Sharing of Charges, Common Potential and Applications

Sharing of Charges, Common Potential and Applications

A \( 2 \, \mu\text{F} \) capacitor charged to \( 400 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \)

**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) × 2 × 10⁻⁶ × (400)² = 0.16 J . Charge: Q = 2 × 10⁻⁶ × 400 = 8 × 10⁻⁴ C . Total C = 2 + 6 = 8 μF , V = (8 × 10⁻⁴/8 × 10⁻⁶) = 100 V . Final energy: U_f = (1/2) × 8 × 10⁻⁶ × (100)² = 0.04 J . Loss: U_i - U_f =

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Sharing of Charges, Common Potential and Applications

Three capacitors \( 2 \, \text{pF} \), \( 4 \, \text{pF} \), and \( 8 \, \text{pF} \) are in parallel. What is the total

**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. C = 2 + 4 + 8 = 14 pF . 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 14 pF follows, reflecting potential-capacitance relations.

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Sharing of Charges, Common Potential and Applications

When two identical capacitors, one charged and one uncharged, are connected in parallel, why does the total energy decre

**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. When a charged capacitor ( C , charge Q , voltage V ) is connected in parallel with an uncharged capacitor ( C ), the total capacitance becomes 2C , and the charge redistributes to a final voltage V' = Q/(2C) = V/2 . Initial energy is U_i = (Q²/2C) , while final energy

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Sharing of Charges, Common Potential and Applications

A \( 6 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 9 \, \mu\text{F} \)

**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) × 6 × 10⁻⁶ × (150)² = 0.0675 J . Charge: Q = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴ C . Total C = 6 + 9 = 15 μF , V = (9 × 10⁻⁴/15 × 10⁻⁶) = 60 V . Final energy: U_f = (1/2) × 15 × 10⁻⁶ × (60)² = 0.027 J . Loss: U_i - U_f =

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Sharing of Charges, Common Potential and Applications

Three capacitors \( 5 \, \text{pF} \), \( 10 \, \text{pF} \), and \( 20 \, \text{pF} \) are in parallel. What is the tot

**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. C = 5 + 10 + 20 = 35 pF . 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 35 pF follows, reflecting potential-capacitance relations.

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Sharing of Charges, Common Potential and Applications

Three capacitors \( 8 \, \text{pF} \), \( 16 \, \text{pF} \), and \( 32 \, \text{pF} \) are in parallel. What is the tot

**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. C = 8 + 16 + 32 = 56 pF . 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 56 pF follows, reflecting potential-capacitance relations.

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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 \( 5 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) is connected to an uncharged \( 5 \, \mu\text{F} \)

**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 = 5 × 10⁻⁶ × 200 = 10⁻³ C . Total capacitance: 5 + 5 = 10 μF . Final voltage: V = (Q/C) = (10⁻³/10 × 10⁻⁶) = 100 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

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Sharing of Charges, Common Potential and Applications

A \( 14 \, \mu\text{F} \) capacitor charged to \( 20 \, \text{V} \) is connected to an uncharged \( 14 \, \mu\text{F} \)

**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 charge: Q = 14 × 10⁻⁶ × 20 = 2.8 × 10⁻⁴ C . Total capacitance: 14 + 14 = 28 μF . Final voltage: V = (Q/C) = (2.8 × 10⁻⁴/28 × 10⁻⁶) = 10 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

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Sharing of Charges, Common Potential and Applications

A \( 4 \, \mu\text{F} \) capacitor is charged to \( 50 \, \text{V} \) and then connected to an uncharged \( 2 \, \mu\tex

**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 = 4 × 10⁻⁶ × 50 = 2 × 10⁻⁴ C . Total capacitance: 4 + 2 = 6 μF . Final voltage: V = (Q/C) = (2 × 10⁻⁴/6 × 10⁻⁶) = 33.33 V . Final energy: U = (1/2) C V² = (1/2) × 6 × 10⁻⁶ × (33.33)² = 3.33 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E =

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Sharing of Charges, Common Potential and Applications

A \( 5 \, \mu\text{F} \) capacitor charged to \( 120 \, \text{V} \) is connected to an uncharged \( 15 \, \mu\text{F} \)

**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) × 5 × 10⁻⁶ × (120)² = 0.036 J . Charge: Q = 5 × 10⁻⁶ × 120 = 6 × 10⁻⁴ C . Total C = 5 + 15 = 20 μF , V = (6 × 10⁻⁴/20 × 10⁻⁶) = 30 V . Final energy: U_f = (1/2) × 20 × 10⁻⁶ × (30)² = 0.009 J . Loss: U_i - U_f =

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Sharing of Charges, Common Potential and Applications

A \( 3 \, \mu\text{F} \) capacitor charged to \( 200 \, \text{V} \) is connected to an uncharged \( 9 \, \mu\text{F} \)

**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 energy: U_i = (1/2) × 3 × 10⁻⁶ × (200)² = 0.06 J . Charge: Q = 3 × 10⁻⁶ × 200 = 6 × 10⁻⁴ C . Total C = 3 + 9 = 12 μF , V = (6 × 10⁻⁴/12 × 10⁻⁶) = 50 V . Final energy: U_f =

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