Electric Charge and Electric Field — Practice Quiz

0.05 N·m

Step 1: Formula for torque on a dipole: τ = pE sin θ. Step 2: Substitute values: τ = (2 × 10⁻⁶) × (5 × 10⁴) × sin 30°. Step 3: sin 30° = 0.5. So τ = (2 × 10⁻⁶) × (5 × 10⁴) × 0.5. Step 4: τ = 10⁻¹ × 0.5 = 0.05 N·m. Wrong options: 0.10 N·m results from forgetting sin 30° (using sin 90°); 0.025 N·m uses sin 15° or halves incorrectly; 1.0 N·m ignores the power-of-10 conversion.

σᵢ / (2πε₀r)

Step 1: Choose a cylindrical Gaussian surface of radius r and length l coaxial with the wire. Step 2: The charge enclosed = σᵢl. Flux through flat ends is zero (E ⊥ area vector at ends). Step 3: Flux through curved surface = E × 2πrl. Step 4: Gauss's Law: E × 2πrl = σᵢl / ε₀, so E = σᵢ / (2πε₀r). Wrong options: σᵢ/(4πε₀r²) is Coulomb's law for a point charge; σᵢ/(ε₀r) misses the 2π factor; σᵢ/(2ε₀) is the result for a plane sheet of charge.

Zero

Step 1: Apply Gauss's Law: draw a spherical Gaussian surface of radius r < R inside the shell. Step 2: The charge enclosed by this Gaussian surface is zero (all charge Q is on the shell at radius R, outside the Gaussian surface). Step 3: By Gauss's Law: E × 4πr² = Q_enclosed / ε₀ = 0. Step 4: Therefore E = 0 everywhere inside a uniformly charged spherical shell. Wrong options: kQ/R² is the field on the surface of the shell; kQ/r² is the external field; kQr/R³ applies to a solid uniformly charged sphere's internal field.

σ / (2ε₀), independent of distance

Step 1: Use a cylindrical Gaussian surface (pill-box) with its axis perpendicular to the sheet. Step 2: Let the cross-sectional area of the cylinder be Δs. Charge enclosed = σΔs. Step 3: Flux through both flat caps = 2EΔs; curved surface contributes zero flux. Step 4: Gauss's Law: 2EΔs = σΔs/ε₀ → E = σ/(2ε₀). Key point: The field is uniform (constant) – it does NOT depend on the distance from the sheet. Wrong options: The 1/r² and 1/r dependencies apply to point charges and line charges, respectively; σ/(4ε₀) has an incorrect numerical factor.