#capacitors

Combination of Capacitors - Series and Parallel

Two capacitors of \( 15 \, \text{pF} \) each are connected in series. What is the equivalent capacitance?

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. (1/C) = (1/15) + (1/15) = (2/15) . C = (15/2) = 7.5 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 7.5 pF follows, reflecting potential-capacitance relations.

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

View Details

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

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

View Details

Combination of Capacitors - Series and Parallel

Four capacitors of \( 5 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. (1/C) = (1/5) + (1/5) + (1/5) + (1/5) = (4/5) . C = (5/4) = 1.25 μF . 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 1.25 µF follows, reflecting potential-capacitance relations.

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

View Details

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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with capacitance \( 80 \, \text{pF} \) has a dielectric (\( K = 5 \), thickness \( d/5 \)) in

**Spherical conductor capacitance** C = 4π ε₀ R, R radius (m), potential V = Q/C = Q/(4π ε₀ R)=k Q/R. For R=2 cm=0.02 m, Q=2×10⁻⁸ C, V=9×10⁹×2×10⁻⁸/0.02=9000 V, showing high voltage for small sphere with modest charge. Potential difference: V = E₀ ( (4d/5) ) + (E₀/K) ( (d/5) ) = E₀ d ( (4/5) + (1/5 × 5) ) . V = E₀ d ( (4/5) + (1/25) ) = E₀ d × (21/25) . C = (Q/V) = (Q/(21/25) V₀) = (25/21) × 80 ≈ 95.24 pF . 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 > Capacitance of Parallel Plate and Spherical Capacitor

View Details

Combination of Capacitors - Series and Parallel

Four capacitors of \( 15 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. (1/C) = (1/15) + (1/15) + (1/15) + (1/15) = (4/15) . C = (15/4) = 3.75 μF . 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 3.75 µF follows, reflecting potential-capacitance relations.

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

View Details

Combination of Capacitors - Series and Parallel

Why does the total energy stored in a system of capacitors connected in series decrease if one of the capacitors is remo

**Charge conservation in series** explains same Q: when battery charges first plate, it induces -Q on second plate of same capacitor, which comes from next capacitor's plate, etc., so all have same magnitude Q, potential divides as per 1/C. In series, the equivalent capacitance Cₑq decreases when a capacitor is removed ( (1/Cₑq) = sum (1/C_i) ). With a battery maintaining constant voltage V , the energy stored is U = (1/2) Cₑq V² . A smaller Cₑq reduces U , as energy is directly proportional to capacitance under constant V . Additionally, removing a capacitor redistributes charges, potentially dissipating energy during the process, further

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

View Details

Combination of Capacitors - Series and Parallel

In a series combination of capacitors with different dielectric materials between their plates, why do capacitors with h

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. In series, the charge Q on each capacitor is the same. Capacitance is C = (K ε₀ A/d) , where K is the dielectric constant. A higher K increases C . Since V = (Q/C) , a larger C (due to higher K ) results in a smaller V . Thus, capacitors with higher dielectric constants

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

View Details

Energy Stored in Capacitor and Energy Density

In a system of two capacitors connected in series across a battery, why does the capacitor with smaller capacitance stor

**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 series, the charge Q on each capacitor is the same. Energy stored in a capacitor is U = (Q²/2C) . For a smaller capacitance C , the denominator 2C is smaller, so U is larger compared to a capacitor with larger C . Alternatively, since V = (Q/C) , the smaller

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

View Details

Combination of Capacitors - Series and Parallel

Two capacitors of \( 60 \, \text{pF} \) and \( 120 \, \text{pF} \) are connected in series. What is the equivalent capac

**Parallel combination** has same voltage V across each, charges Q_i = C_i V, total Q = Σ Q_i = V Σ C_i, so C_eq = Σ C_i, sum of capacitances. For five 10 μF in series, 1/C=5/10=0.5, C_eq=2 μF, much smaller than individual, while parallel would be 50 μF. (1/C) = (1/60) + (1/120) = (2 + 1/120) = (3/120) . C = (120/3) = 40 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 40 pF follows, reflecting potential-capacitance relations.

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

View Details

Sharing of Charges, Common Potential and Applications

Three capacitors \( 10 \, \text{pF} \), \( 20 \, \text{pF} \), and \( 40 \, \text{pF} \) are in parallel. What is the to

**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 = 10 + 20 + 40 = 70 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 70 pF follows, reflecting potential-capacitance relations.

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

View Details

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

View Details