InterviewSolution
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InterviewSolution
| 1. |
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. |
| Answer» <html><body><p></p>Solution :Radiusof thecylinder, `r =5 cm=0.05 m` <br/>`therefore`Areaofeachcircularface,<br/>` S =pi r ^ 2=3.14xx (0.05 ) ^ 2m ^ 2 ` <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CHY_DMB_PHY_XII_P1_U01_C02_SLV_060_S01.png" width="80%"/><br/>Asthelengthofthecylinderis20 cmor0.2 m, thetwocircularfacesareat`<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>=+ 0.1m andx=- 0.1m`.<br/>Theareavectorsrepresentingthe right and the leftcircularfacesare ` vec S _ <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>=hati(pi r ^ 2)andvec S _ 2=-hati(pi r ^ 2 )`respectively.<br/> theelectricfieldsat <a href="https://interviewquestions.tuteehub.com/tag/thepositionsof-3202244" style="font-weight:bold;" target="_blank" title="Click to know more about THEPOSITIONSOF">THEPOSITIONSOF</a> thesecircular facesare,` vec E _ 1=(200 hati)N * <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> ^(- 1andvec E _ 2=(-200 hati)N * C ^(- 1 ) `respectively.<br/> (i)The electricfluxacross therightcircularface, <br/>` phi _1=vec E _ 1 *vec S _ 1=(200 hati)*hati(pi r ^ 2 )` <br/>`=200 xx3.14 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a>(0.05 ) ^ 2=1.57N*m^ 2*C ^( -1 )`<br/> Similarly,for theleftcircularface,<br/>` phi _ 2=vec E _2 *vecS _ 2 =(- 200hati)*(-hatipir ^ 2 ) ` <br/>`= 200xx 3.14xx (0.05) ^ 2=1.57N *m ^ 2C ^ (- 1 ) ` <br/> `thereforephi _ 1= phi _ 2` i.e.,equalfluxpassesacrosseachof thetwofaces.<br/>(ii) The curvedsurface iseverywhereparallel to theelectric fieldvector.<br/>`therefore` Theelectricfluxlinkedwiththecurved surface=0 ` .<br/> (iii)Theentireoutersurfaceconsistsof thetwocircularfacesand thecurvedsurface. so, thefluxlinkedwithentiresurface,<br/> `thereforephi=phi _ 1+phi_ 2+0=1.57 + 1.57+ 0`<br/>` = 3.14 N*m^ 2*C ^(- 1 ) `<br/>(iv)From Gauss' theorem , netflux`phi=( q ) /(in _0) ` . So theenclosedchargeis,<br/>`q=phiin_ 0=3.14xx8.854 xx 10 ^(- 12 )` <br/>`=2.78 xx 10 ^(-11 ) `C</body></html> | |