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.

Explanation

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`