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#identical capacitors

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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 =

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 \( 8 \, \mu\text{F} \) 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,

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 \( 16 \, \mu\text{F} \) capacitor charged to \( 10 \, \text{V} \) is connected to an uncharged \( 16 \, \mu\text{F} \)

**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 charge: Q = 16 × 10⁻⁶ × 10 = 1.6 × 10⁻⁴ C . Total capacitance: 16 + 16 = 32 μF . Final voltage: V = (Q/C) = (1.6 × 10⁻⁴/32 × 10⁻⁶) = 5 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,

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

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