Consider a uniform electric field `E = 3xx10^(3) hat(i) N//C`. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane ? (b) What is the flux through the same square if the normal to its plane makes a `60^(@)` angle with the x-axis ?

Explanation

(a) Electric field intensity, `barE =3 xx 10^(3) hati N//C`
Magnitude of electric field intensity, `|barE| = 3 xx 10 ^(3) N//C`
Side of the square, s = 10 cm = 0.1 m
Area of the square, `A = s^(2)` = `0.01 m^(2)`
The plane of the square is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, `theta = 0^(@)` Flux `(phi)` through the plane is given by the relation,
`phi=|barE|A cos theta`
`= 3 xx 10^(3) xx 0.01 xx cos0^(@)`
`=30 N m^(2) //C`
b) Plane makes an angle of `60^(@)` with the x-axis. Hence, `theta = 60^(@)`
Flux, `phi=|barE|A cos theta`
`= 3 × 10^(3) × 0.01 × cos60^(@)`
`30 xx(1)/(2) = 15 N m^(2)//C`