Chapter 21 The Electric Field I: Discrete Charge Distributions

Transcription

1 Chpte The lectic ield I: Discete Chge Distibutions Conceptul oblems Objects e composed of toms which e composed of chged pticles (potons nd electons); howeve, we el obseve the effects of the electosttic foce. plin wh we do not obseve these effects. Detemine the Concept The net chge on lge objects is lws ve close to zeo. Hence the most obvious foce is the gvittionl foce. A cbon tom cn become cbon ion if it hs one o moe of its electons emoved duing pocess clled ioniztion. Wht is the net chge on cbon tom tht hs hd two of its electons emoved? () e, (b) e, (c) e, (d) e Detemine the Concept If two electons e emoved fom cbon tom, it will is coect. hve net positive chge of e. ( c) You do simple demonsttion fo ou high school phsics teche in which ou clim to dispove Coulomb s lw. You fist un ubbe comb though ou d hi, then use it to ttct tin neutl pieces of ppe on the desk. You then s Coulomb s lw sttes tht fo thee to be electosttic foces of ttction between two objects, both objects need to be chged. Howeve, the ppe ws not chged. So ccoding to Coulomb s lw, thee should be no electosttic foces of ttction between them, et thee clel ws. You est ou cse. () Wht is wong with ou ssumptions? (b) Does ttction between the ppe nd the comb euie tht the net chge on the comb be negtive? plin. Detemine the Concept () Coulomb s lw is onl vlid fo point pticles. The ppe bits cnnot be modeled s point pticles becuse the ppe bits become polized. (b) No, the ttction does not depend on the sign of the chge on the comb. The induced chge on the ppe tht is closest to the comb is lws opposite in sign to of the chge on the comb, nd thus the net foce on the ppe is lws ttctive. You hve positivel chged insulting od nd two metl sphees on insulting stnds. Give step-b-step diections of how the od, without ctull touching eithe sphee, cn be used to give one of the sphees () negtive chge, nd (b) positive chge. 997

2 998 Chpte Detemine the Concept () Connect the metl sphee to gound; bing the insulting od ne the metl sphee nd disconnect the sphee fom gound; then emove the insulting od. The sphee will be negtivel chged. (b) Bing the insulting od in contct with the metl sphee; some of the positive chge on the od will be tnsfeed to the metl sphee. 5 () Two point pticles tht hve chges of nd e septed b distnce d. Use field lines to dw visuliztion of the electic field in the neighbohood of this sstem. (b) Dw the field lines t distnces much gete thn d fom the chges. Detemine the Concept () We cn use the ules fo dwing electic field lines to dw the electic field lines fo this sstem. Two field lines hve been ssigned to ech chge. (b) At distnces much gete thn the seption distnce between the two chges, the sstem of two chged bodies will look like single chge of nd the field ptten will be tht due to point chge of. ight field lines hve been ssigned to the single chge. () (b) 6 A metl sphee is positivel chged. Is it possible fo the sphee to electicll ttct nothe positivel chged bll? plin ou nswe. Detemine the Concept Yes. Becuse metl sphee is conducto, the poimit of positivel chged bll (not necessil conducto), will induce edistibution of chges on the metl sphee with the sufce nee the positivel chged bll becoming negtivel chged. Becuse the negtive chges on the metl sphee e close to the positivel chged bll thn e the positive chges on the metl sphee, the net foce will be ttctive.

3 The lectic ield I: Discete Chge Distibutions A simple demonsttion of electosttic ttction cn be done simpl b dngling smll bll of cumpled luminum foil on sting nd binging chged od ne the bll. The bll initill will be ttcted to the od, but once the touch, the bll will be stongl epelled fom it. plin these obsevtions. Detemine the Concept Assume tht the od hs negtive chge. When the chged od is bought ne the luminum foil, it induces edistibution of chges with the side nee the od becoming positivel chged, nd so it swings towd the od. When it touches the od, some of the negtive chge is tnsfeed to the foil, which, s esult, cuies net negtive chge nd is now epelled b the od. 8 Two positive point chges tht e eul in mgnitude e fied in plce, one t. m nd the othe t. m, on the is. A thid positive point chge is plced t n euilibium position. () Whee is this euilibium position? (b) Is the euilibium position stble if the thid pticle is constined to move pllel with the is? (c) Wht bout if it is constined to move pllel with the is? plin. Detemine the Concept () A thid positive chge cn be plced midw between the fied positive chges. This is the onl loction. (b) Yes. The position identified in () is one of stble euilibium. It is stble in the -diection becuse egdless of whethe ou displce the thid positive chge to the ight o to the left, the net foce cting on it is bck towd the midpoint between the two fied chges. (c) If the thid positive chge is displced in the diection, the net foce cting on it will be w fom its euilibium position. Hence the position midw between the fied positive chges is one of unstble euilibium in the diection. 9 Two neutl conducting sphees e in contct nd e suppoted on lge wooden tble b insulted stnds. A positivel chged od is bought up close to the sufce of one of the sphees on the side opposite its point of contct with the othe sphee. () Descibe the induced chges on the two conducting sphees, nd sketch the chge distibutions on them. (b) The two sphees e septed nd then the chged od is emoved. The sphees e then septed f pt. Sketch the chge distibutions on the septed sphees. Detemine the Concept Becuse the sphees e conductos, thee e fee electons on them tht will eposition themselves when the positivel chged od is bought neb.

4 Chpte () On the sphee ne the positivel chged od, the induced chge is negtive nd ne the od. On the othe sphee, the net chge is positive nd on the side f fom the od. This is shown in the digm. (b) When the sphees e septed nd f pt nd the od hs been emoved, the induced chges e distibuted unifoml ove ech sphee. The chge distibutions e shown in the digm. Thee point chges,, Q, nd Q, e plced t the cones of n euiltel tingle s shown in igue -. No othe chged objects e neb. () Wht is the diection of the net foce on chge due to the othe two chges? (b) Wht is the totl electic foce on the sstem of thee chges? plin. Detemine the Concept The foces cting on point chge e shown in the digm. The foce cting on point chge due to point chge Q is long the line joining them nd diected towd Q. The foce cting on point chge due to point chge Q is long the line joining them nd diected w fom point chge Q. Q Q Q Q () Becuse point chges Q nd Q e eul in mgnitude, the foces due to these chges e eul nd thei sum (the net foce on chge ) will be to the ight. Note tht the veticl components of these foces dd up to zeo. (b) Becuse no othe chged objects e neb, the foces cting on this sstem of thee point chges e intenl foces nd the net foce cting on the sstem is zeo. A positivel chged pticle is fee to move in egion with n electic field. Which sttements must be tue? () The pticle is cceleting in the diection pependicul to.

5 (b) (c) (d) (e) (f) The lectic ield I: Discete Chge Distibutions The pticle is cceleting in the diection of. The pticle is moving in the diection of. The pticle could be momentil t est. The foce on the pticle is opposite the diection of. The pticle is moving opposite the diection of. Detemine the Concept () lse. The onl foce cting on the pticle is in the diection of. (b) Tue. The electicl foce epeienced b the pticle is, b definition, in the diection of. (c) lse. We don t know whethe the pticle is moving o momentil t est. All we know is tht the net foce cting on it is in the diection of. (d) ossibl, Whethe the pticle is eve t est depends on how it ws initill plced in the electic field. Tht is, it depends on whethe its initil velocit ws zeo, in the diection of, o opposite. (e) lse. B definition, the electic foce cting on positivel chged pticle in n electic field is in the diection of the field. (f) Tue. All we know fo sue is tht the electic foce nd, hence, the cceletion of the pticle, is in the diection of. The pticle could be moving in the diection of the field, in the diection opposite the field, o it could be momentil t est. We do know tht, if it is t n time t est, it will not st t est. ou chges e fied in plce t the cones of sue s shown in igue -. No othe chges e neb. Which of the following sttements is tue? () (b) (c) is zeo t the midpoints of ll fou sides of the sue. is zeo t the cente of the sue. is zeo midw between the top two chges nd midw between the bottom two chges. Detemine the Concept is zeo wheeve the net foce cting on test chge is zeo. () lse. A test chge plced t these loctions will epeience net foce.

6 Chpte (b) Tue. At the cente of the sue the two positive chges lone poduce net electic field of zeo, nd the two negtive chges lone lso poduce net electic field of zeo. Thus, the net foce cting on test chge t the midpoint of the sue is zeo. (c) lse. A test chge plced t n of these loctions will epeience net foce. [SSM] Two point pticles tht hve chges of nd e septed b distnce d. () Use field lines to sketch the electic field in the neighbohood of this sstem. (b) Dw the field lines t distnces much gete thn d fom the chges. Detemine the Concept () We cn use the ules fo dwing electic field lines to dw the electic field lines fo this sstem. In the field-line sketch we ve ssigned field lines to ech chge. (b) At distnces much gete thn the seption distnce between the two chges, the sstem of two chged bodies will look like single chge of nd the field ptten will be tht due to point chge of. ou field lines hve been ssigned to ech chge. () (b) Thee eul positive point chges (ech chge ) e fied t the vetices of n euiltel tingle with sides of length. The oigin is t the midpoint of one side the tingle, the cente of the tingle on the is t nd the vete opposite the oigin is on the is t. () pess nd in tems of. (b) Wite n epession fo the electic field on the is distnce fom the oigin on the intevl <. (c) Show tht the epession ou obtined in (b) gives the epected esults fo nd fo.

7 The lectic ield I: Discete Chge Distibutions ictue the oblem () We cn use the geomet of n euiltel tingle to epess nd in tems of the side of the tingle. (b) The electic field on the is distnce fom the oigin on the intevl < is the supeposition of the electic fields due to the point chges,, nd locted t the vetices of the tingle. () Appl the thgoen theoem to the tingle whose vetices e t (,), (, ), nd (,) to obtin: Refeing to the tingle whose vetices e t (,), (, ), nd (,), note tht: Substitute fo tn nd evlute : ( ) tn tn 6 (b) The electic field t is the supeposition of the electic fields due to the chges,, nd : () pess the electic field t due to :,,,, whee ( ) i ( )i, nd, Substituting fo,,, nd, ields: ( ) ( ) ( ) i i

8 Chpte pess the electic field t due to :,,,, whee ( ) i ( ) j i j, nd ( ) ( ), Substituting fo,,, nd, ields: ( ) ( i j ) oceed simill fo to obtin: ( ) ( i j ) Substituting fo,, nd in eution () nd simplifing ields: ( ) ( ) (c) vluting ( ) i ( ) ( ) ields: ( i j ) i ( ) ( i j ) ( ) i ( ) ( ) ( ) i Note tht, becuse the electic fields due to nd cncel ech othe t the oigin nd the esultnt field is tht due to, this is the epected esult. vlute ( 6 ) to obtin: ( ) ( 6 ) ( ) ( ) 6 i i ( ) 6 ( 6 ) Note tht, becuse the point t the cente of the euiltel tingle is euidistnt fom the thee vetices, the electic fields due to the chges t the vetices cncel nd this is the epected esult.

9 The lectic ield I: Discete Chge Distibutions 5 5 A molecule hs dipole moment given b p. The molecule is momentil t est with p mking n ngle θ with unifom electic field. Descibe the subseuent motion of the dipole moment. Detemine the Concept The dipole moment ottes bck nd foth in oscillto motion. The dipole moment gins ngul speed s it ottes towd the diection of the electic field, nd losing ngul speed s it ottes w fom the diection of the electic field. 6 Tue o flse: () (b) (c) (d) The electic field of point chge lws points w fom the chge. The electic foce on chged pticle in n electic field is lws in the sme diection s the field. lectic field lines neve intesect. All molecules hve dipole moments in the pesence of n etenl electic field. () lse. The diection of the field is towd negtive chge. (b) lse. The diection of the electic foce cting on point chge depends on the sign of the point chge. (c) lse. lectic field lines intesect n point in spce occupied b point chge. (d) Tue. An electic field ptill polizes the molecules; esulting in the seption of thei chges nd the cetion of electic dipole moments. 7 [SSM] Two molecules hve dipole moments of eul mgnitude. The dipole moments e oiented in vious configutions s shown in igue -. Detemine the electic-field diection t ech of the numbeed loctions. plin ou nswes. Detemine the Concept igue - shows the electic field due to single dipole, whee the dipole moment is diected towd the ight. The electic field due two pi of dipoles cn be obtined b supeposing the two electic fields. () down up up (b) up ight left (c) down up up (d) down up up

10 6 Chpte stimtion nd Appoimtion 8 stimte the foce euied to bind the two potons in the He nucleus togethe. HINT: Model the potons s point chges. You will need to hve n estimte of the distnce between them. ictue the oblem Becuse the nucleus is in euilibium, the binding foce must be eul to the electosttic foce of epulsion between the potons. Appl to poton to obtin: electosttic binding Solve fo binding : binding electost tic Using Coulomb s lw, substitute fo electosttic : binding Assuming the dimete of the He nucleus to be ppoimtel 5 m, substitute numeicl vlues nd evlute : electosttic binding 9 9 ( N m / C )(.6 C) 5 ( m). kn 9 A popul clssoom demonsttion consists of ubbing plstic od with fu to give the od chge, nd then plcing the od ne n empt sod cn tht is on its side (igue -6). plin wh the cn will oll towd the od. Detemine the Concept Becuse the cn is gounded, the pesence of the negtivel-chged plstic od induces positive chge on it. The positive chges induced on the cn e ttcted, vi the Coulomb intection, to the negtive chges on the plstic od. Unlike chges ttct, so the cn will oll towd the od. Spks in i occu when ions in the i e cceleted to such high speed b n electic field tht when the impct on neutl gs molecules the neutl molecules become ions. If the electic field stength is lge enough, the ionized collision poducts e themselves cceleted nd poduce moe ions on impct, nd so foth. This vlnche of ions is wht we cll spk. () Assume tht n ion moves, on vege, ectl one men fee pth though the i befoe hitting molecule. If the ion needs to cuie ppoimtel. ev of kinetic eneg in ode to ionize molecule, estimte the minimum stength of the electic field euied t stndd oom pessue nd tempetue. Assume tht the coss-sectionl e of n i molecule is bout. nm. (b) How does the stength of the electic field in () depend on tempetue? (c) How does the

11 The lectic ield I: Discete Chge Distibutions 7 stength of the electic field in () depend on pessue? ictue the oblem We cn use the definition of electic field to epess in tems of the wok done on the ionizing electons nd the distnce the tvel λ between collisions. We cn use the idel-gs lw to elte the numbe densit of molecules in the gs ρ nd the sctteing coss-section σ to the men fee pth nd, hence, to the electic field. () Appl consevtion of eneg to elte the wok done on the electons b the electic field to the chnge in thei kinetic eneg: W ΔK Δs om the definition of electic field we hve: Substitute fo nd Δs to obtin: The men fee pth λ of pticle is elted to the sctteing coss-section σ nd the numbe densit fo i molecules n: W λ, whee the men fee pth λ is the distnce tveled b the electons between ionizing collisions with nitogen toms. λ σ n Substitute fo λ to obtin: W σ nw σ n Use the idel-gs lw to obtin: Substituting fo n ields: N n V kt σw () kt Substitute numeicl vlues nd evlute : N (. nm ).5 (.ev) m.6 9 J (.6 C).8 ( K) K 9 J ev. 6 N/C (b) om eution () we see tht: T

12 8 Chpte (c) Also, fom eution (): Chge A plstic od is ubbed ginst wool shit, theeb cuiing chge of.8 μc. How mn electons e tnsfeed fom the wool shit to the plstic od? ictue the oblem The chge cuied b the plstic od is n integl numbe of electonic chges, tht is, n e (e). Relte the chge cuied b the plstic od to the numbe of electons tnsfeed fom the wool shit: ( e) n e n e e Substitute numeicl vlues nd evlute n e : n e.8μc 9 C.6 electon 5. electons A chge eul to the chge of Avogdo s numbe of potons (N A 6. ) is clled fd. Clculte the numbe of coulombs in fd. ictue the oblem One fd N A e. We cn use this definition to find the numbe of coulombs in fd. Use the definition of fd to clculte the numbe of coulombs in fd: fd N 9 ( 6. electons)(.6 C/electon) 9.6 C A e [SSM] Wht is the totl chge of ll of the potons in. kg of cbon? ictue the oblem We cn find the numbe of coulombs of positive chge thee e in. kg of cbon fomq 6nCe, whee n C is the numbe of toms in. kg of cbon nd the fcto of 6 is pesent to ccount fo the pesence of 6 potons in ech tom. We cn find the numbe of toms in. kg of cbon b setting up popotion elting Avogdo s numbe, the mss of cbon, nd the molecul mss of cbon to n C. See Appendi C fo the mol mss of cbon.

13 The lectic ield I: Discete Chge Distibutions 9 pess the positive chge in tems of the electonic chge, the numbe of potons pe tom, nd the numbe of toms in. kg of cbon: Q 6n C e Using popotion, elte the numbe of toms in. kg of cbon n C, to Avogdo s numbe nd the molecul mss M of cbon: n C mc N M A n C N A M m C Substitute fo n C to obtin: 6N AmCe Q M Substitute numeicl vlues nd evlute Q: 66. Q toms 9 (. kg)(.6 C) mol kg. mol.8 7 C Suppose cube of luminum which is. cm on side ccumultes net chge of.5 pc. ()Wht pecentge of the electons oiginll in the cube ws emoved? (b) B wht pecentge hs the mss of the cube decesed becuse of this emovl? ictue the oblem () The pecentge of the electons oiginll in the cube tht ws emoved cn be found fom the tio of the numbe of electons emoved to the numbe of electons oiginll in the cube. (b) The pecentge decese in the mss of the cube cn be found fom the tio of the mss of the electons emoved to the mss of the cube. () pess the tio of the electons emoved to the numbe of electons oiginll in the cube: N N em ini Qccumulted e N N () electons pe tom toms The numbe of toms in the cube is the tio of the mss of the cube to the mss of n luminum tom: N toms m m cube Al tom ρ AlV m cube Al tom

14 Chpte The mss of n luminum tom is its mol mss divided b Avogdo s numbe: m Al tom M N Al A Substituting nd simplifing ields: N toms ρ AlV M N cube Al A ρ AlV M cube Al N A Substitute fo N toms in eution () nd simplif to obtin: N N em ini Qccumulted e ρ AlVcubeN A N electons pe tom M Al Qccumulted M Al N ρ V en electons pe tom Al cube A Substitute numeicl vlues nd evlute N em : N ini N N em ini electons g.7 tom cm.99 5 % (.5 pc) 6.98 g mol 9 (. cm) (.6 C) 6. toms mol (b) pess the tio of the mss of the electons emoved to the mss of the cube: m m em cube N emm ρ V Al electon cube om (), the numbe of electons emoved is given b: Substituting nd simplifing ields: N m m em em cube Q ccumulted Q Q e m e ρ V ccumulted Al ccumulted Al cube m eρ V cube electon electon

15 The lectic ield I: Discete Chge Distibutions Substitute numeicl vlues nd evlute m m em cube m em : m cube (.5 pc)( 9.9 kg) 9 g (.6 C).7 (. cm ) 9 cm Duing pocess descibed b the photoelectic effect, ult-violet light cn be used to chge piece of metl. () If such light is incident on slb of conducting mteil nd electons e ejected with enough eneg tht the escpe the sufce of the metl, how long befoe the metl hs net chge of.5 nc if. 6 electons e ejected pe second? (b) If. ev is needed to eject n electon fom the sufce, wht is the powe ting of the light bem? (Assume this pocess is % efficient.) ictue the oblem () The euied time is the tio of the chge tht ccumultes to the te t which it is deliveed to the conducto. (b) We cn use the definition of powe to find the powe ting of the light bem. % () The euied time is the tio of the chge tht ccumultes to the te t which it is deliveed: Δ Δ t I Δ d dt Substitute numeicl vlues nd evlute Δt: Δt. 6.5 nc electons.6 s 9 C electon 9.6 s h 6 s.6 h (b) The powe ting of the light bem is the te t which it delives eneg: p Δ Δt The eneg deliveed b the bem is the poduct of the eneg pe electon, the electon cuent (tht is, the numbe of electons emoved pe unit time), nd the elpsed time: Substituting fo Δ in the epession fo nd simplifing ields: Δ pe electon I electon Δt pe I electonδt electon p Δt pe electon I electon

16 Chpte Substitute numeicl vlues nd evlute : ev.6. electon ev -9 J. 6 electons s. W Coulomb s Lw 6 A point chge. μc is t the oigin nd point chge 6. μc is on the -is t. m. () ind the electic foce on chge. (b) ind the electic foce on. (c) How would ou nswes fo ts () nd (b) diffe if wee 6. μc? ictue the oblem We cn find the electic foces the two chges eet on ech b ppling Coulomb s lw nd Newton s d lw. Note tht i, becuse the vecto pointing fom to is in the positive diection. The digm shows the sitution fo ts () nd (b).,,.μc 6.μC, m () Use Coulomb s lw to epess the foce tht eets on :,,, Substitute numeicl vlues nd evlute, : 9 ( N m /C )(.μc)( 6.μC) (.m), i ( mn)i (b) Becuse these e ction-ndection foces, we cn ppl Newton s d lw to obtin:,, ( mn)i (c) If is 6. μc, the foce between nd is ttctive nd both foce vectos e evesed: nd 9 ( N m /C )(.μc)( 6.μC) (.m) i,,, ( mn)i ( mn)i

17 The lectic ield I: Discete Chge Distibutions 7 [SSM] Thee point chges e on the -is: 6. μc is t. m,. μc is t the oigin, nd 6. μc is t. m. ind the electic foce on. ictue the oblem eets n ttctive electic foce, on point chge nd eets epulsive electic foce, on point chge. We cn find the net electic foce on b dding these foces (tht is, b using the supeposition pinciple).,, - 6.μC.μC pess the net foce cting on :,, pess the foce tht eets on :, k, 6.μC pess the foce tht eets on k ( i, ) :, i, m Substitute nd simplif to obtin: k k, i k,,, i i Substitute numeicl vlues nd evlute : 9. μc 6. μc ( )( ) i N m /C 6. μc (.5 N)i ( ) ( ).m 6.m 8 A. μc point chge nd. μc point chge e distnce L pt. Whee should thid point chge be plced so tht the electic foce on tht thid chge is zeo? ictue the oblem The thid point chge should be plced t the loction t which the foces on the thid point chge due to ech of the othe two point chges cncel. Thee cn be no such plce ecept on the line between the two point chges. Denote the. μc nd. μc point chges b the numels nd, espectivel, nd the thid point chge b the numel. Assume tht the

18 Chpte. μc point chge is to the left of the. μc point chge, let the diection be to the ight. Then the. μc point chge is locted t L. Appl the condition fo tnsltionl euilibium to the thid point chge:,, o (),, Letting the distnce fom the thid point chge to the. μc point chge be, epess the foce tht the. μc point chge eets on the thid point chge:, ( L ) The foce tht the. μc point chge eets on the thid chge is given b: Substitute in eution () to obtin:, ( L ) o, simplifing, L ( ) Rewiting this eution eplicitl s udtic eutions ields: L L Use the udtic fomul to obtin: L ± L L L ± L The oot coesponding to the negtive sign between the tems is etneous becuse it coesponds to position to the left of the. μc point chge nd is, theefoe, not phsicll meningful oot. Hence the thid point chge should be plced between the point chges nd distnce eul to.l w fom the.-μc chge. 9 A. μc point chge nd. μc point chge e distnce L pt. Whee should thid point chge be plced so tht the electic foce on tht thid chge is zeo? ictue the oblem The thid point chge should be plced t the loction t which the foces on the thid point chge due to ech of the othe two point chges cncel. Thee cn be no such plce between the two point chges. Beond the. μc point chge, nd on the line contining the two point chges,

19 The lectic ield I: Discete Chge Distibutions 5 the foce due to the. μc point chge ovewhelms the foce due to the. μc point chge. Beond the. μc point chge, nd on the line contining the two point chges, howeve, we cn find plce whee these foces cncel becuse the e eul in mgnitude nd oppositel diected. Denote the. μc nd. μc point chges b the numels nd, espectivel, nd the thid point chge b the numel. Let the diection be to the ight with the oigin t the position of the. μc point chge nd the. μc point chge be locted t L. Appl the condition fo tnsltionl euilibium to the thid point chge:,, o (),, Letting the distnce fom the thid point chge to the. μc point chge be, epess the foce tht the. μc point chge eets on the thid point chge:, ( L ) The foce tht the. μc point chge eets on the thid point chge is given b: Substitute in eution () to obtin:, ( L ) ( ) L Rewiting this eution eplicitl s udtic eutions ields: L L Use the udtic fomul to obtin: L ± L L L ± L The oot coesponding to the positive sign between the tems is etneous becuse it coesponds to position to the ight of the. μc point chge nd is, theefoe, not phsicll meningful oot. Hence the thid point chge should be plced distnce eul to.l fom the.-μc chge on the side w fom the.-μc chge. Thee point chges, ech of mgnitude. nc, e t septe cones of sue of edge length 5. cm. The two point chges t opposite cones e positive, nd the thid point chge is negtive. ind the electic foce eeted b these point chges on fouth point chge. nc t the emining cone.

20 6 Chpte ictue the oblem The configution of the point chges nd the foces on the fouth point chge e shown in the figue s is coodinte sstem. om the figue it is evident tht the net foce on the point chge is long the digonl of the sue nd diected w fom point chge. We cn ppl Coulomb s lw to epess,,, nd, nd then dd them (tht is, use the pinciple of supeposition of foces) to find the net electic foce on point chge., cm, 5.. nc. nc,,. nc 5.. nc, cm pess the net foce cting on point chge : (),,, pess the foce tht point chge eets on point chge :,, j Substitute numeicl vlues nd evlute, :, N m C. nc ( ) 5. nc j (.6 N)j ( ).5 m pess the foce tht point chge eets on point chge :,, i Substitute numeicl vlues nd evlute, :, N m C. nc ( ) i 5. nc (.6 N)i ( ).5 m

21 The lectic ield I: Discete Chge Distibutions 7 pess the foce tht point chge eets on point chge : pess, in tems of, nd, :,,,,, whee is unit vecto pointing fom to.,,, (.5 m) i (.5 m)j Convet, to, : (.5m) i (.5m) j (.5m) (.5m),,,.77i.77 j Substitute numeicl vlues nd evlute, :, N m C 5 ( ) i 5. N (. N)j. nc (. nc) (.77 i.77 j ) ( ).5 m Substitute numeicl vlues in eution () nd simplif to find : 5 ( ) 5 j ( ) i 5 ( ) i 5. N. N. N (. N) 5 (. N) i 5 (. N)j A point chge of 5. μc is on the is t. cm, nd second point chge of 5. μc is on the is t. cm. ind the electic foce on point chge of. μc on the is t 8. cm. ictue the oblem The configution of the point chges nd the foces on point chge e shown in the figue s is coodinte sstem. om the geomet of the chge distibution it is evident tht the net foce on the. μc point chge is in the negtive diection. We cn ppl Coulomb s lw to epess, nd, nd then dd them (tht is, use the pinciple of supeposition of foces) to find the net foce on point chge. j

22 8 Chpte, cm. 5.μ C θ. μc θ 8., cm., 5.μC, The net foce cting on point chge is given b: The foce tht point chge eets on point chge is:,, cosθ i sin, θ j is given b: nd N m C ( 5. μc)(. μc) (. m) (.8 m). cm θ tn cm. N The foce tht point chge eets on point chge is: Substitute fo, nd, nd simplif to obtin: cosθ i sin, θ cosθ i sinθ j cosθ i sinθ j sinθ j j Substitute numeicl vlues nd evlute : sin.56 j ( 8.65N)j (. N) A point pticle tht hs chge of.5 μc is locted t the oigin. A second point pticle tht hs chge of 6. μc is t. m,.5 m. A

23 The lectic ield I: Discete Chge Distibutions 9 thid point pticle, nd electon, is t point with coodintes (, ). ind the vlues of nd such tht the electon is in euilibium. ictue the oblem The positions of the point pticles e shown in the digm. It is ppent tht the electon must be locted long the line joining the two point pticles. Moeove, becuse it is negtivel chged, it must be close to the pticle with chge of.5 μc thn to the pticle with chge of 6. μc, s is indicted in the figue. We cn find the nd coodintes of the electon s position b euting the two electosttic foces cting on it nd solving fo its distnce fom the oigin. We cn use simil tingles to epess this dil distnce in tems of the nd coodintes of the electon..5, m 6. μc e,e e e,e.5μc., m pess the condition tht must be stisfied if the electon is to be in euilibium:, e, e Letting epesent the distnce fom the oigin to the electon, epess the mgnitude of the foce tht the pticle whose chge is eets on the electon:, e e (.5m ) pess the mgnitude of the foce tht the pticle whose chge is eets on the electon: e, k e Substitute nd simplif to obtin: (.5 m )

24 Chpte Substitute fo nd nd 6.5 simplif to obtin: ( ).5m Solving fo ields:.6 m nd.86 m Becuse < is unphsicl, we ll conside onl the positive oot. Use the simil tingles in the digm to estblish the popotion involving the coodinte of the electon: e.5m.6 m.m e.99 m Use the simil tingles in the digm to estblish the popotion involving the coodinte of the electon: e. m.6 m.m e.8 m The coodintes of the electon s position e: (, ) (.8m,.9m) e e A point pticle tht hs chge of. μc is locted t the oigin; second point pticle tht hs chge of. μc is locted t,. m; nd thid point pticle tht hs chge of. μc is locted t. m,. ind the totl electic foce on ech of the thee point chges. ictue the oblem Let epesent the chge of the point pticle t the oigin, the chge of the point pticle t (,. m), nd the chge of the point pticle t (. m, ). The digm shows the foces cting on ech of the point pticles. Note the ction-nd-ection pis. We cn ppl Coulomb s lw nd the pinciple of supeposition of foces to find the net foce cting on ech of the point pticles.,.,,, m.μc. μ C. μc, m,,.,

25 pess the net foce cting on the point pticle whose chge is : pess the foce tht the point pticle whose chge is eets on the point pticle whose chge is : The lectic ield I: Discete Chge Distibutions Substitute numeicl vlues nd evlute, :,, 9 N m C,,,,,,,, ( ) (. μc. μc ) ( ) (.m ) j (.8 N).m j, pess the foce tht the point pticle whose chge is eets on the point pticle whose chge is :,,, Substitute numeicl vlues nd evlute, : 9 N m C ( ) (. μc. μc ) ( ) (.m ) i (.899 N) i.m, Substitute to find : (.9 N) i (.8 N)j pess the net foce cting on the point pticle whose chge is :,,,, (.8 N)j, becuse, nd, e ction-nd-ectio foces. pess the foce tht the point pticle whose chge is eets on the point pticle whose chge is :,,,, [(.m) i (.m) j ]

26 Chpte Substitute numeicl vlues nd evlute, :, ( ) ( ) 9 N m. μc C C μ (. m) (.8 N) i (.6 N)j [(.m) i (.m) j ] ind the net foce cting on the point pticle whose chge is :, (.8 N) j (.8 N) i (.6 N) j (.8 N) (. N) i (. N)j j Noting tht, nd, e n ction-nd-ection pi, s e, nd,, epess the net foce cting on the point pticle whose chge is :,,, (. N) i (.6 N)j, [ j ] (.899 N) i (.8N) i (.6 N) A point pticle tht hs chge of 5. μc is locted t, nd point pticle tht hs chge is locted t. cm,. The electic foce on point pticle tht hs chge of. μc t 8. cm, is (9.7 N) î. When this.-μc chge is epositioned t 7.8 cm,, the electic foce on it is zeo. Detemine the chge. ictue the oblem Let epesent the chge of the point pticle t the oigin nd the chge of the point pticle initill t (8. cm, ) nd lte positioned t (7.8 cm, ). The digm shows the foces cting on the point pticle whose chge is t (8. cm, ). We cn ppl Coulomb s lw nd the pinciple of supeposition of foces to find the net foce cting on ech of the point pticles.. μc 5. μc,,, cm. 8.

27 pess the net foce on the point pticle whose chge is when it is t (8. cm, ): Substitute numeicl vlues to obtin: The lectic ield I: Discete Chge Distibutions ( ) i N (. μc) N m C ( 8.cm, ),,,,,,, 5. μc ( ) ( ) ( ) ( ) i.8 m. m i.8 m. m,,, Solving fo ields:. μc Remks: An ltentive solution is to eute the electosttic foces cting on when it is t (7.8 cm, ). 5 [SSM] ive identicl point chges, ech hving chge Q, e eull spced on semicicle of dius R s shown in igue -7. ind the foce (in tems of k, Q, nd R) on chge locted euidistnt fom the five othe chges. ictue the oblem B consideing the smmet of the of chged point pticles, we cn see tht the component of the foce on is zeo. We cn ppl Coulomb s lw nd the pinciple of supeposition of foces to find the net foce cting on. pess the net foce cting on the point chge : pess the foce on point chge due to the point chge Q on the is: Q on is, Q t 5, Q on is, Q i R pess the net foce on point chge due to the point chges t 5 : Q t 5, Q cos5 i R Q i R

28 Chpte Substitute fo Qon is, nd Q i Q i R Q t 5, to obtin: R Q ( )i R 6 The stuctue of the NH molecule is ppoimtel tht of n euiltel tethedon, with thee H ions foming the bse nd n N ion t the pe of the tethedon. The length of ech side is.6 m. Clculte the electic foce tht cts on ech ion. ictue the oblem Let the H ions be in the - plne with H t (,, ), H t (,, ), nd H t,,. The N ion, with chge in ou nottion, is then t,, whee.6 m. To simplif ou clcultions 9 we ll set ke C 8.56 N. We cn ppl Coulomb s lw nd the pinciple of supeposition of foces to find the net foce cting on ech ion. z,, (,,),, pess the net foce cting on point chge :,,, ind :, ( ) C i i,, C,

29 The lectic ield I: Discete Chge Distibutions 5 ind :, j i j i,,, C C Noting tht the mgnitude of point chge is thee times tht of the othe point chges nd tht it is negtive, epess, : k j i k j i k j i,, C C C C Substitute in the epession fo to obtin: k k j i j i i 6 C C C C om smmet considetions: k 6 C pess the condition tht the molecule is in euilibium: Solve fo nd evlute : ( ) k 6 C

30 6 Chpte The lectic ield 7 [SSM] A point chge of. μc is t the oigin. Wht is the mgnitude nd diection of the electic field on the is t () 6. m, nd (b) m? (c) Sketch the function vesus fo both positive nd negtive vlues of. (Remembe tht is negtive when points in the diection.) ictue the oblem Let epesent the point chge t the oigin nd use Coulomb s lw fo due to point chge to find the electic field t 6. m nd m. () pess the electic field t point locted distnce fom point chge : ( ), vlute this epession fo 6. m: (.m) N m C 6 9 ( 6.m) (. μc) i (. kn/c)i (b) vlute t m: 9 N m C ( m) ( m) (. μc) ( i ) (.6 kn/c)i (c) The following gph ws plotted using spedsheet pogm: 5 5 (N/C) (m)

31 The lectic ield I: Discete Chge Distibutions 7 8 Two point chges, ech. μc, e on the is; one point chge is t the oigin nd the othe is t 8. m. ind the electic field on the is t (). m, (b). m, (c) 6. m, nd (d) m. (e) At wht point on the is is the electic field zeo? (f) Sketch vesus fo. m < < m. ictue the oblem Let epesent the point chges of. μc nd use Coulomb s lw fo due to point chge nd the pinciple of supeposition fo fields to find the electic field t the loctions specified. Noting tht, use Coulomb s lw nd the pinciple of supeposition to epess the electic field due to the given chges t point distnce fom the oigin: ( ) ( ) ( ), ( 6kN m /C) ( 8.m ),, () Appl this eution to the point t. m: (.m) ( 6kN m /C) ( ) ( i ).m ( 8.m ), (b) vlute t. m: (.m) ( 6kN m /C) ( ) () i (c) vlute t 6. m:.m (.m) ( 6kN m /C) ( ) () i 6 (d) vlute t m: 6.m m (.m) ( m) ( 6kN m /C) ( ) () i ( ), 8.m, ( ) ( i ) ( 9. kn/c m )i ( ) ( i ) ( 8. kn/c 6.m )i ( ) ( i ) ( 8. kn/c.m )i () i ( 9. kn/c)i

32 8 Chpte (e) om smmet considetions: (.m) (f) The following gph ws plotted using spedsheet pogm: 75 5 ( kn m /C) (m) 9 When.-nC point chge is plced t the oigin, it epeiences n electic foce of 8. N in the diection. () Wht is the electic field t the oigin? (b) Wht would be the electic foce on.-nc point chge plced t the oigin? (c) If this foce is due to the electic field of point chge on the is t. cm, wht is the vlue of tht chge? ictue the oblem We cn find the electic field t the oigin fom its definition nd the electic foce on point chge plced thee using. We cn ppl Coulomb s lw to find the vlue of the point chge plced t cm. () Appl the definition of electic field to obtin: (,).nc 5 (. N/C)j (,) ( 8. N) j (b) The foce on point chge in n electic field is given b: (,) (,) (.nc)( (.6mN) j ) kn/c j (c) Appl Coulomb s lw to obtin: (.nc) (. m) ( j ) (.6 mn)j

33 The lectic ield I: Discete Chge Distibutions 9 Solving fo ields: (.6 mn)(.m) 9 ( N m /C )(.nc) nc The electic field ne the sufce of th points downwd nd hs mgnitude of 5 N/C. () Compe the mgnitude of the upwd electic foce on n electon with the mgnitude of the gvittionl foce on the electon. (b) Wht chge should be plced on ping pong bll of mss.7 g so tht the electic foce blnces the weight of the bll ne th s sufce? ictue the oblem We cn compe the electic nd gvittionl foces cting on n electon b epessing thei tio. Becuse the ping pong bll is in euilibium unde the influence of the electic nd gvittionl foces cting on it, we cn use the condition fo tnsltionl euilibium to find the chge tht would hve to be plced on it in ode to blnce th s gvittionl foce on it. () pess the mgnitude of the electic foce cting on the electon: pess the mgnitude of the gvittionl foce cting on the electon: pess the tio of these foces to obtin: Substitute numeicl vlues nd evlute e / g : (b) Letting the upwd diection be positive, ppl the condition fo sttic euilibium to the ping pong bll to obtin: e e g e g e g m g e e mg 9 (.6 C)( 5 N/C) ( 9.9 kg)( 9.8m/s ).69 o e.69 ( ) g Thus, the electic foce is gete b fcto of.69. o e g mg mg

34 Chpte Substitute numeicl vlues nd evlute : (. kg)( 9.8m/s ).96 mc 5 N/C [SSM] Two point chges nd both hve chge eul to 6. nc nd e on the is t. cm nd. cm espectivel. () Wht is the mgnitude nd diection of the electic field on the is t. cm? (b) Wht is the foce eeted on thid chge. nc when it is plced on the is t. cm? ictue the oblem The digm shows the loctions of the point chges nd nd the point on the is t which we e to find. om smmet considetions we cn conclude tht the component of t n point on the is is zeo. We cn use Coulomb s lw fo the electic field due to point chges nd the pinciple of supeposition fo fields to find the field t n point on the is nd to find the foce on point chge plced on the is t. cm., cm. 6. nc θ. θ, cm. 6. nc () Letting, epess the - component of the electic field due to one point chge s function of the distnce fom eithe point chge to the point of inteest: cos θ i pess fo both chges: cos i θ

35 The lectic ield I: Discete Chge Distibutions Substitute fo cosθ nd, substitute numeicl vlues, nd evlute to obtin: (. cm).m i (. m) i 9 ( N m /C )( 6.nC)( [(. m) (. m) ] (.5kN/C) i ( 5kN/C)i ). m i The mgnitude nd diection of the electic field t. cm is: 5 (b) Appl to find the foce on point chge plced on the is t. cm: (.nc)(.5kn/c) ( 69μN)i i A point chge of 5. μc is locted on the is t. cm, nd second point chge of 8. μc is locted on the i t. cm. Whee should thid chge of 6. μc be plced so tht the electic field t the oigin is zeo? ictue the oblem If the electic field t is zeo, both its nd components must be zeo. The onl w this condition cn be stisfied with point chges of 5. μc nd 8. μc on the is is if the point chge 6. μc is lso on the is. Let the subscipts 5, 8, nd 6 identif the point chges nd thei fields. We cn use Coulomb s lw fo due to point chge nd the pinciple of supeposition fo fields to detemine whee the 6. μc point chge should be locted so tht the electic field t is zeo. pess the electic field t in tems of the fields due to the point chges of 5. μc, 8. μc, nd 6. μc: ( ) 5 μc 8 μc 6 μc Substitute fo ech of the fields to obtin: o i ( 8 i ) ( i )

36 Chpte Divide out the unit vecto obtin: î to Substitute numeicl vlues to 6 8 obtin: (.cm) (.cm) 5 6 Solving fo 6 ields: 6.cm A 5.-μC point chge is locted t. m,. m nd -μc point chge is locted t. m,. m. () ind the mgnitude nd diection of the electic field t. m,. (b) Clculte the mgnitude nd diection of the electic foce on n electon tht is plced t. m,. ictue the oblem The digm shows the electic field vectos t the point of inteest due to the two point chges. We cn use Coulomb s lw fo due to point chges nd the supeposition pinciple fo electic fields to find. We cn ppl to find the foce on n electon t (. m, )., m μ C, m 5.μC () pess the electic field t (. m, ) due to the point chges nd :

37 The lectic ield I: Discete Chge Distibutions Substitute numeicl vlues nd evlute :,, 9 ( N m /C )( 5. μc) ( 5.m) (.m) (.55 N/C)(.98i.7j ) (. kn/c) i (.575kN/C)j Substitute numeicl vlues nd evlute : ( 5.m) i (.m) ( 5.m) (.m) j,, 9 ( N m /C )( μc) (.m) (.m) (.5 N/C)(.77 i.77 j ) ( 9.5 kn/c) i ( 9.5 kn/c)j (.m) i (.m) (.m) (.m) j Substitute fo nd nd simplif to find : (. kn/c) i (.575 kn/c) j ( 9.5 kn/c) i ( 9.5 kn/c) ( 8. kn/c) i (.kn/c)j j The mgnitude of is: ( 8.kN/C) (.kn/c) kn/c The diection of is:.kn/c θ tn 8. kn/c Note tht the ngle etuned b ou.kn/c clculto fo tn is the 8. kn/c efeence ngle nd must be incesed b 8 to ield θ. (b) pess nd evlute the foce on n electon t point : [ j ] 9 (.6 C)( 8. kn/c) i (.kn/c) 5 ( ) i 5. N (.6 N)j

38 Chpte ind the mgnitude of : 5 5 (. N) (.6 N). 5 N ind the diection of : θ 5.6 N tn 5. N 5 Two eul positive chges e on the is; one point chge is t nd the othe is t. () Show tht on the is the component of the electic field is given b /( ) /. (b) Show tht ne the oigin, whee is much smlle thn, /. (c) Show tht fo vlues of much lge thn, /. plin wh peson might epect this esult even without deiving it b tking the ppopite limit. ictue the oblem The digm shows the loctions of point chges nd nd the point on the is t which we e to find. om smmet considetions we cn conclude tht the component of t n point on the is is zeo. We cn use Coulomb s lw fo the electic field due to point chges nd the pinciple of supeposition of fields to find the field t n point on the is. We cn estblish the esults clled fo in ts (b) nd (c) b fctoing the dicnd nd using the ppoimtion α wheneve α <<. θ θ () pess the -component of the electic field due to the point chges t nd s function of the distnce fom eithe chge to point : cos i θ

39 The lectic ield I: Discete Chge Distibutions 5 Substitute fo cosθ nd to obtin: ( ) i i i ( ) i The mgnitude of is: ( ) (b) o <<,, so: ( ) (c) o >>, the chges septed b would ppe to be single chge of mgnitude. Its field would be given b. cto the dicnd to obtin: o <<, nd: [ ] 5 A 5.-μC point chge is locted t. m,. m nd.-μc point chge is locted t. m,. m. () ind the mgnitude nd diection of the electic field t. m,. m. (b) ind the mgnitude nd diection of the foce on poton plced t. m,. m. ictue the oblem The digm shows the electic field vectos t the point of inteest due to the two point chges. We cn use Coulomb s lw fo due to point chges nd the supeposition pinciple fo electic fields to find. We cn ppl to find the foce on poton t (. m,. m).

40 6 Chpte, m 5.μ C, m.μc () pess the electic field t (. m,. m) due to the point chges nd : vlute :. vlute :, 9 ( N m /C )(. μc) ( 5.m) (.m) ( 5.m) i (.m) ( 5.m) (.m) (.6 kn/c)(.857i.5 j ) (.97 kn/c) i (.5 kn/c) j j,, 9 ( N m /C )( 5. μc) (.m) (.m) (.m) i (.m) (.m) (.m) i.kn/c) (.5kN/C)(.89i.7 j ) (.kn/c) ( j j Substitute nd simplif to find : (.98kN/C) i (.5 kn/c) j (.kn/c) i (.kn/c) (. kn/c) i (.55 kn/c)j j The mgnitude of is: (. kn/c) (.55kN/C).9 kn/c

41 The lectic ield I: Discete Chge Distibutions 7 The diection of is:.55kn/c θ tn. kn/c Note tht the ngle etuned b ou.55kn/c clculto fo tn is the. kn/c efeence ngle nd must be incesed b 8 to ield θ. (b) pess nd evlute the foce on poton t point : The mgnitude of is: [ j ] 9 (.6 C)(. kn/c) i (.55 kn/c) 6 ( ) i 6.76 N (.8 N)j (.76 N) (.8 N). N The diection of is: 6.8 N θ tn 6.76 N whee, s noted bove, the ngle etuned b ou clculto fo 6.8 N tn is the efeence 6.76 N ngle nd must be incesed b 8 to ield θ. 6 Two positive point chges, ech hving chge Q, e on the is, one t nd the othe t. () Show tht the electic field stength on the is is getest t the points / nd / b computing / nd setting the deivtive eul to zeo. (b) Sketch the function vesus using ou esults fo t () of this poblem nd the fct tht is ppoimtel / when is much smlle thn nd is ppoimtel / when is much lge thn,. ictue the oblem In oblem it is shown tht the electic field on the is, due to eul positive chges locted t (, ) nd (,), is given b ( ) vlues b setting / eul to zeo.. We cn identif the loctions t which hs it getest

42 8 Chpte () vlute to obtin: d d [ ( ) ] d ( ) d d d ( ) ( ) [ ] 5 ( ) ( ) ( ) 5 [ ( ) ( ) ] Set this deivtive eul to zeo fo eteme vlues: 5 ( ) ( ) Solving fo ields: ± (b) The following gph ws plotted using spedsheet pogm: nd [SSM] Two point pticles, ech hving chge, sit on the bse of n euiltel tingle tht hs sides of length L s shown in igue -8. A thid point pticle tht hs chge eul to sits t the pe of the tingle. Whee must fouth point pticle tht hs chge eul to be plced in ode tht the electic field t the cente of the tingle be zeo? (The cente is in the plne of the tingle nd euidistnt fom the thee vetices.)

43 The lectic ield I: Discete Chge Distibutions 9 ictue the oblem The electic field of th chged point pticle must cncel the sum of the electic fields due to the othe thee chged point pticles. B smmet, the position of the th chged point pticle must lie on the veticl centeline of the tingle. Using tigonomet, one cn show tht the cente of n euiltel tingle is distnce L fom ech vete, whee L is the length of the side of the tingle. Note tht the components of the fields due to the bse chged pticles cncel ech othe, so we onl need concen ouselves with the components of the fields due to the chged point pticles t the vetices of the tingle. Choose coodinte sstem in which the oigin is t the midpoint of the bse of the tingle, the diection is to the ight, nd the diection is upwd. Note tht the components of the electic field vectos dd up to zeo. L 6 L L 6 pess the condition tht must be stisfied if the electic field t the cente of the tingle is to be zeo: i to i Substituting fo,,, nd ields: ( ) k( ) k( ) k L cos6 j L cos6 j L j j Solving fo ields: ± L

44 Chpte Becuse the positive solution coesponds to the th chge being t the cente of the tingle, it follows tht: L The chge must be plced distnce eul to the length of the side of the tingle divided b the sue oot of thee below the midpoint of the bse of the tingle, whee L is the length of side of the tingle. 8 Two point pticles, ech hving chge eul to, sit on the bse of n euiltel tingle tht hs sides of length L s shown in igue -8. A thid point pticle tht hs chge eul to sits t the pe of the tingle. A fouth point pticle tht hs chge is plced t the midpoint of the bseline mking the electic field t the cente of the tingle eul to zeo. Wht is the vlue of? (The cente is in the plne of the tingle nd euidistnt fom ll thee vetices.) ictue the oblem The electic field of th chge must cncel the sum of the electic fields due to the othe thee chges. B smmet, the position of the th chge must lie on the veticl centeline of the tingle. Using tigonomet, one cn show tht the cente of n euiltel tingle is distnce L fom ech vete, whee L is the length of the side of the tingle. The distnce fom the cente point of the tingle to the midpoint of the bse is hlf this distnce. Note tht the components of the fields due to the bse chges cncel ech othe, so we onl need concen ouselves with the components of the fields due to the chges t the vetices of the tingle. Choose coodinte sstem in which the oigin is t the midpoint of the bse of the tingle, the diection is to the ight, nd the diection is upwd. L L L 6 6 '

45 pess the condition tht must be stisfied if the electic field t the cente of the tingle is to be zeo: Substituting fo,, The lectic ield I: Discete Chge Distibutions, nd ields: i to i ( ) k( ) k( ) k L cos6 j L cos6 j L ' j L j Solve fo to obtin: ' 9 Two eul positive point chges e on the is; one is t nd the othe is t. The electic field t the oigin is zeo. A test chge plced t the oigin will theefoe be in euilibium. () Discuss the stbilit of the euilibium fo positive test chge b consideing smll displcements fom euilibium long the is nd smll displcements long the is. (b) Repet t () fo negtive test chge. (c) ind the mgnitude nd sign of chge tht when plced t the oigin esults in net foce of zeo on ech of the thee chges. ictue the oblem We cn detemine the stbilit of the euilibium in t () nd t (b) b consideing the foces the eul point chges t nd eet on the test chge when it is given smll displcement long eithe the o is. (c) The ppliction of Coulomb s lw in t (c) will led to the mgnitude nd sign of the chge tht must be plced t the oigin in ode tht net foce of zeo is epeienced b ech of the thee point chges. () Becuse is in the diection, positive test chge tht is displced fom (, ) in eithe the diection o the diection will epeience foce pointing w fom the oigin nd ccelete in the diection of the foce. Conseuentl, the euilibium t (,) is unstble fo smll displcement long the is. If the positive test chge is displced in the diection of incesing (the positive diection), the chge t will eet gete foce thn the chge t, nd the net foce is then in the diection; i.e., it is estoing foce. Simill, if the positive test chge is displced in the diection of decesing (the negtive diection), the chge t will eet gete foce thn the chge t, nd the net foce is then in the diection; i.e., it is estoing foce. Conseuentl, the euilibium t (,) is stble fo smll displcement long the is.

46 Chpte (b) ollowing the sme guments s in t (), one finds tht, fo negtive test chge, the euilibium is stble t (,) fo displcements long the is nd unstble fo displcements long the is. (c) pess the net foce cting on the t chge t : ( ) Solve fo to obtin: (d) If point chge is given smll displcement long the is, the foce eeted on it b the neest positive point chge will incese nd the foce eeted on it b the positive point chge tht is fthe w will decese. Hence thee will be net foce in the diection of its displcement tht will ccelete the point chge w fom the oigin. The euilibium of the sstem is unstble. Remks: In t (c), we could just s well hve epessed the net foce cting on the chge t. Due to the smmetic distibution of the chges t nd, summing the foces cting on t the oigin does not led to eltionship between nd. 5 Two positive point chges e on the is t nd. A bed of mss m nd chge slides without fiction long tut thed tht uns long the is. Let be the position of the bed. () Show tht fo <<, the bed epeiences line estoing foce ( foce tht is popotionl to nd diected towd the euilibium position t ) nd theefoe undegoes simple hmonic motion. (b) ind the peiod of the motion. ictue the oblem In oblem it is shown tht the electic field on the is, due to eul positive point chges locted t (, ) nd (,), is given b ( ). We cn use T π m k' to epess the peiod of the motion of the bed in tems of the estoing constnt k. () pess the foce cting on the bed when its displcement fom the oigin is : ( ) cto fom the denominto to obtin: