A capacitor that stores energy of 0.5 J, and has capacitance of 1 μF, has potential difference of ______ across it.
Energy U=½CV^2 => V=√(2U/C)=√(1e6)=1000 V.
Ref: NCERT Physics Class 12 Chapter 1 Electric Charges and Fields
#capacitor energy
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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 =
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > 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 =
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications
A \( 7 \, \mu\text{F} \) capacitor is charged to \( 400 \, \text{V} \). 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) × 7 × 10⁻⁶ × (400)² . U = (1/2) × 7 × 10⁻⁶ × 160000 = 0.56 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 0.56 J follows, reflecting potential-capacitance relations.
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > 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
A \( 10 \, \mu\text{F} \) capacitor charged to \( 60 \, \text{V} \) is connected to an uncharged \( 30 \, \mu\text{F} \)
**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) × 10 × 10⁻⁶ × (60)² = 0.018 J . Charge: Q = 10 × 10⁻⁶ × 60 = 6 × 10⁻⁴ C . Total C = 10 + 30 = 40 μF , V = (6 × 10⁻⁴/40 × 10⁻⁶) = 15 V . Final energy: U_f =
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications
A \( 4 \, \mu\text{F} \) capacitor is charged to \( 250 \, \text{V} \). What is the energy stored in it?
**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. U = (1/2) C V² = (1/2) × 4 × 10⁻⁶ × (250)² . U = (1/2) × 4 × 10⁻⁶ × 62500 = 0.125 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
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density
