Electric Charges and Fields — Practice Quiz

1.13 × 10⁵ N m² C⁻¹

Step 1: Apply Gauss's law: φ = q / ε₀ Step 2: q = 1 µC = 1 × 10⁻⁶ C Step 3: φ = (1 × 10⁻⁶) / (8.85 × 10⁻¹²) Step 4: φ = 1.13 × 10⁵ N m² C⁻¹ Key point: The flux depends only on the enclosed charge, NOT on the radius of the Gaussian surface. Option B incorrectly uses the radius. The flux would be the same for any radius sphere enclosing this charge.

360 N/C

Step 1: Formula for field due to infinite line charge: E = λ / (2πε₀r) Step 2: λ = 2 × 10⁻⁸ C/m, r = 1 m Step 3: 2πε₀ = 2 × 3.14 × 8.85 × 10⁻¹² = 5.56 × 10⁻¹¹ Step 4: E = (2 × 10⁻⁸) / (5.56 × 10⁻¹¹) ≈ 360 N/C Alternatively: E = 2kλ/r = 2 × 9×10⁹ × 2×10⁻⁸ / 1 = 360 N/C. Option B (180 N/C) misses the factor of 2 in the formula; Option C doubles incorrectly.

Two electric field lines can cross each other at a point of zero field.

Step 1: Electric field lines NEVER cross each other, at any point including where E = 0. Step 2: If two field lines crossed, there would be two different directions of electric field at that point, which is physically impossible — the electric field at any point has a unique direction. Step 3: The other three statements are correct properties of electric field lines. This is a conceptual trick question — students often think crossing is allowed at special points, but it is NEVER allowed.

1 × 10⁻³ N m

Step 1: Torque on a dipole: τ = pE sinθ Step 2: p = 5 × 10⁻⁸ C m, E = 4 × 10⁴ N/C, θ = 30° Step 3: sin 30° = 0.5 Step 4: τ = 5 × 10⁻⁸ × 4 × 10⁴ × 0.5 = 5 × 10⁻⁸ × 2 × 10⁴ = 10⁻³ N m Option B uses sin 60° instead of sin 30°; Option C uses √3/2; Option D makes a factor error. The torque aligns the dipole with the field (tends to make θ → 0).