Combination of Capacitors - Series and Parallel

Combination of Capacitors - Series and Parallel

In an electrostatic field, if a positive charge is moved along an equipotential surface, what can be said about the work

**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. An equipotential surface has a constant potential at all points. The work done by the electric field when a charge moves along such a surface is zero because the potential difference between any two points on the surface is zero ( W = q Δ V , and Δ V = 0 ). Additionally,

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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.

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Combination of Capacitors - Series and Parallel

A spherical conductor of radius 20 cm has a charge of \( 8 \times 10^{-8} \, \text{C} \). What is the electric field at

**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. For r = 0.5 m > R = 0.2 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (8 × 10⁻⁸/(0.5)²) = 9 × 10⁹ × (8 × 10⁻⁸/0.25) = 2.88 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C =

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Combination of Capacitors - Series and Parallel

Why does the potential due to an electric dipole fall off as \( 1/r^2 \) at large distances, unlike the potential of a p

**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. The potential due to a point charge falls as 1/r because it behaves as a single source of charge. An electric dipole consists of two equal and opposite charges separated by a small distance, so their potentials partially cancel out at large distances. The net potential depends on the dipole moment and the angle

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Combination of Capacitors - Series and Parallel

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (5 × 10⁻⁸/0.05) = 9 × 10⁹ × 10⁻⁶ = 9000 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

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Combination of Capacitors - Series and Parallel

A dipole with \( p = 5 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 3 \t

**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. U = -p E cos θ = -5 × 10⁻⁹ × 3 × 10⁴ × cos 30° . cos 30° = (√(3)/2) ≈ 0.866 , so U = -5 × 10⁻⁹ × 3 × 10⁴ × 0.866 = -1.3 × 10⁻⁴ 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

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Combination of Capacitors - Series and Parallel

An electric dipole with moment \( p = 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at \( (0, 3,

**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. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (10⁻⁹/3²) = 9 × 10⁹ × (10⁻⁹/9) = 1 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

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Combination of Capacitors - Series and Parallel

A dipole \( p = 8 \times 10^{-10} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻¹⁰ × 10⁶ × (cos 0° - cos 90°) = 8 × 10⁻⁴ × (1 - 0) = 8 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq

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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.

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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 +

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Combination of Capacitors - Series and Parallel

A dipole \( p = 2 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \(

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 2 × 10⁻⁹ × 5 × 10⁵ × (cos 90° - cos 0°) = 2 × 10⁻⁹ × 5 × 10⁵ × (0 - 1) = -10⁻³ 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

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Combination of Capacitors - Series and Parallel

Five capacitors of \( 20 \, \mu\text{F} \) 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) = (1/20) + (1/20) + (1/20) + (1/20) + (1/20) = (5/20) . C = (20/5) = 4 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 4 µF follows, reflecting potential-capacitance relations.

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