Practice question

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Along the dipole axis ( θ = 180° ): V = -(1/4 π ε₀) (p/r²) . V = -9 × 10⁹ × (8 × 10⁻⁹/4²) = -9 × 10⁹ × (8 × 10⁻⁹/16) = -4.5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq