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#stored energy

3 public questions tagged with this topic.

Energy Stored in Capacitor and Energy Density

A \( 9 \, \mu\text{F} \) capacitor is charged to \( 600 \, \text{V} \). What is the energy stored in it?

**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) × 9 × 10⁻⁶ × (600)² . U = (1/2) × 9 × 10⁻⁶ × 360000 = 1.62 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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Capacitor with Dielectric and Effect of Inserting Slab

When a dielectric slab is inserted between the plates of a charged parallel plate capacitor (disconnected from the batte

**Effect of dielectric** is to reduce effective field due to polarization, bound surface charges opposite to free charges, net field E = E₀ - E_p = E₀/K. Capacitance C = Q/V = K ε₀ A/d for full filling, partial filling C = ε₀ A/(d - t + t/K) for slab thickness t. When a dielectric slab (with K > 1 ) is inserted into a disconnected capacitor, the charge Q remains constant. The capacitance increases ( C' = K C ), and the potential difference decreases ( V' = V/K ). The energy stored is given by U = (Q²/2C) , so with increased C

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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Energy Stored in Capacitor and Energy Density

A capacitor of \( 5 \, \mu\text{F} \) is charged to \( 100 \, \text{V} \). What is the energy stored in it?

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². U = (1/2) C V² = (1/2) × 5 × 10⁻⁶ × (100)² = (1/2) × 5 × 10⁻⁶ × 10⁴ = 0.025 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 and common potential V = Q_total/C_total, result 0.025 J follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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