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The suface integral of electrostatic field vec(E) produced by any sources over any closed surface S enclosinga volume V in vacumm, i.e.,total electric flux over theclosed S in vacumm is 1//in_(0) times the total charge(Q) contained inside S, i.e, phi_(E) = oint vec(E). vec(ds) = (Q)/(in_(0)) The charge insideS may be pointchargesor evencontinous charge distributions. There is no contribution to total electric flux fromthe charges outside S. Further, the location at Qinside S does not affectthe value of surface integral. Read the above passageand answerthe following questions : (i) what are the SI unitand dimensions of electric flux ? (ii) A closedsurfacein vacumm encloses charge -q, + 3q and +5q. Another charge+4q lies outsidethe surface. What is total electricflux over the surface ? (iii) A point charge q lies insidea sphericalof radius r. How will the electric flux be affected if radius of thesphere is doubled ? (iv) What values of life do yo+-earn from the theorem ? |
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Answer» Solution :(i) SI UNITS of electricflux are `NC^(-1) m^(2)`. Dimensional formula of electric FLUX is `phi = ((MLT^(-2)) L^(2))/(AT) = [M^(1) L^(3) T^(-3) A^(-1)]` (ii) `phi = (Sigma q_(inside))/(in_(0)) = (-q + 3q + 5q)/(in_(0)) = (7q)/(in_(0))` As `phi = (q)/(in_(0))` , therfore, electric flux is not affected by area//shape of the surface. So the electric flux remains unaffected. (iv) This theorem empahsisesthat total normal electric flux formthe surface depends only on algebraic sum of chargesenclosedby the surface, irrespective of their locations. The charges outside the surface do not affect the electricflux. In day to daylife, the theorem implies that the knowledge you can impart depends only on what you havestoredwithin you. You cannotemanite what youhavenot absorbed or what lies outside your reach. In the examnination, you can WRITE what you havelearnt by heart. Extraneous help from outside by UNFAIR means will never serve your purpose. |
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