Question

Given a uniform electric field E = 5?10^3 i N/C, find the flux of this field through a square of 10 cm on a side whose plane is parallel to the YZ-plane. What would be the flux through the same square, if the plane makes an angle of 30 degree with the X-axis ?

Answer 1

given, electric field intensity, E = 5 × 10³ î N/C

so, the magnitude of electric field , |E| = 5 × 10³ N/C

side of square , a = 10cm = 0.1m

area of square , |A|= a² = (0.1)² = 0.01m²

as question says that the plane of square is parallel to Y-Z plane. so, area vector is perpendicular on Y-Z plane. e.g., parallel to electric field intensity. hence, angle between area vector and electric field , $\theta$ = 0°

so, electric flux , $\phi$ = |E||A|cos0° = 5 × 10³ × 0.01 = 50 Nm²/C

again, plane makes an angle 30° with x-axis. so, the area vector makes 60° with x -axis (electric field intensity ) $\theta$ = 60°

now electric flux , $\phi$ = 5 × 10³ × 0.01 × cos60°

= 50 × √3/2 = 25 Nm²/C

Answer 2

Explanation:

given, electric field intensity, E = 5 × 10³ î N/C

so, the magnitude of electric field , |E| = 5 × 10³ N/C

side of square , a = 10cm = 0.1m

area of square , |A|= a² = (0.1)² = 0.01m²

as question says that the plane of square is parallel to Y-Z plane. so, area vector is perpendicular on Y-Z plane. e.g., parallel to electric field intensity. hence, angle between area vector and electric field , \thetaθ = 0°

so, electric flux , \phiϕ = |E||A|cos0° = 5 × 10³ × 0.01 = 50 Nm²/C

again, plane makes an angle 30° with x-axis. so, the area vector makes 60° with x -axis (electric field intensity ) \thetaθ = 60°

now electric flux , \phiϕ = 5 × 10³ × 0.01 × cos60°

= 50 × √3/2 = 25 Nm²/C