#charge redistribution

Energy Stored in Capacitor and Energy Density

When a conductor is placed in an external electric field, why does the potential throughout its volume become constant i

**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. In electrostatic equilibrium, the electric field inside a conductor is zero because free charges rearrange to cancel any internal field. Since the electric field is the negative gradient of potential ( E = -(dV/dr) ), if E = 0 , the potential gradient must be zero, implying the potential V is constant

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

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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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Combination of Capacitors - Series and Parallel

In a system of two identical conductors initially charged differently and then connected by a wire, why does the final e

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. Initially, the conductors have charges Q₁ and Q₂ , with energy U_i = (Q₁²/2C) + (Q₂²/2C) . Upon connection, charge redistributes to equal potentials, total charge Q₁ + Q₂ splits equally ( (Q₁ + Q₂/2) each), so final energy U_f = 2 × (((Q₁ + Q₂/2))²/2C) = ((Q₁ + Q₂)²/4C) . Typically, (Q₁ +

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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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

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 \( 5 \, \mu\text{F} \) 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 =

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 \( 3 \, \mu\text{F} \) 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 =

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 \( 6 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 18 \, \mu\text{F} \)

**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 × (100)² = 0.03 J . Charge: Q = 6 × 10⁻⁶ × 100 = 6 × 10⁻⁴ C . Total C = 6 + 18 = 24 μF , V = (6 × 10⁻⁴/24 × 10⁻⁶) = 25 V . Final energy: U_f = (1/2) × 24 × 10⁻⁶ × (25)² = 0.0075 J . Loss: U_i - U_f =

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

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Electric Potential and Potential Difference

A charged conductor is surrounded by a thin concentric hollow conducting shell. If the shell is grounded, what happens t

**Electric potential** V = k Q/r, k = 1/(4π ε₀)=9×10⁹ N·m²/C², Q charge (C), r distance (m), scalar, potential at surface of spherical conductor radius R, V = k Q/R. Potential difference ΔV = V_B - V_A = -∫ E·dl, work per unit charge to move charge without acceleration. When the outer shell is grounded (potential V = 0 ), the potential on the inner conductor adjusts due to charge redistribution. If the inner conductor has charge Q , the inner surface of the shell induces -Q , and since the shell's potential is zero, the outer surface of the shell acquires +Q . The

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference

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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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Sharing of Charges, Common Potential and Applications

A \( 3 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 3 \, \mu\text{F} \)

**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 = 3 × 10⁻⁶ × 150 = 4.5 × 10⁻⁴ C . Total capacitance: 3 + 3 = 6 μF . Final voltage: V = (Q/C) = (4.5 × 10⁻⁴/6 × 10⁻⁶) = 75 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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Sharing of Charges, Common Potential and Applications

A \( 6 \, \mu\text{F} \) capacitor charged to \( 100 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \)

**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 = 6 × 10⁻⁶ × 100 = 6 × 10⁻⁴ C . Total capacitance: 6 + 6 = 12 μF . Final voltage: V = (Q/C) = (6 × 10⁻⁴/12 × 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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Sharing of Charges, Common Potential and Applications

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

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