Given a uniform electric field →E=5×10³ î 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 could be the flux through the same square if the plane makes a 30° angle with the x-axis?

Flux through a surface is given by: Φ=→E.→dS
Area is S=0.1 × 0.1 = 0.01 m²
When the plane is parallel to the y-z plane, the area vector →dS is along the x-axis (î). Therefore, the angle between →E and →dS is 0°.
Φ = E S cos(0°) = 5 × 10³ × 0.01 = 50 Nm²/C
When the square makes an angle of 30° with the x-axis, the angle made by the normal to the square with the electric field is 30°. Therefore, the angle between →E and →dS is 30°.
Hence, Φ = E S cos(30°) = 5 × 10³ × 0.01 × (√3/2) = 25√3 Nm²/C ≈ 43.3 Nm²/C