Practice question
Question
A \( 6 \, \mu\text{F} \) capacitor is charged to \( 300 \, \text{V} \). What is the energy stored in
it?
Explanation
**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. U = (1/2) C V² = (1/2) × 6 × 10⁻⁶ × (300)² . U = (1/2) × 6 × 10⁻⁶ × 90000 = 0.27 J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V²
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.