A closed triangular box is kept in an electric field of magnitude `E=2xx10^(3)NC^(-1)` as shown in the figure.

Calculate the electric flux through the entire surface.

Explanation

Electric field `E = 2 xx 10^3NC^-1`
(a) Electric fluxthrough vertical rectangular surface.
According to Gauss law electric flux is
`phi = EA costheta`
`phi_(vts) = EA os theta`
`phi_(vts) = -2 xx 10^3 xx (0.15 xx 0.05)cos 0^@`
`= -2 xx 0.15 xx 0.05 = -15 Nm^2//c`
`phi_(vts)= -15 Nm^2//c`
(b) Electric flux through slanted surface is
`phi_(ss) = EA cos60^@`
`:.phi_(ss) = EA xx 1/2`
`"Hypotenuous" = "Opposite side"/sin30`
`= (5 xx 10^-2)/(1/2) = 2 xx 0.05 = 0.1 m`
`:.Area = 0.1 xx 0.15 = 0.015m^2`
`:.phi_(ss) = 1/2EA`
`= 1/2 xx 2 xx 10^3 xx 0.015`
`phi_(ss) = 15 Nm^2 C^-1`
(c) Entire surface of the box
`phi_(entire) = phi_(rs) + phi_(ss) + phi_(ends)`
`phi_(ends) = EA cos 90^@`
= 0
`:.phi_(entire) = -15+15+0`
= 0
`:.phi_(entire) = zero`