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#capacitance calculation

18 public questions tagged with this topic.

Combination of Capacitors - Series and Parallel

Three capacitors \( 6 \, \text{pF} \), \( 12 \, \text{pF} \), and \( 4 \, \text{pF} \) are in series. What is the equiva

**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) = (1/6) + (1/12) + (1/4) = (2/12) + (1/12) + (3/12) = (6/12) = 0.5 . C = 2 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 2 pF follows, reflecting potential-capacitance relations.

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

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Conductors, Electrostatic Shielding and Dielectrics

A parallel plate capacitor has plates of area \( 0.07 \, \text{m}^2 \) and separation 0.35 mm in air. What is its capaci

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.07/0.35 × 10⁻³) = 1.77 × 10⁻⁹ F = 1770 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 1770 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Capacitor with Dielectric and Effect of Inserting Slab

A parallel plate capacitor with \( C = 20 \, \text{pF} \) in air has a dielectric (\( K = 3 \)) inserted fully between p

**Dielectric slab inserted** into capacitor with constant charge Q increases capacitance C' = K C₀, K dielectric constant, so potential difference V' = Q/C' = V₀/K decreases K times. With constant voltage V maintained by battery, capacitance increase causes charge Q' = K Q₀ to increase, extra charge supplied by battery, energy changes. C' = K C = 3 × 20 = 60 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 60 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitor with Dielectric and Effect of Inserting Slab

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Equipotential Surfaces and Relation Between Field and Potential

A parallel plate capacitor has plates of area \( 0.06 \, \text{m}^2 \) and separation 0.3 mm in air. What is its capacit

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.06/0.3 × 10⁻³) = 1.77 × 10⁻⁹ F = 1770 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 1770 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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Combination of Capacitors - Series and Parallel

Two capacitors of \( 40 \, \text{pF} \) and \( 80 \, \text{pF} \) are connected in series. What is the equivalent capaci

**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) = (1/40) + (1/80) = (2 + 1/80) = (3/80) . C = (80/3) ≈ 26.67 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 26.67 pF follows, reflecting potential-capacitance relations.

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

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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

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Sharing of Charges, Common Potential and Applications

Three capacitors \( 4 \, \text{pF} \), \( 8 \, \text{pF} \), and \( 16 \, \text{pF} \) are in parallel. What is the tota

**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. C = 4 + 8 + 16 = 28 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 28 pF follows, reflecting potential-capacitance relations.

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

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Combination of Capacitors - Series and Parallel

Two capacitors of \( 20 \, \text{pF} \) and \( 30 \, \text{pF} \) are connected in series. What is the equivalent capaci

**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. (1/C) = (1/20) + (1/30) = (3 + 2/60) = (5/60) . C = (60/5) = 12 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 12 pF follows, reflecting potential-capacitance relations.

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

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Capacitance of Parallel Plate and Spherical Capacitor

A parallel plate capacitor with \( C = 50 \, \text{pF} \) in air has a dielectric (\( K = 4 \)) inserted fully between p

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. C' = K C = 4 × 50 = 200 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 200 pF follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

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Combination of Capacitors - Series and Parallel

Four capacitors of \( 25 \, \mu\text{F} \) 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) = (1/25) + (1/25) + (1/25) + (1/25) = (4/25) . C = (25/4) = 6.25 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 6.25 µF follows, reflecting potential-capacitance relations.

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

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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

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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

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