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#tilted surface

2 public questions tagged with this topic.

A uniform field \( E = 6 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a rectangle of 40 cm

**Measure of field penetration** depends on both magnitude and projected area. Understanding angle between E and normal vector is crucial, flux zero when field parallel to surface, maximum when perpendicular. Area: A = 0.4 × 0.25 = 0.1 m² . Flux: Φ = E A cos 45° = 6 × 10³ × 0.1 × (√(2)/2) = 424.26 N·m²/C . Substituting values gives 424 N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Flux

A uniform field \( E = 8 \times 10^3 \, \text{N/C} \) is along the y-axis. What is the flux through a circle of radius 1

**Measure of field penetration** depends on both magnitude and projected area. Understanding angle between E and normal vector is crucial, flux zero when field parallel to surface, maximum when perpendicular. Area: A = π (0.15)² = 0.0707 m² . Flux: Φ = E A cos 30° = 8 × 10³ × 0.0707 × (√(3)/2) = 489.7 N·m²/C . Substituting values gives 490 N·m²/C, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Flux