Physics 604 Problem Set 1 Due Sept 16, 2010
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1 Physics 64 Polem et 1 Due ept ) ) Inside good conducto the electic field is eo (electons in the conducto ecuse they e fee to move move in wy to cncel ny electic field impessed on the conducto inside the volume of the conducto) Apply Guss s lw to ny volume wholly inside the conducto inside E nd Thee cn e no [net] chge inside the conducto Electic field cn come up to the sufce of conducto whee it must e teminted y sufce chge on the sufce ) uppose we hve closed conducting shell nd chge outside the shell We would like to compute the electic field inside the shell ecuse of the pesence of the chge Choose closed sufce completely inside the shell enclosing the inteio of the shell Becuse E on the sufce since it is inside the conducto thee is no net chge inside the sufce independent of whee the exteio chge is plced nd no electic field induced inside of the sufce y the exteio chge ecuse the chge inside must emin eo In othe wods the inteio of conducting shell is shielded fom extenl electicl distunces On the othe hnd n unlnced chge inteio to the shell will genete field outside the shell ecuse E nd inside fo ny sufce tht totlly encloses the shell c) A Gussin pillox gument shows Enoml / If the pillox extends inside the conducto to outside nd one consides smll e element PB E nd Enoml A A A This specifies the noml component uppose one hs smll Gussin loop ove the sufce with displcement pllel to the sufce Then E dl Etl A Et A
2 ecuse the ckwd pt of the loop integl inside the conducto must vnish Becuse the loop cn e oiented in ny diection pllel to the sufce the tngentil field (in ll tngent diections) must vnish ) This polem is stightfowd ppliction of Poisson s Eution Note tht fo smll the potentil goes to 1 nucleus 4 The discussion in clss shows nucleus nucleus epesenting the density of the singly-chged nucleus Thee e no othe singulities in the potentil to woy out Using the expession fo the plcin in spheicl coodintes on the ck cove of the text x x 1 e 1 4 e 1 e 4 e 1 e e 4 3 e e electon 4 is the vege electon chge distiution in the tom Note tht it integtes to the pope vlue: 3 3 e electon electond x dd cosd 4 3 e x x dx Becuse of the finite sie of the nucleus in fct the chge density of the poton is sped out nd the potentil isn t 1/ t distnces shot comped to the poton sie This kind of thing is studied with get pecision t Jeffeson
3 3) ) Neglect the edge effects nd ssume tht the chge is unifomly distiuted on the pltes the uppe plte with positive chge the lowe plte with negtive chge Assume the -diection is ligned long d By the usul Gussin pillox gument the field inside is The potentil diffeence is EA A E A d E dl E d E d The cpcitnce is A C d / ) Hee ssume tht the negtive chge is on the inside sphee nd unifomly distiuted y symmety Guss s w gives 4 E 1 E 4 The potentil diffeence is 1 1 Ed 4 The cpcitnce is 4 C c) Assume tht the negtive chge is on the inside cylinde nd unifomly distiuted long the length of the cylinde Guss s w gives The potentil diffeence is / / 1 E E / Ed ln ln
4 The cpcitnce is C / ln / d) One 4) ) Neglect the edge effects nd the coections due to the fct the chge is slightly displced fom unifom distiution on the conducto (coections of ode /d nd /d) nd ssume tht the fields e simply two line chge fields supeposed / / x ln x / ln x / whee e the positions of the cente of the line chges To the fist significnt ode in /d nd /d / / x ln d / ln d / / d / d ln 5) This polem is stightfowd ppliction of Geen s econd Identity Following the sme pocedue s going fom En 135 to En 136 yields (note tht the tem is eo in En 136 ecuse the scl potentil solves the plce Eution) ln x d 4 R n n R 1 1 n d d 4 4 R 1 1 d dx dy d x d whee is the dius of the sphee nd R x x The finl integl is clely the vege of the potentil ove the sufce of the sphee It does not mtte wht dius is chosen fo the sphee in pefoming the vege ut of couse the vlues of the scl potentil on the sufce will depend on the choice of dius 6) The (uppe ound) cpcitnce detemined y the til function is
5 C d 3 d x whee is the inne dius of the cylinde whee is the oute dius of the cylinde nd is the length of the cylinde Evluting the exct nd estimted cpcitnce numeiclly yields this tle: / Exct C / = ln / 1 Til Function C / = / 1 / / The Exct field is moe like the line til function when In the limit clely the two expessions gee y the expnsion ln 1x x fo smll x Notice the til vlues e indeed highe thn the exct vlues
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