Practice question

**Potential due to point charge** V = (1/4π ε₀)Q/r, positive Q gives positive V, negative gives negative. At r=0.3 m from Q=3×10⁻⁸ C, V=9×10⁹×3×10⁻⁸/0.3=900 V. Work done moving charge q from infinity (V=0) to point where V=40 V is W = q V =9×10⁻⁶×40=3.6×10⁻⁴ J. Work done: W = p E (cos θ₀ - cos θ₁) = 6 × 10⁻⁹ × 3 × 10⁵ × (cos 0° - cos 90°) . W = 6 × 10⁻⁹ × 3 × 10⁵ × (1 - 0) = 1.8 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq