Question

Given a uniform electric field ⃗5X10^3 i cap N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the y-z plane. What would be the flux through the same square if the plane makes a 30° angle with the x-axis?

Answer 1

(a) Electric field intensity, E = 3 × 103 î N/C
Magnitude of electric field intensity, |E| = 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=|E| Acosθ
= 3 × 103 × 0.01 × cos 0°
= 30 N m2/C

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

Flux,ϕ =|E|
= 3 × 103 × 0.01 × cos 60°

=30×1/2
= 15 N m2/C