Electric Potential and Gauss s Law, Configuration Energy Challenge Problem Solutions

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1 Poblem 1: Electic Potential an Gauss s Law, Configuation Enegy Challenge Poblem Solutions Consie a vey long o, aius an chage to a unifom linea chage ensity λ a) Calculate the electic fiel eveywhee outsie of this o (ie fin E ( ) ) b) Calculate the electic potential eveywhee outsie, whee the potential is efine to be zeo at a aius > (ie V( ) ) Poblem 1 Solution: (a)this is easily calculate using Gauss s Law an a cylinical Gaussian suface of aius an length l By symmety, the electic fiel is completely aial (this is a vey long o), so all of the flu goes out the sies of the cyline: E A = πle = Q enc ε = λl ε E = λ πε ö (b) To get the potential we simply integate the electic fiel fom to wheeve we want to know it (in this case ): λ λ λ V ( ) = V ( ) V ( ) = E ( ') ' = ' = ln ( ' ) = ln π ' ε πε πε

2 Poblem : Estimate the lagest voltage at which it s easonable to hol high voltage powe lines Then check out this vieo, cae of a Boule City, Nevaa powe company Ai ionizes 6 - when electic fiels ae on the oe of 1 V m 1 Poblem Solution: In oe to answe this question we have to think about what happens if we go to vey high voltages What beaks own? The poblem with high voltages is that they lea to high fiels An high fiels mean beakown You eive the voltage an fiel in poblem : E = λ πε; V = λ πε ln V = E ln ( ) ( ) ( ) ( ) ( ) ( ) ( ) The stongest fiel, an hence beakown, appeas at = ~ 1 cm, the aius of a powe line (that makes the iamete just une 1 inch it might be o 4 times that big but pobably not ten times) The voltage is efine elative to some goun, eithe anothe cable (pobably ~ 1 m away) o at the most the eal goun ( ~ 1 m away) So, V E ( ) ( )( ) ( ) ma = ma ln = 1 V m 1 cm ln 1 m 1 cm 1 V As it tuns out, a typical powe-line voltage is about 5 kv, about as lage as we estimate hee Some high voltage lines can even go up to 6 kv though (o ouble that fo AC voltages) They must use lage iamete cables By the way, you can tell that beakown is a eal concen In humi weathe (uing ainstoms fo eample) you will sometimes hea cackling coming fom the powe lines This is coona ischage, a high voltage, low cuent beakown, simila to the cackling you hea fom the Van e Gaff geneato in class The movie is of an ac ischage, a vey high cuent phenomenon that can be vey angeous

3 Poblem : Consie a unifomly chage sphee of aius an chage Q Fin the electic potential iffeence between any point lying on a sphee of aius an the point at the oigin, ie V() V() Choose the zeo efeence point fo the potential at =, ie V () = How oes you epession fo V( ) change if you chose the zeo efeence point fo the potential at =, ie V ( ) = Poblem Solution: In oe to solve this poblem we must fist calculate the electic fiel as a function of fo the egions < < an > Then we integate the electic fiel to fin the electic potential iffeence between any point lying on a sphee of aius an the point at the oigin Because we ae computing the integal along a path we must be caeful to use the coect functional fom fo the electic fiel in each egion that ou path cosses Thee ae two istinct egions of space efine by the chage sphee: egion I: <, an egion II: > So we shall apply Gauss s Law in each egion to fin the electic fiel in that egion Fo egion I: <, we choose a sphee of aius as ou Gaussian suface Then, the electic flu though this close suface is E I A = E I 4π The sphee has a unifom chage ensity ρ = Q/(4 / ) π Because the chage istibution is unifom, the chage enclose in ou Gaussian suface is given by Qenc ρ(4/) π Q = = ε ε ε Now we apply Gauss s Law: to aive at: E I A = Q enc ε

4 E Q I 4π = ε which we can solve fo the electic fiel insie the sphee Q E = E ˆ = ˆ I, 4πε < < I Fo egion II: > : we choose the same spheical Gaussian suface of aius >, an the electic flu has the same fom E II A = E II 4π

5 All the chage is now enclose, Qenc = Q, then Gauss s Law becomes E Q II 4π = ε We can solve this equation fo the electic fiel Q E = E ˆ = ˆ II, 4πε > II In this egion of space, the electic fiel falls off as 1/ as we epect since outsie the chage istibution, the sphee acts as if all the chage wee concentate at the oigin Ou complete epession fo the electic fiel as a function of is then Q E ˆ ˆ I = EI =,< < 4πε E() = Q E = E ˆ ˆ II II =, > 4πε We can now fin the electic potential iffeence between any point lying on a sphee of aius an the oigin, ie V() V() We begin by consieing values of such that < < We shall calculate the potential iffeence by calculating the line integal We use as integation vaiable = V() V() = E ;< < = I an integate fom = to = : = = Q Q Q V() V() = ˆ ˆ = ; 4πε = < < 4πε 8πε = = Fo > : we ae taking a path fom the oigin though egions I an egions II an so we nee to use both functional foms fo the electic fiel in the appopiate egions The potential iffeence between any point lying on a sphee of aius > an the oigin is given by the line integal epession

6 = = V() V() = E E ; > I = = II Using ou esults fo the electic fiel we get that = = Q Q V() V() = ˆ ˆ ˆ ˆ; > 4πε 4πε = = This becomes = = Q Q V() V() = ; 4πε > 4πε = = Integating yiels = = Q Q V() V() = + ; > 8πε 4πε = = Substituting in the enpoints yiels Q Q 1 1 V() V() = V() V() = + ; > 8πε 4πε A little algeba then yiels Q Q V() V() = ; 4πε 8πε > Thus the electic potential iffeence between any point lying on a sphee of aius an the oigin (whee V () = ) is given by When we set Q ; < < 8πε V() V() = Q Q ; > 4πε 8πε V () =, we have an epession fo the electic potential function

7 Q ; < < 8πε V() = Q Q ; > 4πε 8πε We plot V( ) vs in the figue below Note that the gaph of the electic potential function is continuous at = When we set is =, the potential iffeence between the sphee at infinity an the oigin Q V( ) V() = 8πε If we ha chosen the zeo efeence point fo the electic potential at =, with Q V ( ) = The with that choice, we have that V () = 8πε Theefoe using ou esults above the new fom fo the potential function is V() Q V() ; < < 8πε = Q Q V() + ; > 4πε 8πε This amounts to just aing the constant function V( ) giving Q 8πε to the above esults fo the potential

8 Q Q ;< < 8πε 8πε V() = Q ; > 4πε In the above epession we can easily check that V ( ) = Equivalently we shift ou pevious gaph up by Q/8πε as shown in the gaph below

Poblem 4: An infinite slab of chage caying a chage pe unit volume ρ has a chage sheet caying chage pe unit aea σ 1 to its left an a chage sheet caying chage pe unit aea σ to its ight (see top pat of

9 Poblem 4: An infinite slab of chage caying a chage pe unit volume ρ has a chage sheet caying chage pe unit aea σ 1 to its left an a chage sheet caying chage pe unit aea σ to its ight (see top pat of sketch) The lowe plot in the sketch shows the electic potential V( ) in volts ue to this slab of chage an the two chage sheets as a function of hoizontal istance fom the cente of the slab The slab is 4 metes wie in the -iection, an its bounaies ae locate at = man = + m, as inicate The slab is infinite in the y iection an in the z iection (out of the page) The chage sheets ae locate at = 6man = + 6m, as inicate (a) The potential V( ) is a linea function of in the egion 6m< < m What is the electic fiel in this egion? (b) The potential V() is a linea function of in the egion m< < 6m What is the electic fiel in this egion? (c) In the egion m < < m, the potential V() is a quaatic function of given by the equation V 5 V 5 = V What is the electic fiel in this egion? 16 m 4 ( ) () Use Gauss s Law an you answes above to fin an epession fo the chage ensity ρ of the slab Inicate the Gaussian suface you use on a figue (e) Use Gauss s Law an you answes above to fin the two suface chage ensities of the left an ight chage sheets Inicate the Gaussian suface you use on a figue Poblem 4 Solution: (a) E V ˆ Δ V ˆ 5 V V = = = 15 ˆ i Δ i 4 m m i

10 (b) E V ˆ Δ V ˆ 5 V V = = = = 15 ˆ i Δ i 4 m m i (c) In the egion insie the slab, the electic fiel is E V ˆ 5V = i = ˆ i 8 m () E A = EA = 5 V 8 S m A = q in ε ρ = 5 V ε 8 m = ρa ε (e)solution: The electic fiel vanishes in the egions > 6m potential is zeo an emains zeo so the gaient is zeo) an < 6m (the electic Using Gauss s law with the Gaussian pillboes inicate in the figue, we have E 5V σ1 = ε 4 m 5V In a simila manne, σ = ε 4 m S 5V qin σ1a A = EA= A= = 4 m ε ε A common mistake is to think that the sign must flip because the electic fiel sign flips Note that because the aea vecto of the Gaussian pillbo also flips iection this is NOT tue It is vey impotant to aw pictues an show the vecto iections If the vectos

11 ( E an A ) ae in the same iection then the ot pouct (an the enclose chage) is positive

12 Poblem 5: Thee infinite sheets of chage ae locate at =, =, an =, as shown in the sketch The sheet at = has a chage pe unit aea of σ, an the othe two sheets have chage pe unit aea of σ a) What is the electic fiel in each of the fou egions I-IV labele in the sketch? Clealy pesent you easoning, elevant figues, an any accompanying calculations Plot the component of the electic fiel, E, on the gaph on the bottom of the net page Clealy inicate on the vetical ais the values of E fo the iffeent egions b) Fin the electic potential in each of the fou egions I-IV labele above, with the choice that the potential is zeo at = + ie V ( + ) = Show you calculations Plot the electic potential as a function of on the gaph on the bottom of the net page Inicate units on the vetical ais c) How much wok must you o to bing a point-like object with chage +Q in fom infinity to the oigin =? Poblem 5 Solutions: (a) We begin by choosing a Gaussian cyline with en caps in egions I an IV as shown in the figue below The total chage enclose is zeo an hence the electic flu on the encaps must be zeo Thus the electic fiel must be zeo in egions I an IV

13 This tuns out to be coect but the conclusion epens on an aitional agument base on symmety If the electic fiel is non-zeo on the encaps it must point eithe in the +-iection in both egions I an IV o in the -iection in both egions I an IV Neithe is possible ue to the symmety of the chage istibution Fo eample, if the electic fiel pointe in the +-iection in both egions I an IV Then if we looke at the chage istibution fom the othe sie of the plane of the pape, the fiel shoul point in the -iection Howeve the chage istibution is ientical when looking fom the othe sie of the pape Theefoe the fiel must point in the +-iection accoing to ou oiginal assetion Theefoe by symmety the only possibility is fo the fiels in egions I an IV to point towa = o away fom = In the fist case the flu woul be nonzeo on ou Gaussian suface but it must be zeo because the chage enclose is zeo Hence the electic fiel in egions I an IV is zeo (A simila agument hols if we assume that the fiel points in the -iection in both egions I an IV) Fo egions II an III, we choose a Gaussian cyline with en caps in egions II an III as shown in the figue below The electic flu on the encaps is E A = EA The chage enclose ivie byε is Qenc / ε = σa/ ε Theefoe by Gauss s Law, EA = σ A / ε which implies that the magnitue of the electic fiel is E = σ / ε Thus the electic fiel is given by E ; < σ ˆ i ; < < ε = σ < <+ ˆ i ; ε ; <

14 The gaph of the component of the electic fiel, E vs is shown on the gaph below (b) The electic potential iffeence between infinity an a point by P locate at, is given P V( ) V( ) = E s We shall evaluate this integal fo points in each egion We stat with P anywhee in egion IV, < Because the electic fiel in egion IV is zeo, the integal is zeo, If V( ) V( ) = E IV s = P is anywhee in egion III, < <+ then V( ) V( ) = E s E s IV III σ σ σ σ = E = = ( ) = ε ε ε ε If P is anywhee in egion II, < < then V( ) V( ) = E s E s E s IV III II σ σ σ σ = ε = + ε ε ε

15 If P is anywhee in egion I, < then V( ) V( ) = E s E s E s E s IV III II I σ σ σ σ = ε = = ε ε ε Because the electic fiel is continuous we can wite ou esult as Note this can be witten as V( ) V( ) ; σ σ + ; ε ε ε ε ; = σ σ ; + ; σ σ V( ) V( ) = ; ε ε ; This esult looks goo because the aea une the gaph of the component of the electic fiel, E vs fo the egion < < is zeo The plot of the electic potential as a function of on the gaph is shown below with units of [V] on the vetical ais (c) The wok you must o is equal to the change in potential enegy (assuming the pointlike object begins an ens at est) Theefoe

16 Qσ W = U() U( )) =+ Q( V() V( )) =+ ε

17 Poblem 6: You may fin the following integals helpful in this answeing this question b b n 1 n+ 1 n+ 1 = ( b a ); n n + 1 1, ln( b / a) = a Consie a chage infinite cyline of aius a The chage ensity is non-unifom an given by ρ () = b; <, whee is the istance fom the cental ais an b is a constant a) Fin an epession fo the iection an magnitue of the electic fiel eveywhee ie insie an outsie the cyline Clealy pesent you easoning, elevant figues, an any accompanying calculations b) A point-like object with chage + q an mass m is elease fom est at the point a istance fom the cental ais of the cyline Fin the spee of the object when it eaches a istance fom the cental ais of the cyline Poblem 6 Solutions: (a) Because the chage istibution efines two istinct egions of space, egion I efine by < an egion II efine by >, we must apply Gauss s Law twice to fin the electic fiel eveywhee In egion I, whee <, we choose a Gaussian cyline of aius an length l

18 Because the electic fiel points away fom the cental ais, the electic flu on ou Gaussian suface is I A = EIπ l E Because the chage ensity is non-unifom, we must integate the chage ensity We choose as ou integation volume a cylinical shell of aius, length l an thickness The integation volume is then V = π l Theefoe the chage ivie by ε enclose within ou Gaussian suface is 1 1 πlb Qenc / ε ρπl b πl ε ε ε πlb = = = = ε Theefoe Gauss s Law becomes E πl = πlb / I We can now solve fo the iection an magnitue of the electic fiel when <, E I b = ˆ ε In egion II whee >, we choose a Gaussian cyline of aius an length l

19 Because the electic fiel points away fom the cental ais, the electic flu on ou Gaussian suface is II A = EII π l E We again must integate the chage ensity but this time taking ou enpoints as = an Theefoe the chage ivie by ε enclose within ou Gaussian suface is = πlb Qenc / ε ρπ l b π l ε ε ε 1 1 πlb = = = = ε Theefoe Gauss s Law becomes E πl = πlb / II We can now solve fo the iection an magnitue of the electic fiel when >, E II b = ε 1 ˆ Collecte ou esults we have that E b ˆ; ε = b > 1 ˆ; ε < (b) The change in kinetic enegy when the object moves fom a istance fom the cental ais of the cyline to a istance is given by K ( ) K( ) = ( U( ) U( )) = q( V( ) V( )) Because the paticle was elease at est, K( ) =, an spee of the object is q vf = ( V ( ) V ( )) m K( ) (1/ ) mvf =, the final

20 The electic potential iffeence between two points in egion II is given by b 1 ( ) ( ) = E ˆ II s = s ε V V b 1 b b = = ln = ln(/ ) ε ε ε Theefoe the spee of the object when it eaches a istance the cyline is fom the cental ais of v f qb = ln(/ ) mε

21 MIT OpenCouseWae 8SC Physics II: Electicity an Magnetism Fall 1 Fo infomation about citing these mateials o ou Tems of Use, visit:

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