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The electric field componentsare E_(x)=ax^(-1//2),E_(y)=E_(z)=0in which a =800 N/C m^(1//2)calculate(a) theflux throughthe cube and (b) the chargewithin the cube asume that a=0.1 m |
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Answer» Solution :For any closed surface, all area vectors are to be shown perpendicularly outward. Here, out of six faces of a cube, each having area `a^2`, electric flux gets associated with faces 1 and 2 only (because remaining four faces are parallel to given electric field along X-axis. Electric flux LINKED with FACE-1, `phi_(1) = A_(1)E_(1)cos theta_(1) =A_(1) (alphax_(1)^(1//2)) cos theta_(1)` `=(a^(2)) (ALPHA a^(1//2))cos (180^(@)) (therefore vecA_(1) || -vecE_(1))` `therefore phi_(1) = -alpha a^(5//2)`........(1) `(therefore x_(1) =a)` Electric flux linked with face -2 is: `phi_(2) = A_(2)E_(2) cos theta_(2)` `therefore theta_(2) = A_(2)(alphax_(2)^(1//2)) cos theta_(2)` `phi_(2) = sqrt(2)(alpha a^(5//2))`......... (2) Net electric flux passing through given cube is, `phi = phi_(1) + phi_(2)` `=-alphaa^(5//2) + sqrt(2) alpha a^(5//2)` `= (800)(0.1)^(5//2) (1.414-1)` `=((800)(0.414))/(3.162)^(5)` `phi = 1.048 (Nm^(2))/C` (Here, `phi`comes out to be positive which indicates that this flux comes out of given cube) Now, if net charge enclosed by given cube is q then according to Gauss.s theorem, `phi = q/epsilon_(0)` `therefore q =-phi epsilon_(0) = (1.048) (8.85 xx 10^(-12))` `therefore q = 9.2748 xx 10^(-12)` C |
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