InterviewSolution
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InterviewSolution
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A uniform electricfieldalongthex - axisis givenas,vecE =(200 hati)N* C^ (- 1 ) ,forxgt0=(-200 hati )N *C^ (- 1 ) ,forxlt0Acylinderof length20cmandradius5 cmhas itscentreattheorigin and axisalongthex - axisis placedin vacuum.findout(i)the electricfluxacrosseachof itscircularfaces, (ii)the fluxacrossitscurvedsurface,(iii)the fluxacrossitscentreoutersurfaceand (iv)thenet chargeenclosedby it. |
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Answer» Solution :Radiusof thecylinder, `r =5 cm=0.05 m` `therefore`Areaofeachcircularface, ` S =pi r ^ 2=3.14xx (0.05 ) ^ 2m ^ 2 `  Asthelengthofthecylinderis20 cmor0.2 m, thetwocircularfacesareat`X=+ 0.1m andx=- 0.1m`. Theareavectorsrepresentingthe right and the leftcircularfacesare ` vec S _ 1=hati(pi r ^ 2)andvec S _ 2=-hati(pi r ^ 2 )`respectively. theelectricfieldsat THEPOSITIONSOF thesecircular facesare,` vec E _ 1=(200 hati)N * C ^(- 1andvec E _ 2=(-200 hati)N * C ^(- 1 ) `respectively. (i)The electricfluxacross therightcircularface, ` phi _1=vec E _ 1 *vec S _ 1=(200 hati)*hati(pi r ^ 2 )` `=200 xx3.14 XX(0.05 ) ^ 2=1.57N*m^ 2*C ^( -1 )` Similarly,for theleftcircularface, ` phi _ 2=vec E _2 *vecS _ 2 =(- 200hati)*(-hatipir ^ 2 ) ` `= 200xx 3.14xx (0.05) ^ 2=1.57N *m ^ 2C ^ (- 1 ) ` `thereforephi _ 1= phi _ 2` i.e.,equalfluxpassesacrosseachof thetwofaces. (ii) The curvedsurface iseverywhereparallel to theelectric fieldvector. `therefore` Theelectricfluxlinkedwiththecurved surface=0 ` . (iii)Theentireoutersurfaceconsistsof thetwocircularfacesand thecurvedsurface. so, thefluxlinkedwithentiresurface, `thereforephi=phi _ 1+phi_ 2+0=1.57 + 1.57+ 0` ` = 3.14 N*m^ 2*C ^(- 1 ) ` (iv)From Gauss' theorem , netflux`phi=( q ) /(in _0) ` . So theenclosedchargeis, `q=phiin_ 0=3.14xx8.854 xx 10 ^(- 12 )` `=2.78 xx 10 ^(-11 ) `C |
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