CHAPTER GAUSS'S LAW

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1 lutins--ch 14 (Gauss's Law CHAPTE GAU' LAW 141 This pblem is ticky An electic field line that flws int, then ut f the cap (see Figue I pduces a negative flux when enteing and an equal psitive flux when exiting Its net flux equals ZEO That means that the nly electic field lines that will pduce a net flux ae thse that ente thugh the bdy f the cap while exiting thugh the cap's hle (they pduced negative flux while passing in but pduce n additinal flux as they exit via the hle As the hle has an aea A, the flux thugh the hle, hence thugh the cap, will equal EA flux-in nly FIGUE I E flux-in equals flux-ut 14 The vlume chage density f this chage cnfiguatin will be: Q ρ 4 π Q 4 π Additinally, a diffeential shell f chage f adius b and diffeential thickness db lcated inside the Gaussian suface will have a diffeential vlume dv equal t: dv (suface aea(thickness (4πb db a F a spheical Gaussian suface inside the sphee (see Figue II t the ight, we can wite: q E d Ed encl ρdv b css-sectin f spheical shell dq dv FIGUE II b dv (4πb db db Gaussian suface 49

2 E d E( 4π E b Q b 4π Q E 4 π Q π [ ( 4πb db ] 4 Q ( b db b b b F the field utside the sphee: A Gaussian suface utside the sphee will have Q's wth f chage within it F this situatin, Gauss's Law in tuncated fm becmes: qencl E d Q E( 4π Q E π 4 In the wds, the sphee will appea like a pint chage f > 14 The situatin is sketched t the ight a and b Assume f the mment that the +Q's wth f chage was NOT placed at the cente f the system In that case, the +Q's wth f chage placed n the inside suface f the cnduct wuld migate t the utside suface f the cnduct due t electstatic epulsin Assume nw that the +Q's wth f chage was nt placed n the inside suface f the cnduct, but that the +Q's wth f chage was placed at the cente f the system In that case, -Q's wth f chage wuld migate fm the ute suface f the cnduct t the inne suface f the cnduct, leaving +Q's wth f chage n the utside suface Hw d we knw this? Because the electic field inside the cnduct must be ze f static chage situatins, and the nly way that can happen is if the net 44

3 lutins--ch 14 (Gauss's Law chage intenal t a Gaussian suface inside the cnducting egin is ze That will nly happen if the chage at the cente and the chage induced nt the inside suface f the cnduct ae equal and ppsite The tw situatins tgethe yield the fllwing: The +Q's wth f chage levitating at the cente will induce -Q's wth f chage nt the inne suface f the cnduct, leaving +Q's wth f chage n the ute suface f the cnduct That will be jined by the +Q's wth f chage that was initially placed n the cnduct's inside suface, but that subsequently migated t the ute suface due t electstatic epulsin Bttm line: Thee will be +Q's wth f chage at the cente, -Q's wth f chage n the inside suface f the cnduct, and +Q's wth f chage n the utside suface f the cnduct +Q -Q's wth f chage n inside suface +Q's wth f chage n utside suface c The chage enclsed inside a spheical Gaussian suface whse adius is < 1 is +Q As such, we can wite: FIGUE III qencl E d Q E( 4π Q E π 4 d The egin is a cnduct As such, the electic field in the egin is ze Pf? The chage inside a Gaussian suface within the egin will be -Q + Q e F > the chage enclsed within a Gaussian suface will be +Q As such, we can wite: q E d encl Q E( 4π Q E π 4 441

4 f The suface chage density n a sphee f adius having a ttal f +Q's wth f chage n it is: σ (Q 4π 144 The system is shwn in Figue IV a The units f k must be invese metes as the unit f is metes and the expnent (ie, k must be unitless The units f C must be culmbs pe mete if ρ is t be a vlume chage density b T detemine the Electic Field vesus Psitin gaph, we must detemine the electic field functin f each f the fu egins in which distinct fields exist Using Gauss's Law in cnjunctin with spheical Gaussian sufaces f each egin, we get: Ce k cnstant 1 FIGUE IV --F < 1 : The electic field in this egin is ZEO as thee is n chage enclsed within it --F 1 < < : Assuming the spheical Gaussian suface in this egin has a adius equal t, we need t detemine the amunt f chage esiding inside that suface The vlume chage density b units fm the sphee's cente is: FIGUE V dv (4πb db ρ C b ekb 1 db If we can detemine the diffeential chage inside a spheical shell f adius b and thickness db, then we can integate that between the inside adius f the shell (ie, 1 and the Gaussian adius Figue V highlights the setup dq dv Ce kb (4πb db b b 44

5 lutins--ch 14 (Gauss's Law E d q encl Ed b 1 ρdv E d E( 4π E b 1 π 1 4 C e k 4π C ( π b e kb b 4 db kb 4πC ( e db [ kb ] k1 C k e k e E b 1 b 1 --F < < : Inside the cnduct, the electic field will equal ZEO --F > : The Gaussian suface enclses all the chage within the cnfiguatin including that n the ute suface f the cnduct In tuncated fm, Gauss's Law f this yields: q E d encl E d E( 4π b 1 C [ ( π ]+ π b e kb b db σ 4 ( 4 kb 4π C ( e db + σ( 4π b 1 44

6 E E π 1 [ kb ] 4 C e + σ( 4π k b 1 4π C + k e k e k1 σ( ummaizing what we knw, and emembeing that we can substitute in C 1-1 culmbs pe mete, σ 1-1 culmbs pe squae mete, 1 1 mete, metes, and metes, we have: --F < 1 : E --F 1 < < : C k ek e k 1 E e (1 e (1(1 ( 885x1 1 (11 e 7 whee 1 < < --F < < : E --F < : C k e k e k 1 [ ]+ σ E 1 1 ( ( ( ( [ e e ]+ ( 1 ( x ( whee > The gaph f these functins, each evaluated in its wn egin, is pesented n the next page in Figue VI 444

7 lutins--ch 14 (Gauss's Law Electic Field vesus Psitin f pheical Chage Cnfiguatin E Fld (nts/cul FIGUE VI Psitin (metes c Electic field functins ae evidently nt cntinuus functins, as the gaph shws 145 The system is shwn in Figue VII t the ight a The utside pipe must be an insulat as it has a vlume chage density assciated with it (thee can be n fee chage inside a cnduct as all such chage will migate as fa away fm like chage as pssible, edistibuting itself n a cnduct's utside suface 4 1 FIGUE VII C k e 1 445

8 A Gaussian suface between 1 and will have chage enclsed within it due t the suface chage density σ 1 n the inside pipe's suface That means thee will be an electic flux thugh the Gaussian suface which, in tun, means thee will be an electic field in that egin A cnduct cannt have an electic field within its bunday, hence the mateial making up the inside pipe must be an insulat b As was the case with the spheical-symmety-pblem f the same kind (Pblem 144, we must deive electic field expessins f each egin Ding s yields: --F < 1 : A cylindical Gaussian suface within this egin will have n chage enclsed within it, hence the electic field in this egin will be ze --F 1 < < : Inside the innemst pipe, the chage enclsed within a cylindical Gaussian suface f length L will be the cnsequence f the suface chage density σ 1 As such: qencl E d σ Ed ( cs σ1( πl 1 E ( d σ1( πl 1 E( π L σ11 E ( πl F < < : Outside the inne pipe, the chage enclsed within a cylindical Gaussian suface f length L will be the cnsequence f the suface chage density n the inside and utside sufaces As such: qencl E d σ E( π L σ11 σ E ( πl σ ( πl

9 lutins--ch 14 (Gauss's Law --F < < 4 : Inside the ute pipe, thee is a vlume chage density t cnside emembeing we must include ALL the chage enclsed within the Gaussian suface (ie, including the chage n the inne pipe's sufaces, we can wite: q E d encl E( π L ρdv + σ1( π1l σ( πl a ka [( C/ a e ][( πal da]+ σ1( πl 1 σ( πl a E ( πl ka C ( e da + σ1( 1 σ( a E ( ka [ ( C/ k e ] + σ ( ( a 1 1 σ E ( ( k k C/ k [ e e ( ( E ]+ σ1 1 σ ( --F > 4 : Outside the ute pipe, ALL the chage in the entie system is enclsed within a Gaussian suface As such, we can wite: q E d encl E( π L E 4 a 4 a ρdv + σ ( π L σ ( π L ka π 1 1 [( ]+ σ π σ π ( C/ a e al da ( L ( L + ( πl ka 4 ( C / ke σ ( ( a 1 1 σ E ( k4 k ( C/ k [ e e ]+ σ1( 1 σ( E (

10 ummaizing what we knw and putting in C 1-1 culmbs pe squae mete, σ culmbs pe squae mete, σ x1-1 culmbs pe squae mete (this is the chage's magnitude--the negative sign has aleady been incpated int the pblem via Gauss's Law, 1 1 mete, metes, metes, and 4 4 metes, we get: --F < 1 : E --F 1 < < : E σ 1 1 (1 1 (1 (885x whee 1 < < --F < < : --F < < 4 : E E σ σ ( 1 ( 1 ( x1 ( 1 ( 885 x1 9 whee < < C k e k e k + σ σ ( 1 ( 1( 1 1 [ e e ] + ( 1 ( 1 ( x1 ( 1 ( 885 x1 11e 6 whee < < 4 --F 4 < : 448

11 lutins--ch 14 (Gauss's Law E C k ek 4 e k +σ σ Putting it all tgethe in Figue VIII, we get: [ e (1(4 e (1( ] + (1 1 (1 (x1 1 ( (885x1 1 whee 4 < Electic Field vesus Psitin f Cylindical Chage Cnfiguatin E Fld (nts/cul FIGUE VIII Psitin (metes 146 The easiest pssibility is a vlume chage density functin that is cnstant Assuming that is the case, Gauss's Law yields: 449

12 q E d ρv ( Ed (cs 6 k ρ πl d 6 k ρ 6 k ρ πl ( π L 6 encl 4 A secnd pssibility is t assume that the vlume chage density functin vaies with the distance fm the cental axis In that case, we can use Gauss's Law in cnjunctin with a cylindical Gaussian suface t detemine the electic field utside the d, wite ut the q encl expessin in tems f the vlume chage density functin ρ, and then detemine ρ by cmpaing u integal expessin with the knwn electic field expessin Ding s yields: q E d ( Ed (cs encl E d 6 k ( L 6 6 k 6 a a ρdv ρ[ (( πadal ] π π L ρ[ ( a da] a a ρ ( ada By inspectin, ρ ka 4 Nt clea? Cnside: 45

13 lutins--ch 14 (Gauss's Law --If ρ k, the integal wuld equal ka / Afte evaluating between a and a, the ight-hand side f Gauss's Law wuld equal k / --If ρ ka, the integal wuld equal ka / Afte evaluating between a and a, the ight-hand side f Gauss's Law wuld equal E k / --If ρ ka n, the integal wuld equal ka (n+ /(n+ Afte evaluating between a and a, the ight-hand side f Gauss's Law wuld equal E k (n+ /(n+ --Nting this, the ight-hand side f Gauss's Law will equal the left-hand side (ie, k 6 /6 if the vlume chage density functin is ka a Assuming the disk's thickness is t, the vlume chage density f the disk is: ρ Q/V Q/[π t] Q/[π(1 m ( m] 159Q Using a cylindical Gaussian plug that extends thugh the disk an equal distance n eithe side (see Pat I f this chapte if this is nt clea, we get a nea-field deivatin that yields: qencl E d ρdv EA + EA 15 9Q dv EA 15 9QAt ( EA 1 ( 885 x1 15 9Q( E 1 ( 885 x1 1 E 18 x1 Q b A suface chage density functin f this cnfiguatin must allw us t detemine the amunt f chage undeneath, s t speak, an aeasectin n the disk's suface 451

14 As the ttal chage within the disk is Q while the ttal suface aea is π π(1 m 14 m, we can wite: σ Q/A Q/14 18Q c Teating the disk like a sheet f chage and using a cylindical Gaussian plug that extends thugh the sheet f chage an equal distance n eithe side, we get: qencl E d σda EA + EA 18Q da EA 18QA EA 1 ( 885 x1 1 E 18 x1 Q This is the same nea-field expessin we calculated using the vlume chage density functin in Pat a It is inteesting t nte that this is als numeically equal t σ/ d Cnside a cicula hp f adius with ttal chage dq distibuted unifmly n its suface Define a diffeential chage dq' at sme abitay psitin n the hp That diffeential chage will pduce a diffeential electic field de' at x (see Figue IX When we integate t get the net field de due t all the chage dq in the hp, the y cmpnents will add t ze due t the symmety f the cnfigu- FIGUE IX dq 1/ s ( + x x de cs de nte that: 1/ sin / ( + x 1/ cs x/ ( + x de sin 45

15 lutins--ch 14 (Gauss's Law atin That means we can fget them and fcus slely n the x cmpnents f the field Put anthe way, the net field de fm the hp will equal the x cmpnent f the vect sum f all the de' quantities, : Ding that integal yields: de de' x de de'(cs θ 1 dq' π s csθ 4 dq x 1 ' 4π ( + x + x ( ( x dq ' / 4π ( + x ( xdq 4 π + x 1/ 1 / ( / Again, the diffeential chage dq in this expessin is the ttal chage n the hp Having an expessin f the magnitude and diectin f the electic field f an abitay hp alng its cental axis, we can nw deal with u disk (see Figue X Nting that dq σda, whee da (πd and σ (Q/π, we can wite: d dq x da (πd 1/ s ( + x de cs de de sin disk nte that: 1/ sin / ( + x 1/ cs x/ ( + x FIGUE X 45

16 E actual de ( xdq / 4π ( + x x σda π + / 4 ( x x σ[ ( πd ] π / 4 ( + x xσ d / ( + x σx 1 1 Eactual x ( + x 1/ e As we get away fm the disk, the field stength will decease That means that the cnstant σ/ value we deived in the chapte f the nea-field situatin will becme futhe and futhe ff as ne pceeds ut alng the x axis We want t knw at what x that nea-field value is 95% f the actual value Using the actual field expessin deived in Pat d and the nea-field infmatin deived in Pats a and/ b we can wite: 95E nea fld σx 1 1 x + x ( 1 / Q 1 x [ 1 8x1 Q] π x + x ( 1 8x1 [ ( 8 85x1 ]π( x 5 x ( + x 1 / ( 1 / x ( + x 1 / Evidently, the Gaussian electic field is gd t 5% at 5 metes fm the suface F u case in which 1 mete, the nea-pint Gaussian expessin is within 5% f the actual electic field value when at 5 centimetes fm the disk's suface 454

CHAPTER 24 GAUSS LAW

CHAPTER 24 GAUSS LAW CHAPTR 4 GAUSS LAW LCTRIC FLUX lectic flux is a measue f the numbe f electic filed lines penetating sme suface in a diectin pependicula t that suface. Φ = A = A csθ with θ is the angle between the and