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A closed triangular box is kept in an electric field of magnitude E= 2 xx10^(3)NC^(-1) as shown in the figure . Calculate the electric flux through the (a) vetical rectangular surface (b) slanted surface and (c ) entire surface. |
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Answer» Solution :Electric field of magnitude `E=2xx10^(1)NC^(-1)` (a) Vertical rectangular surface : Rectangular area `A= 5xx10^(-2)xx15xx10^(-2)` `A= 75 xx10^(-24) m^(2)` `theta = 180^(@) implies COS 180^(@)=-1` Electric flux `Phi_(v.s)=EA "cos"theta` `=2xx10^(3)xx75xx10^(-4)xx"cos"180^(@)` `=-150xx10^(-1)` `Phi_(v.s) =-15Nm^(2)C^(-1)` (b) Stanted surface : `cos theta = cos 60^(@) = 0.5 ` `SIN theta = sin^30^(@) = (Opposite )/(hyp)` `hyp = (5xx10^(-2))/(0.5) ` hyp =0.1 m Area of slanted surface `A_(2) = (0.1 xx15xx10^(-2))` `A_(2)= 0.015 M^(2)`  Electric flux `Phi_(v.s) = EA cos theta ` `=2xx10^(3)xx0.015xxcos 60^(@)` `= 2xx10^(3)xx0.015xx0.5 ` `=0.015 x10^(3)` `Phi_(v.s)= 15 Nm^(2)C^(-1)` HORIZONTAL surface `theta= 90^(@), cos90^(@)= 0` Electric flux `Phi_(H.S)= E. A_(3) Cos90^(@)=0` (c ) ENTIRE surface : `Phi_("Total" )= Phi_(V.S)+ Phi_(s.s) + Phi_(H.S)=-15+ 15 +0` `Phi_("Total ")=0` |
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