#capacitor charge sharing

Sharing of Charges, Common Potential and Applications

A \( 2 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) loses how much energy when connected to an uncharged

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial energy: U_i = (1/2) × 2 × 10⁻⁶ × (200)² = 0.04 J . Charge: Q = 2 × 10⁻⁶ × 200 = 4 × 10⁻⁴ C . Total C = 2 + 3 = 5 μF , V = (4 × 10⁻⁴/5 × 10⁻⁶) = 80 V . Final energy: U_f =

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Sharing of Charges, Common Potential and Applications

A \( 8 \, \mu\text{F} \) capacitor charged to \( 50 \, \text{V} \) is connected to an uncharged \( 8 \, \mu\text{F} \) c

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial charge: Q = 8 × 10⁻⁶ × 50 = 4 × 10⁻⁴ C . Total capacitance: 8 + 8 = 16 μF . Final voltage: V = (Q/C) = (4 × 10⁻⁴/16 × 10⁻⁶) = 25 V . 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,

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Sharing of Charges, Common Potential and Applications

A \( 8 \, \mu\text{F} \) capacitor charged to \( 80 \, \text{V} \) is connected to an uncharged \( 24 \, \mu\text{F} \)

**Sharing of charges** when charged capacitor C₁ at V₁ connected to uncharged C₂, total charge Q = C₁ V₁ conserved, common potential V_common = Q/(C₁+C₂) = C₁ V₁/(C₁+C₂), final charges Q₁' = C₁ V_common, Q₂' = C₂ V_common. For 4 μF at 100 V (Q=4×10⁻⁴ C) connected to 4 μF uncharged, V_common=4×10⁻⁴/8×10⁻⁶=50 V. Initial energy: U_i = (1/2) × 8 × 10⁻⁶ × (80)² = 0.0256 J . Charge: Q = 8 × 10⁻⁶ × 80 = 6.4 × 10⁻⁴ C . Total C = 8 + 24 = 32 μF , V = (6.4 × 10⁻⁴/32 × 10⁻⁶) = 20 V . Final energy:

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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Sharing of Charges, Common Potential and Applications

A \( 4 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 12 \, \mu\text{F} \)

**Common potential** after connection is weighted average of initial potentials by capacitances. Energy loss ΔU = ½ C₁ C₂ (V₁-V₂)²/(C₁+C₂) dissipated as heat and spark, always positive unless V₁=V₂, explaining why energy reduces after sharing. Initial energy: U_i = (1/2) × 4 × 10⁻⁶ × (150)² = 0.045 J . Charge: Q = 4 × 10⁻⁶ × 150 = 6 × 10⁻⁴ C . Total C = 4 + 12 = 16 μF , V = (6 × 10⁻⁴/16 × 10⁻⁶) = 37.5 V . Final energy: U_f = (1/2) × 16 × 10⁻⁶ × (37.5)² = 0.01125 J . Loss: U_i - U_f =

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Sharing of Charges, Common Potential and Applications

A \( 4 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 4 \, \mu\text{F} \)

**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial charge: Q = 4 × 10⁻⁶ × 100 = 4 × 10⁻⁴ C . Total capacitance: 4 + 4 = 8 μF . Final voltage: V = (Q/C) = (4 × 10⁻⁴/8 × 10⁻⁶) = 50 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications

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