa is qual to th algbaic sum of th otntials at th cnt u to ach of th chags iniviually. All fou chags a quiistant fom th cnt (a istanc in th figu). If two iagonal chags hav th sam magnitu an sign, thn th lctic fil at th cnt u to ths two chags hav qual magnitu an oosit ictions. Thi sultant is, thfo, zo. Thus, th lctic fil at th cnt will b zo if ach iagonal ai of chags has th sam magnitu an sign. q 4 q 3 Comil by DJJ Pag of 9 8/7/006

2 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E If a iagonal ai of chags has th sam magnitu an sign, thy will giv is to a non-zo otntial at th cnt. Thus, th otntial u to on iagonal ai of chags must cancl th otntial u to th oth iagonal ai of chags. This will b th cas if th two ais of iagonal chags hav oosit signs. Thus, both th lctic fil an th lctic otntial will b zo at th cnt of th squa if all fou chags hav th sam magnitu, q an q 3 hav th sam sign, an q an q 4 hav th sam sign, which is oosit to th signs of q an q An lctic otntial ngy xists whn two otons a saat by a ctain istanc. Dos th lctic otntial ngy in cas, cas, o main th sam whn (a) both otons a lac by lctons, an (b) only on of th otons is lac by an lcton? Justify you answs. 7. REASONING AND SOLUTION An lctic otntial ngy xists whn two otons a saat by a ctain istanc. It is qual to th wok that must b on by an xtnal agnt to assmbl th configuation. Suos that w imagin assmbling th systm, on aticl at a tim. If th a no oth chags in th gion, th a no xisting lctic fils; thfo, no wok is qui to ut th fist oton in lac. That oton, howv, givs is to an lctic fil that fills th gion. Its magnitu at a istanc fom th oton is givn by Equation 8.3, E k /, wh + is th magnitu of th chag on th oton. Sinc th gion contains an lctic fil u to th fist oton, th also xists an lctic otntial, an th xtnal agnt must o wok to lac th scon oton at a istanc fom th fist oton. Th lctic otntial ngy of th final configuation is qual to th wok that must b on to bing th scon oton fom infinity an lac it at a istanc fom th fist oton. Th lctic otntial at a istanc fom th fist oton is Voton + k/ (Equation 9.6). Accoing to Equation 9.3, th lctic otntial ngy of th final configuation is thfo k EPE Voton ( + ) + a. If both otons a lac by lctons, simila agumnts aly. Howv, sinc th lcton cais a ngativ chag ( ), th lctic otntial at a istanc fom th fist lcton is Vlcton k/. Th lctic otntial ngy of th final configuation is now givn by EPE lcton ( ) k k V ( ) + Thfo, if both otons a lac by lctons, th lctic otntial ngy mains th sam. b. Whn only on of th otons is lac by an lcton, w fin that EPE oton ( ) k k V + ( ) Comil by DJJ Pag of 9 8/7/006

3 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E Thus, whn only on of th otons is lac by an lcton, th lctic otntial ngy cass fom + k / to k /. 6. A oton an an lcton a las fom st at th mioint btwn th lats of a chag aalll lat caacito. Exct fo ths aticls, nothing ls is btwn th lats. Igno th attaction btwn th oton an th lcton, an ci which aticl stiks a caacito lat fist. Why? 6. REASONING AND SOLUTION Sinc both aticls a las fom st, thi initial kintic ngis a zo. Thy both hav lctic otntial ngy by vitu of thi sctiv ositions in th lctic fil btwn th lats. Sinc th aticls a oositly chag, thy mov in oosit ictions towa oosit lats of th caacito. As thy mov towa th lats, th aticls gain kintic ngy an los otntial ngy. Using (EPE) 0 an (EPE) f to not th initial an final lctic otntial ngis of th aticl, sctivly, w fin fom ngy consvation that ( EPE) m v + ( EPE) Th final s of ach aticl is givn by 0 aticl f f v f ( ) ( ) EPE EPE m 0 f aticl Sinc both aticls tavl though th sam istanc btwn th lats of th caacito, th chang in th lctic otntial ngy is th sam fo both aticls. Sinc th mass of th lcton is small than th mass of th oton, th final s of th lcton will b gat than that of th oton. Thfo, th lcton tavls fast than th oton as th aticls mov towa th sctiv lats. Th lcton, thfo, stiks th caacito lat fist. CHAPTER 9 Elctic Potntial Engy An Th Elctic Potntial Samls of solutions to Poblms fom chat 9 Cutnll & Johnson 7E 4. A aticl has a chag of an movs fom oint A to oint, a istanc of 0.0 m. Th aticl xincs a constant lctic foc, an its motion is along th lin of action of th foc. Th iffnc btwn th aticl s lctic otntial ngy at A an is. (a) Fin th magnitu an iction of th lctic foc that acts on th aticl. (b) Fin th magnitu an iction of th lctic fil that th aticl xincs. Comil by DJJ Pag 3 of 9 8/7/006

4 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E 4. REASONING Equation 9. inicats that th wok on by th lctic foc as th aticl movs fom oint A to oint is W A EPE A EPE. Fo motion though a istanc s along th lin of action of a constant foc of magnitu F, th wok is givn by Equation 6. as ith +Fs (if th foc an th islacmnt hav th sam iction) o Fs (if th foc an th islacmnt hav oosit ictions). H, EPE A EPE is givn to b ositiv, so w can conclu that th wok is W A +Fs an that th foc oints in th iction of th motion fom oint A to oint. Th lctic fil is givn by Equation 8. as E F/q 0, wh q 0 is th chag. SOLUTION a. Using Equation 9. an th fact that W A +Fs, w fin W + Fs EPE EPE A A 4 EPE A EPE J 3 F N s 0.0 m As iscuss in th asoning, th iction of th foc is fom A towa. b. Fom Equation 8., w fin that th lctic fil has a magnitu of 3 F N 3 E N/C q C 0 Th iction is th sam as that of th foc on th ositiv chag, namly fom A towa. 9. Th otntial at location A is 45 V. A ositivly chag aticl is las th fom st an aivs at location with a s v. Th otntial at location C is 79 V, an whn las fom st fom this sot, th aticl aivs at with twic th s it viously ha, o v. Fin th otntial at 9. REASONING Th only foc acting on th moving chag is th consvativ lctic foc. Thfo, th total ngy of th chag mains constant. Alying th incil of consvation of ngy btwn locations A an, w obtain mva + EPE A mv + EPE Sinc th chag aticl stats fom st, v A 0. Th iffnc in otntial ngis is lat to th iffnc in otntials by Equation 9.4, EPE EPE A qv ( VA). Thus, w hav qv ( V ) mv () A Similaly, alying th consvation of ngy btwn locations C an givs Comil by DJJ Pag 4 of 9 8/7/006

5 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E C Diviing Equation () by Equation () yils VA V VC V 4 This xssion can b solv fo V. qv ( V ) m ( v ) () SOLUTION Solving fo V, w fin that 4 VA VC 4(45 V) 79 V V 339 V 3 3. Two chags A an a fix in lac, at iffnt istancs fom a ctain sot. At this sot th otntials u to th two chags a qual. Chag A is 0.8 m fom th sot, whil chag is 0.43 m fom it. Fin th atio q /q A of th chags.. REASONING Th otntial of ach chag q at a istanc away is givn by Equation 9.6 as V kq/. y alying this xssion to ach chag, w will b abl to fin th si atio, bcaus th istancs a givn fo ach chag. SOLUTION Accoing to Equation 9.6, th otntials of ach chag a V A kqa kq an V A Sinc w know that V A V, it follows that kqa kq q 0.43 m o.4 A qa A 0.8 m 6. Th awing shows six oint chags aang in a ctangl. Th valu of q is, an th istanc is 0.3 m. Fin th total lctic otntial at location P, which is at th cnt of th ctangl. Comil by DJJ Pag 5 of 9 8/7/006

6 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E 6. REASONING Th lctic otntial at a istanc fom a oint chag q is givn by Equation 9.6 as V kq/. Th total lctic otntial at location P u to th six oint chags is th algbaic sum of th iniviual otntials. +7.0q +3.0q +5.0q P 5.0q 3.0q +7.0q SOLUTION Stating at th u lft con of th ctangl, w oc clockwis an a u th six contibutions to th total lctic otntial at P (s th awing): V ( ) ( ) ( ) ( ) ( 3.0 ) ( 5.0 ) k q k q k q k q k q k q k ( + 4.0q) + Substituting q C an 0.3 m givs 9 N m C k( + 4.0q) C 6 V V 0.3 m + ( 0.3 m) + ( )( ) 37. Th mmban that suouns a ctain ty of living cll has a sufac aa of an a thicknss of. Assum that th mmban bhavs lik a aalll lat caacito an has a ilctic constant of 5.0. (a) Th otntial on th out sufac of th mmban is mv gat than that on th insi sufac. How much chag sis on th out sufac? (b) If th chag in at (a) is u to K + ions (chag +), how many such ions a snt on th out sufac? 37. SSM REASONING Th chag that sis on th out sufac of th cll mmban is q CV, accoing to Equation 9.8. fo w can us this xssion, howv, w must fist tmin th caacitanc of th mmban. If w assum that th cll mmban bhavs lik a aalll lat caacito fill with a ilctic, Equation 9.0 (C κ ε 0 A / ) alis as wll. SOLUTION Th caacitanc of th cll mmban is Comil by DJJ Pag 6 of 9 8/7/006

7 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E 9 κε 0A (5.0)( F/m)(5.0 0 m ) C. 0 F m a. Th chag on th out sufac of th mmban is, thfo, 3 q CV (. 0 F)( V).3 0 C + b. If th chag in at (a) is u to K ions with chag + ( C), th numb of ions snt on th out sufac of th mmban is Numb of.3 0 C K ions C 4. Two caacitos a intical, xct that on is mty an th oth is fill with a ilctic (k 4.50). Th mty caacito is connct to a.0-v batty. What must b th otntial iffnc acoss th lats of th caacito fill with a ilctic such that it stos th sam amount of lctical ngy as th mty caacito? 4. REASONING Th ngy us to chag u a caacito is sto in th caacito as lctical ngy. Th ngy sto ns on th caacitanc C of th caacito an th otntial iffnc V btwn its lats; Engy CV (Equation 9.b). Insting a ilctic btwn th lats of a caacito incass its caacitanc by a facto of κ, wh κ is th ilctic constant of th matial. W will us ths two ics of infomation to fin th otntial iffnc acoss th lats of th caacito fill with th ilctic. SOLUTION Th ngy sto in th mty caacito is Engy CV 0 0, wh C 0 is its caacitanc an V 0 is th otntial iffnc btwn its lats. Similaly, th ngy sto in th caacito fill with th ilctic is Engy CV, wh C is its caacitanc an V is th otntial iffnc btwn its lats. Sinc th two ngis a qual, CV 0 0 CV Sinc C κc 0 (s Equation 9.0 an th iscussion that follows), w hav ( κ ) CV 0 0 C 0 V Solving fo th otntial iffnc V, givs V0.0 V V 5.66 V κ 4.50 Comil by DJJ Pag 7 of 9 8/7/006

8 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E 59. Th otntial iffnc btwn th lats of a caacito is 75 V. Miway btwn th lats, a oton an an lcton a las. Th lcton is las fom st. Th oton is ojct niculaly towa th ngativ lat with an initial s. Th oton stiks th ngativ lat at th sam instant that th lcton stiks th ositiv lat. Igno th attaction btwn th two aticls, an fin th initial s of th oton. 59. REASONING If w assum that th motion of th oton an th lcton is hoizontal in th +x iction, th motion of th oton is tmin by Equation.8, x vt 0 + at, wh x is th istanc tavl by th oton, v 0 is its initial s, an a is its acclation. If th istanc btwn th caacito lacs is, thn this lation bcoms 0 v t + a t, o 0 v t+ a t () W can solv Equation () fo th initial s v 0 of th oton, but, fist, w must tmin th tim t an th acclation a of th oton. Sinc th oton stiks th ngativ lat at th sam instant th lcton stiks th ositiv lat, w can us th motion of th lcton to tmin th tim t. Fo th lcton, a t, wh w hav takn into account th fact that th lcton is las fom st. Solving this xssion fo t w hav t / a. Substituting this xssion into Equation (), w hav a v0 + a a () Th acclations can b foun by noting that th magnitus of th focs on th lcton an oton a qual, sinc ths aticls hav th sam magnitu of chag. Th foc on th lcton is F E V /, an th acclation of th lcton is, thfo, Nwton's scon law quis that ma ma, so that a F V m m (3) a a m (4) m Combining Equations (), (3) an (4) las to th following xssion fo v 0, th initial s of th oton: Comil by DJJ Pag 8 of 9 8/7/006

9 Samls of conctual an analytical/numical qustions fom cha9, C&J, 7E v 0 m m V m SOLUTION Substituting valus into th xssion abov, w fin kg (.60 0 C)(75 V) 6 v m/s kg 9. 0 kg Comil by DJJ Pag 9 of 9 8/7/006