The electric field in a region is given by `vecE=3/5 E_0veci+4/5E_0vecj` with `E_0=2.0 xx 10^3 N C^(-1)`. Find the flux of this field through a recatngular surface of area `0.2m^2` parallel to the y-z plane.

Explanation

Given, E = (3/5 ` E_0` I + 4/5 `(E_0)` j)
` E_0 = 5 xx (10^3)N/C `
The plane is parallel to yx plane hence only
`(3/5 E_0 i)` passes perpendicular to the plane `
Where as `(4/5 E_0 i)` goes parallel,
` Area = 0.2 (m^2)(given)`
` So, flux = E xx a = 3/5 xx 2 xx (10^3) xx 0.2`
` = 0.24 xx (10^3)N (m^2)/C`
` = 240 N(m^2)/C.`