Physics Courseware Electromagnetism
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1 Pysics Cousewae lectomagnetism lectic field Poblem.- a) Find te electic field at point P poduced by te wie sown in te figue. Conside tat te wie as a unifom linea cage distibution of λ.5µ C / m b) Find te electic field at point P poduced by te squae of unifom suface cage distibution σ.µc / m
2 Solution: Pat a) We conside te electic field poduced by a diffeential of wie as sown in te figue. Te diffeential of cage is dq λdy and te distance to point P is d + magnitude of te electic field is: y, so te d kλdy + y Te two components can be found by multiplying by te cosine and sine of te angle wit te oizontal: kλdy d + y + y and kλdy d y + y + y y Now, all we need to do is integate te epessions above to find te electic field: Fo te -component te integal is: Cange of vaiable y tanθ kλ kλdy + y + y dy ( + y ) ( + y ) / sec θ, dy sec θdθ, y θ, y θ tan / tan ( / ) tan ( / ) sec θdθ kλ cos d sec θ θ θ kλ kλ sinθ kλ + tan ( y / )
3 And fo te y-component te integal is: kλdy y y kλ + y + y kλ + kλ y + kλ + ydy ( + y ) / kλ + y Wit te values of te poblem: 6.5.5, V/m y , V/m Pat b) If you conside te contibution of a diffeential of aea of te squae, te electic field will be: kσddy d + y + Ten, to find te z-component we multiply by te cosine of te angle between te z-ais and te electic field vecto: d kσddy z + y + + y +
4 You sould ealize tat te epession ddy + y + + y + solid angle poduced at point P by ddy, so te electic field will be: z kσ S.A. is te diffeential of Wee S.A. is te solid angle poduced by te squae. You can use te epession L / deived elsewee: S. A. cos to calculate te electic field: + L / z kσ cos L / + L / Wit te values of te poblem: 6 z. cos. /.5 +. / 6, V/m Poblem.- Two identical conducting spees S and S ave adii.m and ae sepaated by a long distance dm. Initially one spee is caged wit Q µc and te ote is neutal (Q ). Ten you bing a tid identical conducting spee (initially neutal) and make contact wit te caged spee fist and ten wit te neutal one. Afte tat you take te tid spee vey fa away. Wat is now te foce between S and S? Solution: Wen two identical spees touc tey will ecange cage until te potential is te same on bot spees, so until te cage is saed by in equal quantities. Afte te tid spee touces S, te cage on S will be Q µc (te ote µc will be in te tid spee). Afte te tid spee touces S, te cage on S will be Q /µc Te foce is: F 6 6 QQ ( )(.5 ).5 ε o d N
5 Poblem.- Find te foce on te positive cage Q5µC at te oigin of coodinates due to te seven negative cages q-µc located aound te cicle of adius.5m at te positions sown in te figue. [Hint: Tee is an easy way and a ad way of doing tis] Solution: Notice tat all te foces cancel eac ote ecept fo te one due to te cage on te positive -ais. Te foce is: 6 6 Qq (5 )( ) F. N ε o d.5 Poblem.- Find te electic field at point P, wic is at a distance d.m fom te cente of two squaes of side L.m and unifom suface cage density σ.µc / m Solution: Te electic field poduced by eac squae is: kσ 6, V/m 6 Adding te two vectos we get: +,6 V/m
6 Poblem 5.- A spee of adius m is located wit its cente at te oigin of coodinates and as a cage of Q nc unifomly distibuted ove its suface. Anote spee of adius.m as its cente on te -ais at a distance of D.m fom te oigin of coodinates and as a cage of Q -nc also unifomly distibuted ove its suface. Calculate te electic field at point B(.5m,.5m,) Solution: Tee ae two vectos tat contibute to te total electic field, te magnitude of electic field due to te lage spee is kq 8V/m and te one due to te smalle spee kq 7.8V/m. +.5 To add te vectos it sould be done component by component, wit te esult: o o (8cos 5,8sin 5 ) o o ( 7.8cos 5.7, 7.8sin 5.7 ) (.,-.66) V/m
7 Poblem 6.- A spee of adius cm is located wit te cente at te oigin of coodinates and as a cage of nc unifomly distibuted ove its suface. Anote spee of adius 6cm as its cente on te -ais at a distance of D5cm fom te oigin of coodinates and as a cage of -nc also unifomly distibuted ove its suface. Calculate te electic field at point A(,,) Solution: Te electic field at point A(,,): Notice tat te contibution to te electic field fom te cage of te lage spee is zeo. Tis is because te cage is unifomly distibuted on its suface and point A is inside te spee. On te ote and, te smalle spee does poduce an electic field at point A. Since te cage distibution is speical and point A is outside te smalle spee, its electic field is te same as if all te cage wee located at its cente, so: q k d (.) 6 V/m Notice tat in te calculation of electic field we ignoed te sign of te cage. Wat te sign indicates is wete te vecto is towad te cage o away fom it. In tis case te electic field vecto is diected in te positive -diection. Poblem 7.- Calculate te electic field at point C due to te unifomly distibuted cage Q on te semicicle of adius.
8 Solution: In poblems wee tee is a continuous distibution of cages te standad pocedue to find electic field o potential is: - Divide te continuous distibution in small pieces (diffeentials) tat can be teated as point cages. - Calculate te field o potential poduced by te diffeential. Be caeful tat in te case of electic field you will need to calculate te components independently. - All distances, cages and angles sould be witten in tems of te vaiables cosen. - Integate ove tose vaiables. Te poblem above falls pecisely in te categoy tat we just mentioned. So let s divide te cage in small pieces. To do tis, notice tat te cage is unifomly distibuted, so te linea density of cage is: cage Q λ lengt Ten, we divide te ac in diffeentials as sown in te figue: Te lengt of te diffeential of ac is Q Q dq λdl λd θ dθ dθ d l dθ and te diffeential of cage is We fist calculate te magnitude of te electic field poduced by te diffeential of Q k dθ kdq kqdθ cage, wic is as if it wee a point: d d But tis is te magnitude of te vecto. It as an -component and a y-component. By te symmety of te poblem we notice tat we only need to calculate te -component.
9 Te -component of te electic field is: kqdθ d dcosθ cosθ Now we need to integate tis epession to find te electic field: / d / kqdθ cosθ kq Poblem 8.- A speical sell as intenal adius and etenal adius and contains a unifom distibution of cage wit density ρ Calculate te electic field fo a point at a distance fom te cente of te sell. Conside cases: a) < b) << c) > Solution: Inside te sell, wen < te electic field is zeo because if you conside a speical suface and apply Gauss law, te cage enclosed is zeo. At a point in te sell, wee << we can use Gauss s law, but te cage enclosed will only be te one in te sell between and. Te cage can be calculated by multiplying te volume of te sell (between and ) by te density of cage. Tis woks because te density is constant, if it wee a function
10 of te adius we would ave to divide te sell in layes (like an onion) and integate laye by laye. So, te cage enclosed is: Q enclosed ρ And using Gauss s law: ( ) / kq k k enclosed ρ ρ Finally, outside te spee you can calculate te electic field as if all te cage wee located at te cente of te sell. Te total cage is: Q ρ And te electic field: ( ) kq k ρ
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