Consider a uniform electric field E** **= 3 × 10^{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?

#### Solution

**(a) **Electric field intensity, `vec"E"` = 3 × 10^{3 }`hat"I"` N/C

Magnitude of electric field intensity, `|vec"E"|` = 3 × 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 yz plane. Hence, the angle between the unit vector normal to the plane and electric field, θ = 0°

Flux (Φ) through the plane is given by the relation,

Φ = `|vec"E"|"A"costheta`

= 3 × 10^{3} × 0.01 × cos 0°

= 30 N m^{2}/C

**(b) **Plane makes an angle of 60° with the x-axis. Hence, θ = 60°

Flux, Φ = `|vec"E"|"A"costheta`

= 3 × 10^{3} × 0.01 × cos 60°

= `30 xx 1/2`

= 15 N m^{2}/C