#dipole

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

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

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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

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

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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

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

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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

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

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Combination of Capacitors - Series and Parallel

A dipole with \( p = 4 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 5 \t

**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. U = -p E cos θ = -4 × 10⁻⁹ × 5 × 10⁴ × cos 30° . cos 30° = (√(3)/2) ≈ 0.866 , so U = -4 × 10⁻⁹ × 5 × 10⁴ × 0.866 = -1.732 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

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

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Equipotential Surfaces and Relation Between Field and Potential

A dipole \( p = 7 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \(

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. Work done: W = p E (cos θ₀ - cos θ₁) = 7 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 0°) . W = 7 × 10⁻⁹ × 2 × 10⁵ × (0 - 1) = -1.4 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq =

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

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Equipotential Surfaces and Relation Between Field and Potential

A dipole with \( p = 8 \times 10^{-9} \, \text{C m} \) makes an angle of \( 60^\circ \) with a uniform field \( E = 1 \t

**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. U = -p E cos θ = -8 × 10⁻⁹ × 1 × 10⁵ × cos 60° . cos 60° = 0.5 , so U = -8 × 10⁻⁹ × 1 × 10⁵ × 0.5 = -4 × 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 C

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

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Equipotential Surfaces and Relation Between Field and Potential

A dipole with \( p = 3 \times 10^{-9} \, \text{C m} \) makes an angle of \( 60^\circ \) with a uniform field \( E = 2 \t

**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. U = -p E cos θ = -3 × 10⁻⁹ × 2 × 10⁵ × cos 60° . cos 60° = 0.5 , so U = -3 × 10⁻⁹ × 2 × 10⁵ × 0.5 = -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 C

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

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Equipotential Surfaces and Relation Between Field and Potential

A dipole \( p = 4 \times 10^{-10} \, \text{C m} \) is in a uniform field \( E = 10^5 \, \text{N/C} \) at \( 60^\circ \).

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. U = -p · E = -p E cos θ = -4 × 10⁻¹⁰ × 10⁵ × cos 60° = -4 × 10⁻⁵ × 0.5 = -2 × 10⁻⁵ 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

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

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Potential Energy of System of Charges

Why does the potential energy of a dipole in a uniform field increase when rotated from alignment (\( \theta = 0^\circ \

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. The potential energy of a dipole is U = -p E cos θ . At θ = 0° , cos 0 = 1 , U = -p E , the minimum. At θ = 90° , cos 90 = 0 , U = 0 . From θ = 0° to 90° , U increases from -p E to 0

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

A dipole with \( m = 0.5 \, \text{A m}^2 \) in \( B = 0.1 \, \text{T} \) at \( 60^\circ \) has torque:

**Soft ferromagnetic materials** have low coercivity and retentivity, narrow hysteresis loop, lose magnetism when external field removed, ideal for electromagnets and transformer cores. Energy loss per cycle proportional to loop area, explaining why soft materials minimize loss. tau = m B sinθ . Given: m = 0.5 A m² , B = 0.1 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . tau = 0.5 × 0.1 × 0.866 = 0.0433 N m ≈ 0.043 N m . Substituting values gives 0.043 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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Hysteresis, Retentivity, Coercivity and Permanent Magnets

The magnetic potential energy of a dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \(

**Permanent magnet requirement** is high retentivity to maintain field and high coercivity to resist demagnetization. Ability to retain magnetism after field removal is property of hard ferromagnets, related to domain wall pinning and anisotropy. U_m = -m B cosθ . Given: m = 0.7 A m² , B = 0.2 T , θ = 0° , cos 0° = 1 . Substitute: U_m = -0.7 × 0.2 × 1 = -0.14 J . Substituting values gives -0.14 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Hysteresis, Retentivity, Coercivity and Permanent Magnets

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