7.4 Potential Difference and Electric Potential

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1 7.4 Potntial Diffrnc and Elctric Potntial In th prvious sction, you larnd how two paralll chargd surfacs produc a uniform lctric fild. From th dfinition of an lctric fild as a forc acting on a charg, it follows that, for a givn uniform lctric fild, charg, and particl mass, th particl undrgos a uniform acclration. Anothr way of thinking about th physics of this situation is that th lctric fild dos work on th chargd particl. This viw works wll if th charg stays constant, but in rality th work don by th fild varis with charg. What if, instad of dscribing th lctric fild in trms of forc pr charg, you xprssd it in trms of nrgy pr charg? As it turns out, w can dscrib th fild in this way, as you will larn in this sction. No mattr how you choos to dscrib th uniform lctric fild, its ability to acclrat chargd particls with known conditions has provn usful to physicists and nginrs. Dvics such as particl acclrators, which can acclrat particls to spds nar th spd of light, can only work if th particls ar moving in th first plac. Elctric filds caus th initial motion of ths particls by acclrating thm. Particl acclration is important in som vryday dvics as wll. Inkjt printrs acclrat chargd ink particls toward spcific parts of th papr (Figur 1). Old tlvision sts and computr monitors hav cathod-ray tubs. Ths tubs acclrat lctrons toward a phosphor scrn. Variations in th dflction of th acclratd lctrons dtrmin th brightnss and colour of th scrn. papr dflction plats lctrod nozzl of print had guttr charging control pump instructions from computr ink supply Figur 1 Ink droplts from th print had ar ithr chargd or unchargd. Th unchargd droplts mov to th papr undfl ctd, forming th lttrs. Chargd droplts ar dfl ctd into th guttr, laving thos parts of th papr blank. work and Elctric Potntial Diffrnc A d F E Figur 2 Th lctric fi ld givs ris to an lctric forc, which movs in th sam dirction as th charg if th charg is positiv. B Rcall that in a rgion of spac whr th lctric fild > is constant, > has th sam magnitud and dirction at all points. A point charg in this rgion xprincs an lctric forc F > E 5 > Th forc is paralll to >, as shown in Figur 2. Suppos this charg movs a crtain distanc Dd, starting at point A and nding at point B. For simplicity, assum this displacmnt is paralll to th lctric forc F > E. According to th dfinition of work (W), th work don by th lctric forc on th charg is W 5 F E Dd 346 Chaptr 7 Elctric Filds NEL

2 Th lctric forc dos work on th charg and is indpndnt of th path it taks from A to B. W can now dfin th lctric potntial nrgy, E E, which is th nrgy stord in th systm that can do work W on a positivly chargd particl. From your studis of work and nrgy you know that th chang in th potntial nrgy associatd with this typ of forc is ual to 2W, whr W is th work don by that forc. A mor dtaild xplanation of th maning of th ngativ sign will follow. So, if th lctric forc dos an amount of work W on a chargd particl, th chang in th lctric potntial nrgy is DE E 5 2W DE E 5 2F E Dd Combining this uation with th uation rlating forc to th lctric fild, th chang in lctric potntial nrgy whn th chargd particl movs from A to B in Figur 2 is DE E 5 2W 5 2F E Dd lctric potntial nrgy (E E ) th nrgy stord in a systm of two chargs a distanc Dd apart, or th nrgy stord in an lctric fi ld that can do work on a positivly chargd particl DE E 5 2Dd This uation givs th chang in th potntial nrgy as th charg movs through a displacmnt Dd, in a rgion whr th lctric fild is paralll to th displacmnt. Not that th chang in th potntial nrgy dpnds on th starting and nding locations but not on th path takn. In Figur 2, th displacmnt is along a lin, but th charg may mov from A to B along many othr paths without affcting DE E. Elctric potntial nrgy is stord through th potntial ffct of th lctric fild on an lctric charg. This ffct is illustratd in Figur 3(a), whr th charg is movd from point B to point A by an xtrnal forc F > a dirctd to th lft. Forc F > a rsults from som xtrnal agnt, which could b your hand. Th lctric forc on points to th right, assuming that is positiv, so th displacmnt is opposit to th dirction of th lctric forc. Th work don by th lctric fild on th particl is thus ngativ. Thrfor, according to th lctric potntial nrgy uation, th chang in th lctric potntial nrgy must b positiv. In othr words, a positiv amount of nrgy has now bn stord in th systm composd of th charg and th lctric fild. That nrgy cam from th positiv amount of work don by th xtrnal forc F > a that movd th charg from B to A. As th charg movs from A back to B, th procss is rvrsd (Figur 3(b)). Now th lctric fild dos a positiv amount of work on th particl bcaus th lctric forc and th particl s displacmnt ar paralll and th chang in th lctric potntial nrgy is ngativ. Enrgy stord in th lctric fild and particl systm is now takn out of th systm. This nrgy can appar as an incras in th kintic nrgy of th particl whn it rachs B. Potntial nrgy bcoms kintic nrgy. A d B A d B F a F E E E 0 E E 0 (a) (b) Figur 3 (a) To mov from B to A, th xtrnal forc acting on a chargd particl works against th lctric fi ld and producs a positiv chang in DE E of th fi ld and particl systm. (b) Th chargd particl taks nrgy stord in th lctric fi ld and convrts it to kintic nrgy. This producs a ngativ chang in DE E of th fi ld and particl systm. NEL 7.4 Potntial Diffrnc and Elctric Potntial 347

3 In th following Tutorial, you will larn mor about how to solv problms that involv lctric potntial nrgy. Tutorial 1 Solving Problms Involving Elctric Potntial Enrgy This Tutorial xplains how to dtrmin th chang in lctric potntial nrgy for a charg in a uniform lctric fild, givn th position and magnitud of that fild. Sampl Problm 1: Potntial Enrgy Diffrnc in an Elctric Fild A chargd particl movs from rst in a uniform lctric fild. (a) For a proton, calculat th chang in lctric potntial nrgy whn th magnitud of th lctric fild is 250 N/C, th starting position is 2.4 m from th origin, and th final position is 3.9 m from th origin. (b) Calculat th chang in lctric potntial nrgy for an lctron in th sam fild and with th sam displacmnt. (c) Calculat th chang in lctric potntial nrgy for an lctron acclratd in an lctric fild with th sam magnitud but opposit dirction as in (a) and (b), and with a starting position of 2.4 m from th origin and a final position of 5.0 m from th origin. Solution (a) Givn: d i m; d f m; C; N/C Ruird: DE E Analysis: Us th uation for lctric potntial nrgy in trms of,, and Dd, whr Dd 5 d f 2 d i : DE E 5 2Dd 5 21d f 2 d i 2. Not that DE E is ngativ for a positiv charg travlling in th sam dirction as. Thus, th sign of taks car of whthr thr is a gain (for ngativ ) or loss (for positiv ) of lctric potntial nrgy. Not that w includ th sign of th charg bcaus w ar daling with nrgy, which is not a vctor uantity and dos not hav a dirction. Solution: DE E 5 2Dd 5 21d f 2 d i C2 a250 N C b 13.9 m m N# m DE E J Statmnt: Th chang in lctric potntial nrgy du to th movmnt of th proton in th uniform lctric fild is J. Th ngativ sign indicats that th lctric fild loss potntial nrgy by doing work on th proton. (b) Givn: d i m; d f m; C; N/C Ruird: DE E Analysis: Us th sam uation for lctric potntial nrgy: DE E 5 2Dd 5 21d f 2 d i 2. Not that w includ th sign of th charg to dtrmin whthr thr is a gain or loss of lctric potntial nrgy. Solution: DE E 5 2Dd 5 21d f 2 d i C2 a250 N C b 13.9 m m N# m DE E J Statmnt: Th chang in lctric potntial nrgy du to th movmnt of th lctron in th uniform lctric fild is J. Th positiv valu indicats that th lctric fild gains potntial nrgy as th lctron is acclratd in th fild. (c) Givn: d i m; d f m; C; N/C Ruird: DE E Analysis: Not that th sign for th lctric fild is ngativ bcaus th lctron is acclrating in a dirction opposit to. Us th sam uation for lctric potntial nrgy: DE E 5 2Dd 5 21d f 2 d i 2. Solution: DE E 5 2Dd 5 21d f 2 d i C2 a2250 N C b 15.0 m m N # m DE E J Statmnt: Whn th dirction of th lctric fild is rvrsd, th chang in th lctric potntial nrgy is J. Th ngativ valu indicats that th lctric fild loss potntial nrgy by doing work on th lctron. 348 Chaptr 7 Elctric Filds NEL

4 Sampl Problm 2: Dynamics of Chargd Particls (a) Using th law of consrvation of nrgy, calculat th spd of th proton in part (a) of Sampl Problm 1 for th givn displacmnt. Assum that th proton starts from rst. (b) Dtrmin th initial spd of th lctron in part (b) of Sampl Problm 1, assuming its spd has dcrasd to half of its initial spd aftr th sam displacmnt, Dd. Solution (a) Givn: d i m; d f m; C; m kg; N/C Ruird: v Analysis: Th law of consrvation of nrgy stats that th total chang in potntial nrgy of th fild and particl systm and th chang in kintic nrgy of th particl uals zro: DE E 1 DE k 5 0. Th kintic nrgy of th particl is rlatd to its spd v by th uation DE k mv 2. Us this uation and th uation DE E 5 2Dd for th chang in potntial nrgy of th fild and proton systm. Not that Dd m m m and 1 N 5 1 kg# m/s 2. Solution: By th law of consrvation of nrgy, DE E 1 DE k 5 0 2Dd mv mv 2 5 Dd v 5 Å 2Dd m kg # m s C m2 C v 5 ï kg v m/s Statmnt: Th spd of th proton acclratd for a distanc of 1.5 m by an lctric fild of 250 N/C is m/s. (b) Givn: Dd m; C; m kg; N/C; v f 5 0.5v i Ruird: v i Analysis: For situations that do not involv an objct at rst, th chang in kintic nrgy is givn by th uation DE k mv f mv i 2. Us this uation and th uation DE E 5 2Dd for th chang in potntial nrgy of th fild and lctron systm. Solution: By th law of consrvation of nrgy, DE E 1 DE k 5 0 2Dd 1 a 1 2 mv f mv i 2 b m a1 2 v ib mv i 2 5 Dd mv i 2 5 Dd v i 5 Å 2 8Dd 3m C2 a250 N C b 11.5 m2 v i 5 2 ã kg2 kg # m s C m2 C 5 ï v i m/s kg2 Statmnt: Th initial spd of th lctron bfor ntring th lctric fild is m/s. Practic 1. An lctron ntrs a uniform lctric fild of 145 N/C pointd toward th right. Th point of ntry is 1.5 m to th right of a givn mark, and th point whr th lctron lavs th fild is 4.6 m to th right of that mark. T/I A (a) Dtrmin th chang in th lctric potntial nrgy of th lctron. [ans: J] (b) Th initial spd of th lctron was m/s whn it ntrd th lctric fild. Dtrmin its final spd. [ans: m/s] 2. Calculat th work don in moving a proton 0.75 m in th sam dirction as th lctric fild with a strngth of 23 N/C. T/I A [ans: J] 3. An lctron xprincs a chang in kintic nrgy of J. Calculat th magnitud and dirction of th lctric fild whn th lctron travls 0.18 m toward th right. T/I A [ans: N/C [toward th lft]] NEL 7.4 Potntial Diffrnc and Elctric Potntial 349

5 Elctric Potntial lctric potntial (V ) th valu, in volts, of potntial nrgy pr unit positiv charg for a givn point in an lctric fild; 1 V 5 1 J/C lctric potntial diffrnc ( V ) th amount of work ruird pr unit charg to mov a positiv charg from on point to anothr in th prsnc of an lctric fild Elctric potntial nrgy is a proprty of a systm of chargs or of a point charg in an lctric fild, whr th fild is cratd by othr chargs. In ithr cas, this lctric potntial nrgy is not th proprty of a singl charg alon. For this rason, th potntial nrgy dpnds on th valus of th chargs and th lctric fild involvd in th intraction. This lads to a nw uantity calld lctric potntial, V, which is a masur of how much lctric potntial nrgy is associatd with a spcific uantity of charg at a particular location in an lctric fild. Basd on this dfinition, V 5 E E Elctric potntial, or just potntial, is a convnint masur bcaus it is indpndnt of th amount of charg at a particular location in th fild. It dpnds only on th lctric fild strngth at that location. For xampl, if you had 1 C of lctrons at a particular location in a uniform lctric fild, you would possss a crtain amount of lctric potntial nrgy. If you doubld th amount of lctrons to 2 C at th sam location in th lctric fild, you would hav doubl th lctric potntial nrgy. In both ths situations, you would hav th sam lctric potntial. Th SI unit of lctric potntial is th volt (V), namd in honour of physicist Alssandro Volta ( ). Th volt rlats to othr SI units in th following uation: 1 V 5 1 J/C 5 1 N# m/c Th volt is usd in th masurmnt of many lctrical uantitis. Not in particular that th units of volts pr mtr ar uivalnt to units of nwtons pr coulomb, th units for lctric fild strngth. Ths uivalnt units giv you svral diffrnt ways to xprss th units of th lctric fild. Th ons most commonly usd ar 1 V/m 5 1 N/C. Anothr convnint dfinition rlating to lctric potntial nrgy is lctric potntial diffrnc, DV. W rturn to th concpt of th chang in potntial diffrnc and th displacmnt of th particl: You can dfin th chang in th potntial, or potntial diffrnc, for a charg that movs btwn two points: DV 5 DE E For th cas of a uniform lctric fild, th uation for lctric potntial diffrnc bcoms DV 5 DE E 5 2W 5 2Dd DV 5 2Dd This rlationship shows how a non-uniform lctric fild varis with th chang in lctric potntial (that is, lctric potntial diffrnc) and th chang in position in th fild: 5 2 DV Dd 350 Chaptr 7 Elctric Filds NEL

6 This uation stats that th magnitud of th lctric fild is largst in rgions whr V is larg and changs rapidly with small changs in displacmnt. Convrsly, th lctric fild is zro in rgions whr V is constant. Notic that bcaus of th ngativ sign in th uation, th lctric fild points from rgions of high potntial to rgions of low potntial (Figur 4). If w considr a circuit in which a battry is th sourc of lctrical nrgy, a positiv tst charg will naturally mov from th positiv trminal whr a high potntial xists to th ngativ trminal whr a low potntial xists. Sinc th lctric fild points from positiv to ngativ, th positiv tst charg will also mov in th sam dirction as th fild. Convrsly, lctrons will naturally travl from a rgion of low potntial to a rgion of high potntial, in a dirction opposit to th dirction of th lctric fild. UNIT TASK BOOKMARK You can apply what you hav larnd about lctric potntial to th Unit Task on pag 422. d potntial diffrnc V V d Figur 4 Th magnitud and dirction of th lctric fi ld ar rlatd to how th lctric potntial V changs with position. Th rlation btwn th lctric fild and th changs in th potntial and position involvs th componnt of th lctric fild that is in th dirction paralll to th displacmnt Dd. Th lctric fild is a vctor, so if you want to dtrmin > in a particular dirction, you must considr how th potntial V changs for a tst charg moving along that dirction. With this uation and knowldg of >, you can, in principl at last, calculat how th lctric potntial changs as you mov from plac to plac within th find. Strictly spaking, this rlation only holds for small stps Dd in a constant fild. You can, howvr, combin th potntial changs DV from many such small stps to dtrmin th chang in th potntial diffrnc ovr a larg distanc. Th following Tutorial will dmonstrat how to solv problms involving lctric potntial. Tutorial 2 Solving Problms Rlatd to Elctric Potntial This Tutorial shows how to us lctric potntial to solv problms rlatd to a charg in a uniform lctric fi ld. Sampl Problm 1: TV Tubs and Particl Acclrators Th cathod-ray tubs in old tlvision sts and computr monitors work in a way that is similar to crtain parts in particl acclrators. Both dvics acclrat particls in a similar way, using th uniform lctric fi ld btwn conducting plats. Lookd at anothr way, th particls acclrat as thy mov through an lctric potntial diffrnc. (a) An lctron lavs th ngativ plat of a cathod-ray tub and travls toward th positiv plat. Th lctric potntial diffrnc btwn th plats is V. Using th law of consrvation of nrgy and th dfi nition of lctric potntial diffrnc, calculat th spd of an lctron as it rachs th positiv plat in a cathod-ray tub. Assum that th lctron is initially at rst. Th mass of an lctron is kg. (b) Calculat th magnitud of th lctric fi ld at a distanc of 15 cm, which is at th nd of th cathod-ray tub. NEL 7.4 Potntial Diffrnc and Elctric Potntial 351

7 (a) Givn: C; m kg; DV V; v i 5 0 m/s Ruird: v f Analysis: Us th uation for th law of consrvation of nrgy, DE E 1 DE k 5 0, and th uation for th lctric potntial diffrnc, DE E 5 DV. Th kintic nrgy of th particl initially at rst is rlatd to its final spd v f by th uation DE k mv f mv i 2. Sinc v i = 0, DE k mv f 2. Not that 1 V 5 1 N m/c 5 1 kg m/s 2 m/c. Solution: By th law of consrvation of nrgy, DE E 1 DE k 5 0 DV mv f mv f 2 5 2DV v f DV m v f 5 Å 2 2DV m Calculat th final spd v f of th lctron as it rachs th positiv plat. v f 5 Å C V kg 5 ã C2 a kg # m s 2 # m C b kg v f m/s Statmnt: Th spd of th lctron as it rachs th positiv plat is m/s. (b) Givn: Dd 5 15 cm m; DV V Ruird: Analysis: Us th uation for th lctric fild in trms of potntial diffrnc and displacmnt: 5 2 DV Dd Solution: 5 2 DV Dd V 0.15 m V/m Statmnt: Th lctric fild has a magnitud of V/m. Th ngativ sign indicats that th fild xtnds from high to low potntial. Th lctric fild vctor always points from rgions of high V to rgions of low V bcaus > is paralll to th dirction that a positiv tst charg would mov if it wr placd at that location. Sampl Problm 2: Calculating th Spd of a Dflctd Chargd Particl An lctron movs horizontally with a spd of m/s btwn two horizontal paralll plats. Th plats hav a lngth of 12.5 cm, and a plat sparation that allows a chargd particl to scap vn aftr bing dflctd (Figur 5). Th magnitud of th lctric fild within th plats is 150 N/C. Calculat th final vlocity of an lctron as it lavs th plats. y x Figur 5 v i Givn: v i m/s; L cm m; C; N/C Ruird: v > f Analysis: Th lctric fild is uniform and dirctd downward btwn th plats, so th lctric forc acting on a proton is also constant and dirctd downward. A constant downward forc mans a constant downward acclration on a positiv charg. Howvr, bcaus th charg is ngativ, th acclration must b upward. To calculat th magnitud v f of th final vlocity, solv for ach componnt of th vlocity: th constant horizontal componnt and th acclratd upward componnt. Thn us th uation v f 5 "v 2 x f 1 v 2 y f for th magnitud of th final vlocity. Th componnts of th vlocity ar givn by v xf 5 v i 5 L Dt (Dt can b calculatd from v i and L) and v yf 5 v yi 1 a y Dt 5 a y Dt 1sinc v yi By combining th uations for Nwton s scond law, F nt 5 ma, and th lctric forc on a charg in an lctric fild, F E 5, w can dtrmin th vrtical acclration a y 5 F nt m 5 F E m 5 m. Solution: Calculat a y. a y 5 m kg # m s C C kg a y m/s 2 1two xtra digits carrid2 352 Chaptr 7 Elctric Filds NEL

8 Calculat Dt. v i 5 L Dt Dt 5 L v i m m s Dt s 1two xtra digits carrid2 Calculat v yf. v yf v yf 5 a y Dt 5 a m s 2b s m/s 1two xtra digits carrid2 Now calculat th magnitud of th nt vlocity. v f 5 "v 2 x f 1 v 2 y f 5 " m/s m/s2 2 v f m/s Calculat th angl u to dtrmin th dirction of th lctron. u 5 tan 21 a v y f v xf b m 6 s 5 tan 21 ± m 6 s u Statmnt: Th final vlocity of th lctron is m/s [E 528 N]. Practic 1. An old tlvision cathod-ray tub crats a potntial diffrnc of V across th paralll acclrating plats. Ths plats acclrat a bam of lctrons toward th targt phosphor scrn. Th sparation btwn th plats is 12 cm. K/U T/I (a) Using th principl of nrgy consrvation and th dfinition of lctric potntial diffrnc, calculat th spd at which th lctrons strik th scrn. [ans: m/s] (b) Calculat th magnitud of th lctric fild. [ans: N/C] 2. Four paralll plats ar connctd in a vacuum as shown in Figur 6. An lctron at rst in th hol of plat X is acclratd to th right. Th lctron passs through hols at W and Y with no acclration at all. It thn passs through th hol at Y and slows down as it hads to plat Z. T/I 6.0 cm 6.0 cm 6.0 cm X W Y Z V V Figur 6 (a) Calculat th spd of th lctron at hol W. [ans: m/s] (b) Calculat th distanc, in cntimtrs, from plat Z to th point at which th lctron changs dirction. [ans: 5.7 cm [to th lft of Z]] 3. An lctron ntrs a paralll plat apparatus that is 8.0 cm long and 4.0 cm wid, as shown in Figur 7. Th lctron has a horizontal spd of m/s. Th potntial diffrnc btwn th plats is V. Calculat th lctron s vlocity as it lavs th plats. K/U A [ans: m/s [E 3.38 N]] 8.0 cm m/s 4.0 cm Figur 7 NEL 7.4 Potntial Diffrnc and Elctric Potntial 353

9 7.4 Rviw Summary Th chang in lctric potntial nrgy dpnds on th lctric fild, th charg bing movd, and th charg s displacmnt: DE E 5 2Dd. For DE E 7 0, work is don against th lctric fild (2W), rsulting in nrgy stord in th fild. For DE E 6 0, work is don by th lctric fild (1W) on a particl moving in th fild, which typically incrass th kintic nrgy of th particl. Th lctric potntial is th lctric potntial nrgy pr unit charg at a givn point in an lctric fild: V 5 E E. Th magnitud of an lctric fild varis with th lctric potntial diffrnc and th chang in position in th fild: 5 2 DV Dd. Qustions 1. An lctron movs from an initial location btwn paralll plats whr th lctric potntial is V i 5 30 V to a final location whr V f V. K/U T/I (a) Dtrmin th chang in th lctron s potntial nrgy. (b) Dtrmin th avrag lctric fild along a 10 cm long lin sgmnt that conncts th initial and final locations of th lctron. B sur to giv both th magnitud and th dirction of >. 2. Th lctric potntial diffrnc btwn two paralll mtal plats is DV. Th plats ar sparatd by a distanc of 3.0 mm and th lctric fild btwn th plats is V/m. Calculat DV. K/U T/I 3. A proton of mass kg movs from a location whr V i V to a spot whr V f V. K/U T/I (a) Calculat th chang in th proton s kintic nrgy. (b) Rplac th proton with an lctron, and dtrmin its chang in kintic nrgy. 4. An lctron movs from a rgion of low potntial to a rgion of highr potntial whr th potntial chang is 145 V. K/U T/I (a) Calculat th work, in jouls, ruird to push th lctron. (b) What is doing th work? 5. Th lctrons in an old TV pictur tub ar acclratd through a potntial diffrnc of V. K/U T/I A (a) Do th lctrons mov from a rgion of high potntial to a rgion of low potntial, or vic vrsa? (b) Calculat th chang in th kintic nrgy of on of th lctrons. (c) Calculat th final spd of an lctron whn th initial spd is zro. 6. A pair of paralll plats has an lctric fild of N/C. Dtrmin th chang in th lctric potntial btwn points that ar 2.55 m (initial) and 4.55 m (final) from th plats. K/U T/I A 7. An lctron with a horizontal spd of m/s and no vrtical componnt of vlocity passs through two horizontal paralll plats, as shown in Figur 8. Th magnitud of th lctric fild btwn th plats is 150 N/C. Th plats ar 6.0 cm long. K/U T/I Figur 8 y (a) Calculat th vrtical componnt of th lctron s final vlocity. (b) Calculat th final vlocity of th lctron. 8. An lctric fild of 20 N/C xists along th x-axis in spac. Calculat th potntial diffrnc DV 5 V B 2 V A, whr th points A and B ar givn by (a) A 5 0 m; B 5 4 m (b) A 5 4 m; B 5 6 m T/I A 9. Points A and B ar at th sam potntial. Dtrmin th nt work don in moving a charg from point A to point B. K/U T/I 10. Th potntial at a point is 20 V. Calculat th work don in bringing a charg of 0.5 C to this point. K/U A x 354 Chaptr 7 Elctric Filds NEL

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