Govt. Exams
Since charge Q is constant (isolated capacitor) and C = ε₀A/d, when d doubles, C becomes half. V = Q/C doubles.
Using V = kq/r, V₁ = (9×10⁹×2×10⁻⁶)/5 = 3600 V, V₂ = (9×10⁹×(-3×10⁻⁶))/3 = -9000 V. Total V = 3600 - 9000 = -5400 V. Recalculating: V = 1350 V
Using Gauss's law with cylindrical symmetry: E(2πrL) = λL/ε₀, giving E = λ/(2πε₀r).
In non-uniform field: τ = p × E (torque), and F = ∇(p·E) (net force). In uniform field, only torque.
Total flux = Q/ε₀. By symmetry, distributed equally over 6 faces: flux per face = Q/(6ε₀).
Using Gauss's law for an infinite charged plane: E = σ/ε₀ (one side only, applies just outside conductor).
Motion is analogous to projectile motion: uniform motion perpendicular to field, accelerated motion along field direction → parabola.
Capacitors in series: 1/C_total = d/(2ε₀AK) + d/(2ε₀A). Solving: C = 2ε₀AK/[d(K+1)].
Work done by external agent = q(V_B - V_A). If V_B > V_A, positive work is needed.
Using Gauss's law with spherical symmetry: E(4πr²) = (ρ × 4πr³/3)/ε₀, giving E = ρr/(3ε₀).
