Electric Charges and Fields — Exam Practice

Answer The fields from both positive charges point away from their respective charge, so at the midpoint they point in opposite directions along the line joining them and are equal in magnitude — they cancel exactly, giving a net field of zero at the midpoint.

Answer Gauss's Law: the net electric flux through any closed surface equals q_enclosed/ε₀.

Derivation: Choose a cylindrical Gaussian surface of radius r and length L, coaxial with the wire. By symmetry, E is radial and constant in magnitude over the curved surface, with zero flux through the flat end caps (E is parallel to them).
Flux through curved surface: Φ = E · (2πrL).
Charge enclosed: q = λL (λ = linear charge density).
By Gauss's Law: E(2πrL) = λL/ε₀ → E = λ / (2πε₀r).

Answer (a) For a dipole (charges +q, −q separated by 2a) with a point P on the equatorial line at distance r from the centre, the fields due to +q and −q have equal magnitude kq/(r²+a²) each, and their components along the dipole axis cancel while components perpendicular add. Working through the geometry and taking r ≫ a gives E_eq = kp/r³, directed opposite to the dipole moment p = q(2a).

(b) E = kp/r³ = (9×10⁹ × 4×10⁻⁹) / (0.2)³ = 36 / 0.008 = 4500 N/C.

Solution Original: F = k(2)(6)/r² = 12k/r². New charges: 0μC and 4μC — since one charge becomes zero, the new force is zero, not attractive. (This flags a classic trap: recompute rather than assume — with these specific values the new interaction force is actually zero, so double-check the option set against your own working before selecting an answer in a real exam.)

Solution Electric flux Φ = E·A, so its unit is (N/C) × m² = N·m²/C.

Solution Total flux through the closed cube (Gauss's Law) = Q/ε₀. By symmetry, this splits equally across all 6 faces, giving Q/(6ε₀) per face.

Solution Field line direction gives field direction at any point; line density (how closely packed they are) indicates relative field strength — denser regions mean a stronger field.

Solution At equilibrium, each sphere is in equilibrium under gravity (mg), tension (T), and Coulomb repulsion (F_e). Separation between spheres: r = 2 × L sin30° = 2 × 0.5 × 0.5 = 0.5 m.

From force balance: tan30° = F_e/mg → F_e = mg tan30° = (0.01 × 10) × (1/√3) ≈ 0.0577 N.

Using F_e = kq²/r²: 0.0577 = (9×10⁹ × q²) / (0.5)² → q² = (0.0577 × 0.25)/(9×10⁹) ≈ 1.60×10⁻¹² → q ≈ 1.27×10⁻⁶ C ≈ 1.27 μC.

Solution For a conductor, all excess charge resides on the outer surface (electrostatics of conductors), so a Gaussian surface inside the sphere (r < R) encloses zero charge — the field is exactly zero everywhere inside. Note this differs from a uniformly charged non-conducting sphere, where the interior field is nonzero and proportional to r.

Solution τ = pE sinθ = 2×10⁻⁸ × 5×10⁴ × sin30° = 2×10⁻⁸ × 5×10⁴ × 0.5 = 5×10⁻⁴ N·m.