ELETROSTATI ENERGY AND APAITORS EXERISES Section. Electrosttic Energy. INTERPRET We re to find the work required to ssemble liner sequence of chrges. DEVELOP We use the technique described for ssembling the three chrges in Figure.. For this problem, we hve 4 chrges, to be rrnged s shown in the figure below. Number the chrges qi 75 μ, i,,, 4, s they re spced long the line t 5. cm intervls. There re six pirs, so which we cn evlute to find the work W. kqiq W r pirs ij j EVALUATE Evluting the expression bove gives 4 4 4 W k + + + + + 9 ( )( ) kq kq 9. N m / 75 μ ( + + + + + ) 4.4 kj..5 m The work required does not depend on how the chrge configurtion is ssembled, only on its finl stte. 4. INTERPRET This problem is similr to the preceding problem, except tht the finl geometry of the ssembled point chrges is different (nd the point chrges do not ll hve the sme chrge). We cn thus pply the sme strtegy s in Problem. to find the electrosttic energy for this collection of point chrges. DEVELOP We gin use the strtegy outlined in the discussion ccompnying Figure.. The point chrges re rrnged on the corners of squre of side length. Three of them hve chrge q q q + q, nd the fourth hs chrge q4 q. Mke sketch of the finl chrge distribution nd lbel the chrges (see the figure below). Agin there re six pirs of chrges, so the work W required to ssemble them is which we cn evlute. kqiq W r pirs ij j - opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
- hpter EVALUATE Evluting the expression for work gives 4 4 4 W k + + + + + kq kq + + + This is not prticulrly convenient method of clculting energies. Fortuntely, the chrge on single electron is smll enough tht we cn usully pproximte rel chrge distributions s continuous nd integrte. 5. INTERPRET We re to repet the preceding problem, with the finl chrge chnged from q 4 q/ to q 4 q. DEVELOP Use the sme strtegy s for Problem.4, but with q 4 q EVALUATE Evluting the expression for work required to ssemble collection of point chrges gives 4 4 4 W k + + + + + kq + + Although it tkes work to bring the second nd third chrges into plce, tht energy is regined by the negtive work done in bringing in the fourth chrge. 6. INTERPRET We re to find the speed of the chrges in n electrosttic explosion nd cn use conservtion of energy to ddress this problem. DEVELOP The three chrges q re initilly in symmetric rrngement, s shown in Figure.. The work done to ssemble the configurtion is given: kqq kqq kqq kq W + + This is the initil energy of the configurtion. From the symmetry of the initil configurtion, we cn rgue tht they will ech gin of this energy when they fly prt, so we cn find the speed v from their kinetic energy. EVALUATE The initil energy is Ei kq nd the finl energy is Ef ( ) mv. By conservtion of energy, these energies must be equl, so kq k mv v q m onservtion of energy pplies to more thn just mechnicl systems. This technique is used throughout ll res of physics nd is fundmentlly relted to time symmetry. 7. INTERPRET For this problem, we re to find the work required to ssemble crude model of wter molecule. Note tht if the work is negtive, then energy is relesed in forming the molecule. DEVELOP In this pproximtion, electrosttic potentil energy of the wter molecule (i.e., the work required to ssemble the molecule) is kqiq j U W r pirs ij The two oxygen-hydrogen pirs hve seprtion cos 7.5.59. ( ) m, while the hydrogen-hydrogen pir hs seprtion opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
EVALUATE Evluting this expression gives e e.7ke U ke ke +.59 ( )( ) Electrosttic Energy nd pcitors - 9 9 8.7 9. N m /.6 7.76 J 48.5 ev. Becuse the potentil energy is negtive, ssembling this molecules releses energy (or does work). Note tht the electrosttic potentil energy of the ssembled molecule is with respect to the constituents being infinitely fr prt, so the work done equtes to the chnge in potentil energy cused by bringing the chrges together from infinity. Section. pcitors 8. INTERPRET This problem is bout prllel-plte cpcitor. We re given the plte seprtion nd the chrges on the pltes nd re sked to find the electric field between the pltes, the potentil difference between the pltes, nd the energy stored in the cpcitor. DEVELOP The electric field between two closely spced, oppositely chrged, prllel conducting pltes is pproximtely uniform (directed from the positive to the negtive plte), with strength (see Eqution.8) σ q E ε εa Since the electric field is uniform, the potentil difference between the pltes is given by Eqution.b, V Ed, where d is the plte seprtion. Finlly, the energy stored in the cpcitor cn be clculted using Eqution.: U V qv where q V. EVALUATE () Using the eqution bove, the electric field is (b) The potentil difference is (c) The energy stored is q. μ E. MV/m ε ( ) ( ) A 8.85 / N m.5 m ( )( ) V Ed.99 MV/m.5 m 9.9 kv. μ 9.94 kv 5.5 mj ( )( ) U V qv Note tht the finl results re given to two significnt figures, s wrrnted by the dt. When used s intermedite results, however, significnt figures re retined. For completeness, the cpcitnce of the cpcitor is d The vlue is typicl of cpcitor. ( ) ( ) ε 8.85 / N m.5 m A. F.nF 5. m 9. INTERPRET We re to find the work required to chrge cpcitor with the given chrge, then find the dditionl work required to double the chrge. DEVELOP The seprtion between cpcitor pltes is much smller thn the liner dimensions of the pltes, so the discussion in Section. pplies. From Eqution., we see tht the work is W V where V is the finl voltge nd my be expressed using Eqution., Q/V. This gives Q Q W opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
-4 hpter The cpcitnce cn be expressed in terms of the geometry of the cpcity (Eqution., ε A d, which leds to EVALUATE Q Q d W ε A () The work required to trnsfer Q 7. μ is ( 7. μ)(.m) Qd ε A 8.85 / N m.5 m W.4 J (b) The dditionl work required to double the chrge on ech plte is fces. ( Q) ( ) ( ) d Δ W W W 4. J ε A This energy is stored in the cpcitor nd cn be relesed by electriclly connecting the two cpcitor. INTERPRET This problem is bout the energy stored in prllel-plte cpcitor. We re to find the chrge on ech plte (oppositely chrged, of course) needed to store the given energy, nd the resulting electric potentil between the pltes. DEVELOP Using the expression from Problem.9 for the work (i.e., chnge in potentil energy) required to chrge the pltes, we cn solve for the chrge: Qd ε AU U Q ε A d Once Q is known, the potentil difference V between the pltes is U U V ( V) V QV V Q EVALUATE () Using the vlues given in the problem sttement, we find the chrge to be ( ) ( ) ( ) ε 8.85 / N m.5 m 5 J AU Q.74 μ d. m (b) The potentil difference is U 5 ( J) V 4 kv 6 Q.744 Since the electric field between the pltes is E σ ε the potentil cn lso be found using E σ Qd V 4 kv d ε d ε A. INTERPRET We re given the chrge nd voltge of cpcitor nd re to find the cpcitnce. DEVELOP Apply Eqution., Q/V. EVALUATE From Eqution., Q V ( ) ( ).μ 6V nf. The cpcitnce is the chrge per unit voltge.. INTERPRET We re to derive n lternte form of the units of ε, the permittivity constnt. DEVELOP The vlue of ε is 8.85 /( N m ) with units / ( N m ). However, becuse QV /, F (frd) is equivlent to /V. EVALUATE From the bove, we see tht the units of ε re ( ) ( ) ( ) ( )( ) ( ) / N m / N m /m /J /m / V m F/m opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
To see tht the result mkes sense, we cn use Eqution. nd write The units of re F, while the units of da / re m. ε d A Electrosttic Energy nd pcitors -5. INTERPRET We re given the seprtion of prllel-plte cpcitor, its chrge, nd its voltge nd re to find its cpcitnce. DEVELOP Becuse the plte seprtion d r, the rdius of the cpcitor, nd pply Eqution., ε A d, with A πr. EVALUATE Inserting the given quntities gives ( ) π ( ) ε 74 pf d.5 m to two significnt figures. This is typicl vlue for cpcitnce. 8.85 pf/m. m π r 4. INTERPRET For prllel-plte cpcitor, we re given the plte spcing, the chrge, nd the voltge nd re to find the plte re. DEVELOP ombining Equtions. nd. for cpcitnce gives Q ε A Qd A V d εv EVALUATE Inserting the given quntities gives 6 Qd (. )(. m) A.9 m εv (8.85 /N m ) ( 5 V) If we tke the plte to be squre, then ech side hs length of.8 m, which indeed is much greter thn the distnce of. mm between the pltes. Note expressing voltge in its SI units of N m/ gives the proper units. 5. INTERPRET We re given the cpcitnce nd the voltge of cpcitor nd re to find the stored energy. DEVELOP Apply Eqution., U V /. EVALUATE From Eqution., ( )( ) U V 5 μf 5 V.5 J This is the energy it would tke to lift. liter of wter through height of U mgh U.5 J h 5 cm mg (. kg)( 9.8 m/s ) 6. INTERPRET We re given the energy stored in cpcitor nd its voltge nd re to find its cpcitnce. DEVELOP Eqution., U V /, provides the connection between the stored energy U, the cpcitnce, nd the potentil V. EVALUATE Inserting the given vlues gives ( ) ( V) U 5 μj 7 nf V 6 The vlue is within the typicl rnge of cpcitnce ( F to F). Section. Using pcitors 7. INTERPRET This problem involves clculting the equivlent cpcitnce for the two given cpcitors connected in series or in prllel. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
-6 hpter DEVELOP Apply Eqution.5 to find the equivlent series cpcitnce nd Eqution.6b for the equivlent prllel cpcitnce. EVALUATE In prllel, the cpcitnce is In series, the cpcitnce is series. +. μf+. μf. μf (. μf)(. μf ) +. μf +. μf / μf onnecting the cpcitors in prllel results in higher equivlent cpcitnce thn connecting them in 8. INTERPRET This involves connecting two cpcitors in series nd finding their reltive cpcitnce, given the voltge cross ech cpcitor nd the voltge cross the cpcitor pir. DEVELOP For two cpcitors connected in series, the chrge on ech cpcitor must be the sme (see discussion ccompnying Figure.8), Q Q V V Given tht V V, we cn find the rtio of the cpcitnces. EVALUATE Becuse V V, the expression bove gives The equivlent cpcitnce of two cpcitors connected in series is + + In our cse, / since, nd the voltge cross is Q Q Q V V / which is precisely the condition given in the problem sttement. 9. INTERPRET This problem requires us to find the equivlent cpcitnce of the given rrngement of individul cpcitors, which combines series nd prllel connections. DEVELOP For prt (), compute the equivlent cpcitnce of nd (which re in prllel), then combine this in series with to find the overll cpcitnce. For prt (b), note tht the chrge on nd must be the sme s on, which must be the sme s the totl chrge (see discussion ccompnying Figure.8), so Q T Q Q + Q In ddition, the totl chrge nd cpcitnce must stisfy Eqution., T VT where V T. V. Finlly, the voltge drop cross must be the sme s cross, so V Q/ V Q/ Q Q EVALUATE () pcitors nd combined in prllel give n equivlent cpcitnce of, +. μf +. μf. μf ombining this in series with gives, (. μf)(. μf ) T. μf +. μf +. μf, Q T opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
(b) Knowing T nd V T, we cn find Q T, which must be the sme s Q (see Exmple.) QT Q T VT VT Q TVT (. μf )(. V) 4.4 μ Knowing Q, we cn solve for Q nd Q using the expressions bove. The result is Q Q + Q Q + Q Q + Q 4.4 μ Q 9.6 μ Finlly, Q is + + Q 9.6 μ 4.8μ ( ) Q (c) The voltge on cn be found using Eqution.. The result is Q 4.4 μ V 7. V. μf V V V V. V 7. V 4.8 V T The voltge cross the prllel cpcitors my lso be found using Eqution.: Q Q V V 4.8 V Electrosttic Energy nd pcitors -7. INTERPRET This problem requires us to find ll possible equivlent cpcitnces tht cn be obtined connecting three cpcitors in different wys (seril, prllel, nd combintions thereof). DEVELOP The equivlent cpcitnce is mximized when ll cpcitors re connected in prllel nd is given by (Eqution.5) + + On the other hnd, is minimized when the cpcitors re connected in series. For three cpcitors connected in series, the equivlent cpcitnce is given by (Eqution.6) + + + + Intermedite vlues re obtined when one cpcitor is in prllel with the other two in series, or one in series with the other two in prllel. EVALUATE () When ll cpcitors re in prllel, we hve + +. μf +. μf +. μf 6. μf (b) When ll re in series, the equivlent cpcitnce is (. μf)(. μf)(. μf ) 6. μf.55 μf + + (. μf)(. μf) + (. μf)(. μf) + (. μf)(. μf) (c) When one is in prllel with the other two in series, the possible vlues re ( )( ). μf. μf +. μf + μf. μf +. μf +. μf 5. (. μf)(. μf ). μf μf.8 μf + + +. μf +. μf 4. ( )( ). μf. μf. μf μf.7 μf + + +. μf +. μf. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
-8 hpter Similrly, when one is in series with the other two in prllel, the equivlent cpcitnce is i( j + k) + i j + k i + j + k Therefore, the possible vlues re ( + ) (. μf)(. μf +. μf ) 5. μf.9 μf + +. μf +. μf +. μf 6. ( + ) (. μf)(. μf +. μf ) 4. μf. μf + +. μf +. μf +. μf. ( + ) (. μf)(. μf +. μf ). μf.5 μf + +. μf +. μf +. μf. With three cpcitors, ech hving two options (prllel or series), there re eight possible outcomes ( ). Section.4 Energy in the Electric Field. INTERPRET This problem involves finding the uniform electric field tht crries the given energy density. DEVELOP Apply Eqution.7, ( E u ε ), which reltes the field strength nd the electric energy density. EVALUATE Inserting the given energy density into Eqution.7 gives ( ). J/m E u ε 8.85 F/m 5 8. V/m The mnipultion of units is fcilitted by the reltions V J/ nd F /V. Thus, V/m (J/m )/(F/m) V/m / V m ( ). INTERPRET This problem is bout the volume required for storing given mount of electrosttic energy. DEVELOP For uniform electric field, Eqution.8 cn be written s U ε E volume Knowing U nd E llows us to find the volume. EVALUATE Substituting the vlues given in the problem sttement gives the volume s U 4 ( MJ) 9 volume m km εe ( 8.85 pf/m)( kv/m ) This is very big volume occupied by cr bttery. In relity, not ll the stored energy goes into creting the field.. INTERPRET This problem involves finding the mximum electricl energy density possible in ir. DEVELOP 6 From Appendix, we find tht the energy content of gsoline is 44 J/kg, nd the density of gsoline is 67 kg/m, so the equivlent energy density is 6 u gs (44 J/kg)(67 kg/m ).95 J/m EVALUATE From Eqution.7, the field strength giving the sme electrosttic energy density is ( ).95 J/m E u ε 8.6 V/m 8.85 F/m which gretly exceeds the brekdown field in ir. Gsoline is ctully very dense form of energy storge, which is one reson it is hrd to replce! opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
Electrosttic Energy nd pcitors -9 4. INTERPRET In this problem we re sked to find the electric energy stored in proton by ssuming it to be uniformly chrged sphere. DEVELOP For this model of the proton, the field strength t the surfce is E ke/ R (from sphericl symmetry nd Guss s lw). Thus, the energy density in the surfce electric field is PROBLEMS ke ke E 4 u ε 4πk R 8πR 5 EVALUATE With R fm m, the energy density is 9 9 ke (9 N m / )(.6 ) u 9.7 J/m 57 kev/fm 4 5 4 8πR 8 π( m) The energy density is enormous, given the smll size of the proton. 5. INTERPRET This problem involves finding n expression of the work required to ssemble the given chrge configurtion, nd using this expression to find the reltive chrge on one of the chrges with respect to the initil chrge. DEVELOP The work necessry to position Q x is while the work necessry to bring up Q y is kq Q kq Wx kq Q kq Q W Given tht Wy Wx, we cn solve for Q y in terms of Q. EVALUATE W W gives y x x y x y y + ( + ) kq Q y ( + ) kqqy 4kQ 4Q Qy + More explicitly, we cn write Q y.66q, so Q y is little less thn twice Q. 6. INTERPRET This problem involves finding the work required to crete the given chrge distribution. The work is equl to the energy stored in the system. DEVELOP When chrge q (ssumed positive) is on the inner sphere, the potentil difference between the spheres is (Eqution.) b b kqdr V E dr kq r b where we hve used the fct (from Guss s lw; see Exmple.) tht the field outside sphericl chrge distribution is the sme s point chrge t the center of the sphere. To trnsfer n dditionl chrge dq from the outer sphere requires work dw V dq. EVALUATE The totl work required to trnsfer chrge Q (leving the spheres oppositely chrged) is Q Q W V dq kq dq kq b b Since U Q /, this shows tht the cpcitnce of this sphericl cpcitor is b k( b ) k( b ) Note tht cpcitnce depends only on the geometry of the system, nd is independent of V nd Q. 7. INTERPRET This problem requires us to find the chrge on pir of prllel, squre conducting pltes (i.e., prllel-plte cpcitor) given the energy density in the electric field between the pltes. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
- hpter DEVELOP ombine Eqution.7, which reltes the electric field to the energy density, E ± u ε with Eqution.8, which gives the electric field ner the surfce of chrged conducting plte: E σ ε EVALUATE Eliminting the unknown electric field E nd solving for the surfce chrge density σ gives σ ±ε u ε Using the given surfce re, the totl chrge on plte is ( ) ( ) ( ) q σ A± A u ε ±. m 4.5 kj/m 8.85 / N m ±.8 μ Notice tht we do not know the sign on the chrge becuse of the chrge symmetry involved. 8. INTERPRET The problem involves the cpcitnce of cell membrne, given the potentil difference nd chrge cross the membrne. DEVELOP 6 The cpcitnce is in generl Q/ V (Eqution.). The chrge in this cse is Q.5 e. EVALUATE Given the vlues for the cell membrne, the cpcitnce is (.5 6 )(.6 9 ) Q.7 pf V 65 mv Does this mke sense? Wht if we ssumed the cell membrne ws prllel-plte cpcitor with dielectric. Then Eqution.4 pplies: κε A/ d. For the membrne thickness, let d nm. For the cell surfce re, let A ( μm ). And for the dielectric constnt, let s use the one for wter since the cell is lrgely mde up of wter: κ 8. With these choices, the cpcitnce comes out s 7 pf, so our nswer seems to mke sense. 9. INTERPRET In this problem, we re sked to compre the mount of energy stored in two different cpcitors. DEVELOP The energy stored in cpcitor cn be clculted using Eqution.: U V /. EVALUATE The energy stored in ech cpcitor is U V (. μf )( 5 V ) mj V ( 47 pf )(. kv ). mj U Thus, the energy stored in cpcitor is bout 5 times less thn tht stored in cpcitor. The generl expression for the rtio of the energies stored in two cpcitors is U V / V U V / V so the energy stored is liner in cpcitnce but qudrtic in voltge. 4. INTERPRET In this problem we re sked to compre the mount of chrge nd energy stored in three different cpcitors. DEVELOP The chrge stored in cpcitor is Q V (Eqution.), nd the energy stored is U V / (Eqution.). EVALUATE () The chrge stored is Q V (. μf)( V) μ for the first cpcitor Q V (. μf )( V) μ for the second Q V ( μf)( 5 V) 5μ μ to one significnt figure for the third opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
Electrosttic Energy nd pcitors - (b) To one significnt figure, the energy stored in ech cpcitor is U V ( μ )( V ) 45 μj 5 μj U V (. μf )( V ) 5 μj U V ( μf )( 5 V ) 75 μj 4 μj (c) The cost effectiveness eff, mesured in J/ nd to single significnt figure, is 45 μj 6 5 eff 8 J/ J/ 5 5 μj 6 5 eff 4 J/ J/ 5 75 μj 6 5 eff 4. J/.4 J/ 88 Notice tht, for prt (c), more significnt figures were retined for the results of prt (b) becuse these re intermediry results for prt (c). The first cpcitor is the most cost effective of the three. 4. INTERPRET We re to find the voltge cross the given cpcitor nd the power it dischrges if ll its energy is dischrged in the given time. DEVELOP Solve Eqution. for voltge to find the voltge cross the cpcitor. Use the definition of verge power, P E/ Δ t to find the verge power. EVALUATE () From Eqution., V U ( ) ( ) (b) P ΔU Δ t ( J) (.5 ms) kw. / 95 J μf 4.4 kv. This is like connecting yourself to 6-W light bulbs for.5 ms. re to try? 4. INTERPRET This problem involves finding the power consumption of cmer flshtube per flsh nd the verge power consumed if the flsh is used every s. From this power, we re to find the cpcitor required to power the flsh. DEVELOP Eqution 6.5 gives the verge power P ΔW Δ t, where ΔW is the work done (or the energy used). Apply this to find the power consumed by the flsh for the different time intervls Δt. Once we find the verge power required, we pply Eqution., U V / to find the cpcitnce required. EVALUATE () The power delivered to the tube during the flsh is ΔW 5. J Pflsh 5. kw Δt. ms (b) The energy stored in cpcitor is U V Thus, the cpcitnce needed to supply the flsh energy is. U (5. J) 5 μf V ( V) P 5. J s.5 W. 4 The verge power P is times P flsh, which is the sme rtio s for the time intervls ( ms to s). 4. INTERPRET In this problem, we re given the dt of the energy stored in cpcitor t different voltges. From the dt we re sked to deduce the cpcitnce. DEVELOP The energy stored in cpcitor cn be clculted using Eqution.: U V /. Thus, plotting U versus V should give stright line with slope /. (c) The verge power consumption during the -s intervl is ( ) ( ) opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
- hpter EVALUATE The plot is shown below. The slop is.644 kj/v 64.4 /V 64.4 F, from which we find the cpcitnce to be (64.4 F) 9 F. The energy stored increses qudrticlly with voltge. The generl expression for the rtio of the energies stored in two different voltges is U V/ V U V/ V 44. INTERPRET In this problem we wnt to connect cpcitors of known cpcitnce nd voltge rting to obtin the desired equivlent cpcitnce nd voltge rting. DEVELOP In prllel, the voltge cross ech element is the sme, so to increse the voltge rting of combintion of equl cpcitors, series connections must be considered. The generl result of Problem shows tht the voltge cross n cpcitors in series, ech of which is rted t voltge V, is simply nv. The voltge cross cpcitors in prllel does not chnge, so we cn combine cpcitors in series to djust the voltge nd cpcitors in prllel to djust the overll cpcitnce (without chnging the voltge rting). EVALUATE () To obtin the desired voltge rting, we must use two cpcitors in series so tht the voltge becomes Vseries pir V + V + 5V+ 5V V. However, the overll cpcitnce is now (. μf )(. μf ) series pir. μf +. μf +. μf nd we need to increse the totl cpcitnce to twice tht of just two in series, without ltering the voltge rting. This cn be ccomplished with prllel combintion of two series pir. This gives voltge rting of V series pir nd cpcitnce of series pir. μf. Thus, the equivlent cpcitnce would be s shown in prt () of the figure below. (b) For n equivlent cpcitnce of.5 μf nd voltge of 5 V, four cpcitors in series re required, which gives rting V tot 4V 4(5 V) V, nd cpcitnce ( ) 4. μf.5 μf Schemticlly the connections described look like the following: One my use Equtions.5 nd.6 to verify tht the configurtions indeed hve the desired cpcitnce nd voltge rting s specified in the problem sttement. 45. INTERPRET For this problem, we re to find the equivlent cpcitnce for the given cpcitor. DEVELOP Number the cpcitors s shown in the figure below. Reltive to points A nd B,, 4, nd the combintion of nd re in series, so the cpcitnce is given by Eqution.6:. AB + 4 + opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
is prllel combintion, so +, nd we cn find AB. Electrosttic Energy nd pcitors - EVALUATE Inserting the expression for into tht for AB gives AB AB (. μf) (. μf) (. μf. μf ) μf.86 μf 6. 7. + + + The result is given to two significnt figures, s justified by the dt. 46. INTERPRET This problem involves n ssemblge of cpcitors in circuit, nd we re interested in the energy stored in one of the cpcitors. To do this, we will need to find the voltge cross tht cpcitor when the given voltge is pplied cross the entire circuit. DEVELOP Number the cpcitors s shown below. Reltive to points A nd B,, 4, nd the combintion of nd re in series. The energy of the. μf cpcitor is U V (Eqution.), where the voltge drop cross is V V, since nd re in prllel. In ddition, since,, nd re in series, we hve 4 V V+ V4 + V nd Q V. V 4 4 V Once we know V, U cn be clculted. EVALUATE From the equtions bove, we find 7 VAB V + + V + + V V VAB 7 4 This gives (. μf ). U V 5 V μj 7. to two significnt figures. Since the combintion of nd re in series with nd 4, does not feel the full 5 V pplied cross AB, so its working voltge is lower. In fct we find it to be V (/7) V. 47. INTERPRET This problem involves two cpcitors in series. We re to find n expression for the voltge cross ech cpcitor. DEVELOP From Eqution.5, we hve + Furthermore, the chrge on ech cpcitor must be the sme, s rgued in the text ccompnying Figure.8, so, from Eqution., Q V V V. In ddition, the voltges cross ech cpcitor must sum to the voltge cross the equivlent cpcitor, so V V + V. EVALUATE ombining the expressions bove to solve for V in terms of V,, nd gives V V V V V V V Q V V V + + + ( ) opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
-4 hpter Solving for V likewise gives V V V V V V V Q V V V + + + ( ) These expressions gree with tht given in the problem sttement. 48. INTERPRET You wnt to know the working voltge for two cpcitors connected in series. DEVELOP The voltge cross two cpcitors in series is just the sum of the voltge cross ech cpcitor. Your compny s new hire pplied this sme logic to the working voltge, which is the mximum tht the cpcitors cn hndle. However, the working voltge bsiclly tells you the mximum chrge tht cpcitor cn hndle: Q. mx V work Since the chrge on ech cpcitor in the series is the sme (see Figure.8), the working voltge will be set by the smllest mximum chrge of the cpcitors in the series. EVALUATE The mximum chrges on the two cpcitors re Q V (. μf)( 5 V) 5. μ Q V. μf V 4 μ ( )( ) Thus, the most chrge tht the series cn hndle is 5. μ, which mens the working voltge is Q Q Vwork + ( 5. μ) + 75 V. μf. μf The new hire ws wrong: 75 V is the sfe limit. Wht this tells you is tht you should try to mtch the mximum llowble chrge of cpcitors tht you put in series. 49. INTERPRET For this problem, we re to find the cpcitnce nd working voltge of cpcitor given its geometry nd dielectric mteril. DEVELOP Apply Eqution.4 nd use Tble. for the dielectric constnt of polyethylene. EVALUATE () From Eqution.4, nd Tble., one obtins ( ) ( ) 4 ε 8.85 pf/m (5 m ) A κ. 4. nf 6 d 5 m (b) Dielectric brekdown in polyethylene occurs t field strength of 5 kv/mm, corresponding to mximum voltge for this cpcitor of 6 6 V Ed 5 V/m (5 m). kv ( ) The results re given to two significnt figures, s wrrnted by the dt. 5. INTERPRET This problem involves cpcitor using polystyrene s the dielectric between the conductor pltes. DEVELOP From Eqution.4, the cpcitnce of two prllel pltes with dielectric in between is κ κε A/ d. As for prt (b), the mximum voltge for prllel-plte cpcitor is the brekdown field times the distnce between the pltes: Vmx Ebrek d. EVALUATE () The dielectric constnt of polystyrene is given in Tble., so its thickness is: ( )( )( ( ) ) κε.6 8.85 5 cm A π Nm d.46 mm.5 mm 47 pf opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
(b) The brekdown field of polystyrene is given in Tble., so the mximum voltge is: V mx ( 5 MV/m)(.46 mm) 87 kv Electrosttic Energy nd pcitors -5 These vlues seem resonble for the thickness nd working voltge of cpcitor. Notice tht to increse the working voltge for given geometry one would like mteril with both lrge dielectric constnt nd lrge brekdown field. 5. INTERPRET In this problem, we re given the cpcitnce per unit re nd the dielectric strength for cpcitor nd re sked to find the plte seprtion. DEVELOP If we ssume tht the inner nd outer surfces of the membrne ct like prllel-plte cpcitor, with the spce between the pltes filled with mteril of dielectric constnt κ, then we cn use Eqution.4 to find the seprtion d. EVALUATE The cpcitnce per unit re is A κ ε d. Thus, d ( 8.85 pf/m ) ( μ F/cm ).7 nm. This result is bout n order of mgnitude lrger thn the Bohr rdius, which gives n ide of the thickness of the membrne in terms of toms (biologicl molecules being lrgely hydrogen nd crbon). 5. INTERPRET You wnt to see if trimmer cpcitors cn provide the extr cpcitnce you re looking for. DEVELOP Plcing the trimmer cpcitors in prllel with the chep -μf cpcitors will result in combined cpcitnce of. trim μf EVALUATE Since you wnt. μf of cpcitnce nd the chep cpcitors only provide between.7 nd.9 μf, the trimmer cpcitors hve to supply between.7 μf nf trim μf. μf.9 μf nf Since these vlues re both within the vrible limits of the trimmer cpcitors (5 nd 5 nf), they will work for this ppliction. One wy to mke vrible cpcitor is to llow the plte seprtion, d, to be chnged, sy, with smll screw. Another wy is to vry the re, A, by shifting the two pltes so tht there is greter or less overlp. 5. INTERPRET We re sked to find the totl electric-field energy in cubicl region with vrible field. DEVELOP The electric-field energy for vrible field is given by Eqution.8: U ε E dv. For the cubicl region in this cse, dv dxdydz, nd the integrtion for ech vrible is from to m. EVALUATE Performing the integrtion, the energy stored in the field is m m m x U ε dz dy E dx x ( ) ( m 8.85 )( m) ( 4 kv/m) 4 μj Nm ( 6 m) The vlue seems resonble. One cn check tht the units work out by using the fct tht V J N m. 54. INTERPRET This problem involves electrosttic energy contined within chrged sphere. Our system hs sphericl symmetry. DEVELOP From Exmple., we see tht the sphericlly symmetric field inside such sphere is so the energy density is kqr kq r Er 6 u( r) ε (4 πk) R 8πR The energy within the sphere cn be found by integrting this expression over the volume. Er kqr R opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
-6 hpter EVALUATE With thin sphericl shells of rdius r for volume elements, dv 4 π r dr, the integrl for the energy is U udv kq r kq r dr kq r dr R R 4 4 6 π 6 sphere 8π R R R (This is just the energy stored inside the sphere. For the energy outside the sphere, nd the totl energy, see Problems 58 nd 59.) The result shows tht U is inversely proportionl to R. This mens tht the stored energy decreses if the sme mount of chrge Q is distributed over greter volume. Our result cn be compred to the sitution (Problem 55) where the totl chrge Q is distributed over its surfce. In tht cse, the totl energy stored in its electric field is U kq R. 55. INTERPRET This problem involves sphericl chrge distribution outside of which we re to find the totl energy contined in the electric field. DEVELOP This problem is the sme Exmple.5 if we let R R nd R (i.e., we cn consider tht we re compressing n infinitely seprted chrge distribution to one tht is on the surfce of the sphere of rdius R ). EVALUATE The energy is thus kq U R The energy is qudrtic in chrge so, for exmple, it would tke 4 times the energy to ssemble twice the chrge. 56. INTERPRET This problem is bout the electrosttic energy of chrged sphere, both within nd outside the sphere. Our system hs sphericl symmetry. DEVELOP The field outside sphericlly symmetric distribution of rdius R is the sme for the chrge Q uniformly spred over the volume or the surfce (thnks to Guss s lw). Thus, Problem 55 gives the energy in the electric field outside uniformly chrged sphericl volume. The result of Problem 54 gives the energy inside the sphere. The totl energy is the sum of the two: EVALUATE kq kq kq U + R R 5R Applied to U 5 nucleus, the expression gives 9 9 ( )( ) kq 9. N m / 9.6 U 5R 5(. fm) This oulomb energy is bout % of the mss energy mc of the U 5 nucleus..6 J. GeV 57. INTERPRET This problem involves finding the chnge in electrosttic potentil energy between two seprted, chrged, wter drops nd single drop with the sme chrge. DEVELOP The initil electrosttic energy of two isolted sphericl drops, with chrge Q on their surfces nd / rdii R, U i ( kq / R ) (see Problem 55 nd Exmple.5). Together, drop of chrge Q, rdius R, nd / / energy Uf k ( Q) ( R ) kq R is creted. The difference in potentil energy is W Uf Ui EVALUATE Inserting the given quntities gives ( )( 9.587 9. N m / )(.5 8 ) / ( ) kq Δ U R. m 4 6. J In ev, this is 4 6. J 9.6 J/eV Δ U 5.8 ev 58. INTERPRET This problem involves finding the electrosttic energy contined within volume round chrged wire. This system hs line symmetry. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
Electrosttic Energy nd pcitors -7 DEVELOP If the wire length is much, much greter thn its rdius, nd if the volume we re considering is mny rdii wy from either end, then the system hs line symmetry (i.e., we cn consider it to be n infinite wire). In this cse, the electric field outside the wire is rdilly wy from the xis (for λ > ) with mgnitude Er kλ r (see Eqution.6). The energy density in cylindricl shell of rdius r, length L, nd volume dv π rldr, is kλ kλ E u( r) ε (4 πk) r πr Integrte this expression over the volume of the cylinder to obtin the energy. EVALUATE Thus, the energy in the spce mentioned in this problem is R kλ U u( r) dv ln πrldr kλ L ( ) cylinder R π r 9. N m / 8 /m. m ln 7.8 J 9 ( )( ) ( ) ( ) The energy density (energy/volume) decreses s / r. This mens tht more energy is concentrted in the volume closer to the wire. 59. INTERPRET For this problem, we re sked to find the time required for lightening strikes to deplete reservoir of energy, given the chrge trnsferred, the electric potentil energy difference (per chrge), nd the frequency of the lightening strikes. DEVELOP The energy in the thunderstorm of Exmple.4 is bout.4 J, while the energy in lightning 8 8 flsh is ( )( MV) 9. J. EVALUATE At rte of one flsh every 5 s, there is enough energy to lst 56 ( 5 s) min. qv Thus, there is energy for bout ( ) ( ) This seems like resonble time frme for summer thunderstorm..4 9. 56 flshes. 6. INTERPRET We re to find the cpcitnce of the given pir of coxil cylinders. DEVELOP From Exmple.4, we cn see tht the mgnitude of the potentil difference between two points distnces nd b from cylinder with chrge per unit length λ is λ λ b Δ V ln ln πε b π ε for b >. The totl chrge on the inner cylinder is Q λl, nd the definition of cpcitnce is Q/V. EVALUATE ombining these expressions to find the cpcitnce gives Q λl πε L V λ b b ln ( ) ln πε ( ) The cpcitnce depends only on the geometry, s expected. Note tht this result ssumes tht, b L, which is used in the derivtion of the expression for the voltge difference. 6. INTERPRET We re to find the cpcitnce of sphere surrounded by concentric shell. DEVELOP This geometry is the sme s for Problem.8, for which we found the potentil between the spheres to be ( ) V kq b from which we cn find the cpcitnce using Eqution., Q/V. EVALUATE The cpcitnce is Qb ( ) Q V kq( b ) 4π εb 4π εb ( b ) The cpcitnce depends only on the geometry of the cpcitor, s expected. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
-8 hpter 6. INTERPRET The object of interest is sphericl cpcitor, nd we re to explore the limit when the seprtion between the concentric spheres is much, much smller thn the rdii of the spheres (i.e., b << ). DEVELOP The cpcitnce for sphericl cpcitor cn be derived by noting tht the potentil difference between two concentric conducting spheres is ( ) kq b Q b V kq b b V k( b ) Tke the limit d b << to show tht reduces to tht of prllel-plte cpcitor. EVALUATE With d b <<, we find b 4 b 4 ( d) 4 A π ε π ε + π ε ε k ( b ) d d d d which is the result of Eqution., with A 4π being the re of the sphericl pltes. The limit d b << mens tht the rdius of the sphere is much greter thn the seprtion between the two spheres. Thus, the interfce between the spheres is very similr to two prllel pltes. 6. INTERPRET This problem involves sphericl, uniform chrge density. We re to find the frction of the totl energy contined within the sphere. DEVELOP In Problem.6, we found tht the energy within just such chrged sphere is kq Uinside R In Problem.6, we found tht the energy outside such sphere is kq Uoutside R Divide U inside by the sum to find the frction of energy inside the sphere. EVALUATE The frction of energy inside the sphere is U kq inside ( R) finside U + U kq R + kq R + 5 6 inside outside ( ) ( ) Ignoring grvity, this result is independent of the size of the sphere. 64. INTERPRET This problem involves chrged cpcitor into which we insert dielectric mteril so tht it occupies hlf the volume of the cpcitor. We re to find the new cpcitnce, the stored energy, nd the force on the dielectric in this configurtion. DEVELOP Use the coordinte system defined in the figure below. In so fr s fringing fields cn be neglected, the electric field between the pltes is uniform E V/d (but when the dielectric is inserted, V V nd E depends on x). In fct, on the left side, where the slb hs penetrted, E ( κ)( σl ε ) nd on the right, E σ R ε, where σ L nd σ R re the chrge densities on the left nd right sides, respectively. Thus, σ κε E nd L σ R ε E, nd the chrge cn be written (in terms of geometricl vlues tken from Fig..9) s ( ) ε ( ) ε ( ) ( ) Q σ wx+ σ w L x Ew κx+ L x V d w κx+ L x L R. When the bttery is disconnected, the cpcitor is isolted nd the chrge on it is constnt, Q Q, nd we cn use Eqution. U V / to find the energy stored in the cpcitor. The force on prt of n isolted system is relted to the potentil energy of the system by Eqution 8.9. The force on the slb is therefore F x UL ( κ ) ( κ ) du d U L dx dx κ x + L x x + L x opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
in the direction of incresing x (so s to pull the slb into the cpcitor). Electrosttic Energy nd pcitors -9 EVALUATE () From Eqution., Q V where ε A d nd A Lw. Inserting x L/, we find (b) The stored energy is ( κ x + L x) L κ + Q ql UL U x+ L x x+ L x ( κ ) κ where U Q ( ) V. For x L/, the energy is ( ) (c) For x L/, the mgnitude of the force is V κ +. ( κ ) ( κ + ) V Fx L Notice tht the results depend only on geometricl fctors nd the dielectric strength of the mteril. It turns out tht if we rewrite the force, for ny vlue of x, in terms of the voltge for tht x, using q V V V ( κ x+ L x)/ L, the expression cn be used in the succeeding problem. Thus, κ L( κ ) V L( ) V V ( κ ) Fx ( κx+ L x) L L 65. INTERPRET This problem is bout the effect of inserting dielectric mteril into prllel-plte cpcitor, which is connected to bttery nd so remins t constnt voltge. DEVELOP We first note tht the cpcitnce depends on the configurtion nd electricl properties of the pltes nd insulting mterils, not on the externl connections. Thus, we cn use the result from the previous problem: κ x + L x L If the cpcitor remins connected to bttery, the voltge is constnt, V V. EVALUATE (b) The energy is For x L/, we get U V ( κ ) V κ x+ L x U V L + 4. (c) When the cpcitor is connected to bttery, Eqution 6.8 ( F du dx) for the force does not pply becuse the bttery does work, which chnges the energy of the system. However, for prticulr vlues of chrge nd voltge on the cpcitor, the force on the slb considered here is the sme, regrdless of the externl connections. In the preceding problem we found tht Fx V ( κ ) L where V is the prticulr voltge (nd, becuse of the specil form (x) of the cpcitnce, the prticulr chrge q does not pper). Since V V in this problem, V ( κ ) Fx L Note tht the results re different from the preceding problem, becuse the bttery does work. The force on the slb is to the right, drwing the dielectric between the pltes. x opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
- hpter 66. INTERPRET We re to find the cpcitnce per unit length of two prllel wires whose seprtion is much, much greter thn their rdii, so we cn ssume line symmetry (i.e., infinite wires). The wires crry opposite chrge density. DEVELOP Use the coordinte system shown in the figure below. Apply Eqution.: B Δ VAB E dr The totl electric field is the superposition of the electric fields due to the two wires, ech of which is given by Eqution.6. Summing the two contributions gives λ ( ˆ λ ) ( ˆ λ E i + i) + ( i ˆ ) πεr πεr+ πε x b x Insert this into the integrl nd perform the integrtion to find the voltge difference between the wires. The wire cpcitnce per unit length cn then be found using Eqution., Q V, which we cn trnsform into per unit length by dividing ech side by n rbitrry length L: A where L is the cpcitnce per unit length. Q V λ V L L L EVALUATE Evluting the integrl gives B b λ λ b λ b Δ VAB E dr + dx ln( x) ln( b x) ln A πε x b x π π ε ε Inserting this into the expression for cpcitnce gives /ln b L πε The cpcitnce per unit length depends only on geometricl prmeters, nd is positive. 67. INTERPRET We re to find the energy per unit length stored in the electric field within uniformly chrged rod. We will use Guss s lw to determine the electric field within the rod, nd then integrte the energy density to find the totl energy. DEVELOP We use Guss s lw (Eqution.) qenclosed E d ε exploiting the cylindricl symmetry, to find the electric field. Once we hve this field, we integrte the energy density u ε E over the cylinder to find the totl energy U udv. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
Electrosttic Energy nd pcitors - EVALUATE qenclosed π r Lρ rρ E d E( π rl) E ε ε ε L π R R rρ ρ U ε rdrd dz L r dr φ π ε 4ε 4 U πρ 4 πρ R R L ε 4 8ε Incresing the rdius increses the energy per length drmticlly. 68. INTERPRET We re to find the electrosttic energy stored between two prllel pltes of prllel-plte cpcitor, nd then differentite to find the force between the pltes. DEVELOP Using Eqution.8, find the electric field between the pltes: σ Q E ε Aε This is constnt, so totl energy stored in this field is then U uv, where u is the energy density (energy per unit volume). We cn find the force by using Fx du dx. EVALUATE () The electrosttic potentil energy is Q U uv εe ( Ax) x Aε (b) The force between the pltes is du Q Fx dx Aε This is hlf the vlue you would obtin by multiplying the chrge on one plte by the field between the pltes. The nswer we get for (b) is hlf the field times the chrge on one plte: but we must remember tht the field between the pltes is creted by both chrged pltes. A chrge is not ffected by the field it cretes. Only the field creted by the other plte cuses force on ech plte, nd the other plte cretes hlf the field. 69. INTERPRET We re to use the energy stored in cpcitor network t given voltge to find the cpcitnce of n unknown cpcitor in the network. We shll use the eqution for the energy stored in cpcitor, nd the rules for dding cpcitors in series nd in prllel. DEVELOP We first find the cpcitnce network of the entire network in terms of the given vlues nd the unknown cpcitnce. Next, we use Eqution., U networkv with U 5.8 mj nd V V nd work bckwrd through the network to solve for. EVALUATE () See the circuit digrm in the circuit below. (b) U U networkv network.6 μf V + left.76 μf network left. μf left top +. μf top.76 μf + 4. μf. μf top opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.
- hpter This trick of unrveling the network is often useful. We will use the sme trick in deling with resistor networks lter. Note lso tht three significnt figures were retined for the intermedite results, wheres only two significnt figures were retined for the finl result, s wrrnted by the dt. 7. INTERPRET We re considering the energy used by the Ntionl Ignition Fcility. DEVELOP The energy stored in cpcitor is given by U V (Eqution.). EVALUATE We re told tht the NIF cpcitor system stores 4 MJ t kv, so the cpcitnce is U ( ) 4 MJ F V ( kv) The nswer is (d). This is n impressively lrge cpcitnce. The dvntge of using cpcitors in this ppliction is tht they cn dischrge rpidly nd thus supply lrge mount of power over short time. 7. INTERPRET We re considering the energy used by the Ntionl Ignition Fcility. DEVELOP If the voltge is chnged, the cpcitor system still needs to store the sme energy, so the cpcitnce will chnge by fctor of / ( ) V / V. EVALUATE Doubling the voltge mens the cpcitnce could drop by ¼ nd still store the sme 4 MJ. The nswer is (). There s trdeoff between cpcitnce nd voltge. One cn think of the cpcitor system s the equivlent of spring storing mechnicl energy ( U kx ) The cpcitnce is like the spring constnt, k, nd. the voltge is like the displcement, x. The sme energy cn be stored with weker spring constnt if the spring is compressed or stretched further. 7. INTERPRET We re considering the energy used by the Ntionl Ignition Fcility. DEVELOP The power is energy divided by time. EVALUATE We re told tht the lsers deliver MJ of energy in ns. So the power is 6 Δ E J 5 P W PW 9 Δ t s The nswer is (d). This is over times the world s verge power consumption, but it only lsts for frction of second. 7. INTERPRET We re considering the energy used by the Ntionl Ignition Fcility. DEVELOP The energy stored in cpcitor is given by U V (Eqution.). EVALUATE One of the cpcitors t NIF stores U V μf kv 6 kj ( )( ) The nswer is (c). If there re of these cpcitors in prllel, they cn store 7 MJ. To rech 4 MJ with these type of cpcitors would require more thn 5 times this number. opyright 6 Person Eduction, Inc. All rights reserved. This mteril is protected under ll copyright lws s they currently exist. No portion of this mteril my be reproduced, in ny form or by ny mens, without permission in writing from the publisher.