Practice question
Question
In a system of two identical capacitors in series, why does removing one capacitor while maintaining
the same total voltage across the system increase the energy stored?
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. Two identical capacitors in series ( Cₑq = (C/2) ) with total voltage V each have voltage V/2 , storing energy Utₒtₐl = (1/2) (C/2) V² = (1/4) C V² . Removing one, the remaining capacitor has capacitance C and voltage V , storing U = (1/2) C V² , which is
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