General Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface

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1 Genel Physics II Chpte 3: Guss w We now wnt to quickly discuss one of the moe useful tools fo clculting the electic field, nmely Guss lw. In ode to undestnd Guss s lw, it seems we need to know the concept of electic flux. We cn imge the flux s volume te of flow. We lso cn imgine the light come out to the oom is the sme s the light come out fom the light bulk. lectic Flux Flux comes fom the tin wod mening flow. Field lines oughly descibe the electic field stength. The stength of electic field is epesented by the eltive numbe of field lines pssing though unit e: numbe of field lines/e φi = Ai = Ai cosθ fo whole sufce: φ = A i A i smll da fo continuous sufce Now we hve φ = da is whole sufce whee integl. xmple Find the flux tht exits sphee centeed t the oigin due to point chge lso t the oigin. A smll sufce e on sphee is, da = sin θ dθ dφ ˆ. Using the field due to the point chge, = k q ˆ, the flux cn be clculted fom its definition,φ d A, Φ k q ˆ sin θ dθ dφ ˆ π Φ=k q sinθ dθ dφ = 4πkq = q o π xmple Find the flux due to point chge ove the sufce of cube. Agin stt with the definition of flux, Φ da, nd the field of point chge, = k q ˆ. This time howeve, the electic field nd the e element e not lwys pllel nd the integl is vey difficult. But, we know tht the numbe of field lines tht leve the chge is not ffected by the shpe of some imginy sufce tht suounds the chge. θ da da Gzintep Univesity Fculty of ngineeing Deptment of ngineeing Physics 1

2 Genel Physics II Tht is, the sme numbe of field lines exit the cube s exit the sphee in pevious exmple so the flux must gin equl Φ = q. o (Optionl) In ode to clculte flux though the cube due to point chge we cn wite the electic field in Ctesin coodinte system: q xxˆ+ yyˆ+ zzˆ = 4π ( ) 3/ x + y + z et us clculte the flux though the uppe sufce. The sufce e is da = dxdyzˆ, the the flux is q zdxdy Φ uppe = 4π x + y + z ( ) 3/ I hve evluted this integl by using Mthemtic( A compute pogm) nd the esult is q q Φ uppe =. We hve six sufce the the totl flux is Φ=. 6 Guss w A foml sttement of the eltion between chge nd flux is known s Guss lw. The totl electic flux emeging fom n bity volume equls the net chge enclosed within the volume divided by. φ da = Qenclosed Guss lw is poweful tool to clculte electic field. Howeve, it s only useful when cn be tken outside of integl. Tht mens electic field is constnt nd symmeticl fo the chosen sufce (e.g., the line is vey long o the plne is vey big, so tht it s no effect fom the edges). Fo exmple, if pllel da nd constnt then Q da = da = enclosed so now we cn find electic field esily by Q = enclosed da Guss w So the chosen sufce is vey impotnt to mke da so cn be pulled out of the integl! Fo continuous chge distibutions the enclosed chge cn be witten s: Qenc = ρd fo line, Qenc = ρsds fo sufce, Qenc = ρvdv fo vulume chge distibutions. In summy, Guss s w is usully used in eithe of two wys: 1) Given the field nd the sufce then enclosed chge cn be found. ) Given the enclosed chge nd sufficient symmety to choose convenient sufce, then the field cn be found. Gzintep Univesity Fculty of ngineeing Deptment of ngineeing Physics

3 Genel Physics II A note: Conducto: A mteil in which electons e completely fee to move in esponse to pplied electic fields. In conducto the electons will move until they find plce whee the field is zeo. Theefoe, the field inside conducto will lwys be zeo, if you wit long enough fo the electons to find these plces. Typiclly, this tkes micoseconds. xmples Spheicl symmety ) Given conducto sphee of dius with totl chge, Q, find the electic field inside nd outside the sphee. Guss sufce (imginy) Chge Q distibuted ove the sufce of the conducto! Imgine sufce (not el(physicl) sufce it is mthemticl sufce) just inside the el sufce of the conducto ove which we will pply Guss w. Since this "Gussin sufce" is inside conducto the chge inside the conducto is zeo, theefoe da = q encl = q encl =. = ; < o The gussin sufce cn be mde bitily close to the el sufce. Guss sufce (imginy) Noml of the sufce plel to the electic field lectic field outside the sphee cn be clculted: Q Q Q ds. = 4 π = = ; >. S 4π b) A totl chge, Q, is unifomly distibuted thoughout non-conducting sphee of dius,. Find the electic field inside nd out. Sketch vs.. The spheicl symmety mens tht will be constnt ove ny concentic gussin sphee nd will point dilly (pllel with da). Theefoe the flux integl Guss sufce (imginy) Gzintep Univesity Fculty of ngineeing Deptment of ngineeing Physics 3

4 Genel Physics II in Guss w is, da = A = (4π ). Fo gussin sphee with < the chge enclosed is popotionl to the fction of the volume of the el sphee tht the gussin sphee occupies. 4 3 π3 q encl = 4 Q = 3 π3 3 3 Q. Applying Guss w, (4π ) = 3 3 Q o = k Q 3 < q da = enclosed, o Fo gussin sphee with > the chge enclosed is just the totl chge, Q. Applying Guss w, da = q enclosed (4π ) = Q = k Q o o > This field is just the field due to point chge Q t the oigin. Notice the equtions fo the fields inside nd outside gee t = s they must. Q k α α 1 Ty the poblems including electic fields of concentic sphees with vious chge distibutions. i.e is chge distibution of sphee is ρ (1 ) v = ρ + cn you clculte electic field inside nd outside the sphee?? Cyclindicl symmety ) A vey long thin wie hs line chge density, ρ, unifomly distibuted thoughout its length. Find the electic field s function of the distnce. By symmety the electic field must point dilly outwd nd it cn only depend on. Theefoe, the best gussin sufce is concentic cylinde of dius,. Using Guss w q 1 enc ds. = π= ρ d S ρ = π b) A vey long non-conducting cylinde of dius,, hs chge density,, unifomly distibuted thoughout its volume. Find the electic field s function of the distnce fom the xis,, nd sketch gph of vs.. ρ Gzintep Univesity Fculty of ngineeing Deptment of ngineeing Physics 4

5 Genel Physics II By symmety the electic field must point dilly outwd nd it cn only depend on. Theefoe, the best gussin sufce is gin concentic cylinde of dius,. The flux integl in Guss w cn be boken up into thee pts. One fo ech sufce shown, da = da + da + d s 3 s A s 1 s 1 s s 3 The integls ove s 1 nd s e both zeo becuse the field is dil which mens tht no flux exits these fces ( is pependicul to da). Ove s 3 is constnt nd pllel to qenc qenc da so tht integl is, ds. = π=. Be ceful when you clculte the s3 enclosed chge. The totl chge in length of the cylinde is: < the chge enclosed in the gussin sufce is, ρ field will be = ; < π Fo > the chge enclosed in the gussin sufce is, qenc = ρ. Applying Guss w, qenc ds. = ρ ρ π ; > s3 = = π Note tht the field flls off s 1 not 1 s you might expect. The equtions fo the fields inside nd outside gee t = s they must. Ty youself clcultion of the electic field k λ inside nd out side the concentic cylindes! Also ty to the clcultion of electic field of conducto cylinde! enc q π q = ρ = ρ = ρπ. Then fo π α s 1 s. The electic α 1 s 3 Ctesin symmety A lge metl plte of thickness, t, hs unifom chge density, σ. Find the electic field s function of the distnce fom the cente of the plte, z. By symmety the -field cn only be function of z nd it must point diectly wy fom the plte. Inside the plte the field is zeo becuse it is conducto. Outside we cn choose the gussin sufce shown. The sufces s 1 nd s e the sme bity shpe, pllel with the sufce of the plte, nd equl distnces fom it. The gussin sufce is completed with the sufce s 3 which is eveywhee pependicul to the sufce of the plte. The flux leving the gussin sufce is, da = da + d A + da. s 1 s s 3 Gzintep Univesity Fculty of ngineeing Deptment of ngineeing Physics 5

6 Genel Physics II Since the field is only long the z-xis the integl ove s 3 is zeo. The mgnitude of the field cn only depend on z, so the fields t s 1 nd s e equl nd constnt so the integls e stightfowd, da = A + A + = A The chge enclosed in the gussin sufce is, q encl =σa. Applying Guss w, qenc σa σ da. = A = = A A Ty electic fields of the pllel pltes with diffeent chge distibutions. emembe gin The Definition of lectic Flux Φ da Guss w da = q enclosed o The Behvio of Conductos, 1) = eveywhee inside. ) xcess chge stys on the sufce. σ 3) = nd pependicul to the sufce just outside. o z Gzintep Univesity Fculty of ngineeing Deptment of ngineeing Physics 6