Electrostatics — Practice Quiz

(Q + q)/(4πε₀R²) directed outward

Step 1: Apply Gauss's law using a spherical Gaussian surface of radius r just greater than R. Step 2: The enclosed charge = charge on shell (Q) + point charge at centre (q) = Q + q. Step 3: By Gauss's law: E × 4πr² = (Q + q)/ε₀. Step 4: As r → R: E = (Q + q)/(4πε₀R²). Step 5: The shell does NOT shield the field outside — Gauss's law considers all enclosed charges. Option (a) is wrong because σ/ε₀ ignores the interior charge. Option (c) ignores the shell charge.

(-12î + 4ĵ) NC⁻¹

Step 1: Use E⃗ = -(∂V/∂x î + ∂V/∂y ĵ). Step 2: ∂V/∂x = 12x. At x=1: ∂V/∂x = 12. Step 3: ∂V/∂y = -4 (constant, independent of y). Step 4: E⃗ = -(12î + (-4)ĵ) = -12î + 4ĵ NC⁻¹. Step 5: Option (b) is wrong because it forgets the negative sign. Option (c) is wrong in the ĵ component sign. Electric field always points in the direction of decreasing potential.

The net flux depends only on enclosed charges; E on the surface is due to all charges.

Step 1: Gauss's law states ∮E⃗·dA⃗ = Q_enclosed/ε₀. Step 2: The E in the surface integral is the TOTAL electric field, contributed by ALL charges (inside and outside the surface). Step 3: However, the NET flux (the closed surface integral) depends ONLY on the enclosed charge — contributions from outside charges cancel perfectly. Step 4: Option (a) is a very common misconception — E itself is from all sources. Step 5: Option (c) is wrong — Gauss's law is universal, but it's most USEFUL for symmetric charge configurations. Option (d) is wrong — the shape doesn't affect total flux.

V = 133.3 V, ΔU = 8.9×10⁻² J

Step 1: Initial charge Q = C₁V = 8×10⁻⁶ × 200 = 1600×10⁻⁶ C = 1.6×10⁻³ C. Step 2: Common potential V_common = Q/(C₁+C₂) = 1.6×10⁻³/(8+4)×10⁻⁶ = 1.6×10⁻³/12×10⁻⁶ = 133.3 V. Step 3: Initial energy U_i = ½C₁V² = ½×8×10⁻⁶×(200)² = 0.16 J. Step 4: Final energy U_f = ½(C₁+C₂)V_c² = ½×12×10⁻⁶×(133.3)² ≈ 0.1067 J. Step 5: Energy loss ΔU = 0.16 - 0.1067 ≈ 0.053 J... Recalculation: ΔU = C₁C₂(V₁-V₂)²/[2(C₁+C₂)] = 8×4×10⁻¹²×(200)²/(2×12×10⁻⁶) = 6400×10⁻¹²×40000/24×10⁻⁶ = 8.89×10⁻² J. Energy is lost as heat during charge redistribution.