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Answer questions on the basis of your understanding of the followingparagraph and the related studied concept. Electric flux, in general , through any surface is defined as per relation: phi_E= int oversetto E. oversetto (ds), whereintegration has to be performed over the entire surface through which flux is required. The surface under consideration may be a closed one or an open surface. Whenflux through a closed surface is required we use a small circular sign on the integrationsymbol. Thus flux over a closed surface oint E= oint oversetto E. oversetto (ds) .it is customary to take the outward normal as positive in this case. A German physicist Gauss established a fundamental law to find electric flux over a closed surface. As per Gauss' law , the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by in_0 . Mathematically oint oversetto (E) . oversetto (ds) =(1)/( in_0)[q_(en) ], where q_(cn)is the net charge enclosedwithin the surface. It is possible to derive Gauss' lawfrom Coulomb's laws. Gauss' lawcan be applied to obtain electric field at a point due to continuous charge distribution for a number of symmetric charge configurations. Consider a closed surface having certain charges both within and outside as shown. The total electric flux of the given closed surface is :(##U_LIK_SP_PHY_XII_C01_E03_011_Q01.png" width="80%">

Answer»

` + 2.26 xx 10^(5)V m `
` +7.91 xx 10 ^(5)V m`
` -2.26 xx 10 ^(5)V m `
` +1.36 xx 10 ^(-6) V m`

Solution :`phi_E=(1)/(in_0) sum q_(en) =(-2MU C)/(8.85xx10^(-12) )= )(-2XX10^(-6))/(8.85 xx 10 ^(-12)) =-2.26 xx 10 ^(5) V m ]`


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