Conisder an electric field `vecE=E_0hatx` where `E_0` is a constant .
The flux through the shaded area (as shown in the figure) due to this field is

Options

A. `2 E_(0) a^(2)`
B. `sqrt2 E_(0) a^(2)`
C. `E_(0) a^(2)`
D. `(E_(0) a^(2))/(sqrt2)`

Answer

C

Explanation

Here, `vec E = E_(0) hat x`
Let `hat x, hat y` and `hat z` be the unit vectors along x, y and z axis respectively.
InFig. The shaded area,
`vecA = vec(PQ) xx vec(PS)`
`:. vecA = (0 hatx + a haty + 0 hatz) xx (a hatx + 0 haty + a hatz)`
`= (a^(2) hatx - a^(2) hatz`)
Electric flux through the shaded area,
`phi = vecE . vecA = (E_(0) hatx). (a^(2) hatx - a^(2) hatz) = E_(0) a^(2)`