#### Question

The electric field in a region is given by

`vec"E"= 3/5"E"_0 vec"i" + 4/5 "E"_0 vec "i" "with" " E"_0 = 2.0 xx 10^3 "N""C"^-1.`

Find the flux of this field through a rectangular surface of area 0⋅2 m^{2} parallel to the y-z plane.

#### Solution

Given:

Electric field strength `vec"E" = 3/5 "E"_0 hat"i" + 4/5 "E"_0` \[ \stackrel\frown{j}\],

where E_{0} = 2.0 10^{3} N/C

he plane of the rectangular surface is parallel to the *y-z* plane. The normal to the plane of the rectangular surface is along the *x* axis. Only `3/5 "E"_0`\[ \stackrel\frown{i}\], passes perpendicular to the plane; so, only this component of the field will contribute to flux.

On the other hand, `4/5 "E"_0`\[ \stackrel\frown{j}\] moves parallel to the surface.

Surface area of the rectangular surface, *a *= 0⋅2 m^{2}

Flux,

`phi = vec"E" . vec"a" = "E" xx "a"`

`phi = (3/5 xx 2 xx 10^3) xx ( 2 xx 10^-1) "N""m"^2 //"C"`

`phi = 0.24 xx 10^3 "N""m"^2//"C"`

`phi = 240 "N""m"^2//"C"`