Consider a uniform electric field E = 3 × 10³ î 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?

Ans (a).

Electric field intensity, E = 3 x 10³ î N/C

Magnitude of electric field intensity, |E| = 3 x 10³ N/C

Side of the square, s = 10 cm = 0.1 m

Area of the square, A = s² = 0.01 m²

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, Θ = 0°

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

Φ = |E| Acos Θ = 3 x 10³ x 0.01 x cos 0° = 30 Nm²/C

Ans (b).

Plane makes an angle of 60° with the x – axis. Hence, Θ = 60°

Flux, Φ = |E|Aco5 6 = 3 x 10³ x 0.01 x cos 60°

= 30 x 1/2 = 15 Nm²/C