Question

Consider a uniform electric field E = 3 × 10³ îN/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?

Answer 1

(a) Electric field intensity, = 3 × 10³ î N/C

Magnitude of electric field intensity, = 3 × 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 =

= 3 × 10³ × 0.01 × cos0°

= 30 N m²/C

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

Flux, Φ =

= 3 × 10³ × 0.01 × cos60°

= 15 N m²/C