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#dipole moment

38 public questions tagged with this topic.

Energy Stored in Capacitor and Energy Density

An electric dipole with moment \( p = 8 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Along the dipole axis ( θ = 180° ): V = -(1/4 π ε₀) (p/r²) . V = -9 × 10⁹ × (8 × 10⁻⁹/4²) = -9 × 10⁹ × (8 × 10⁻⁹/16) = -4.5 V . 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 > Energy Stored in Capacitor and Energy Density

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Conductors, Electrostatic Shielding and Dielectrics

An electric dipole with moment \( p = 7 \times 10^{-9} \, \text{C m} \) lies along the x-axis. What is the potential at

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (7 × 10⁻⁹/4²) = 9 × 10⁹ × (7 × 10⁻⁹/16) = 3.9375 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

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

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Conductors, Electrostatic Shielding and Dielectrics

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

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. In the equatorial plane ( θ = 90° ): V = (1/4 π ε₀) (p cos θ/r²) . Since cos 90° = 0 , V = 0 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, result 0 V follows, reflecting potential-capacitance relations.

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

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Equipotential Surfaces and Relation Between Field and Potential

An electric dipole with moment \( p = 2 \times 10^{-10} \, \text{C m} \) is at the origin, aligned along the x-axis. Wha

**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. Position vector r = (0, 2, 0) , r = 2 m , hatr = (0, 1, 0) . Dipole moment p = (2 × 10⁻¹⁰, 0, 0) . V = (1/4 π ε₀) (p · hatr/r²) = 9 × 10⁹ × ((2 × 10⁻¹⁰) · (0)/2²) = 0 V (since cos θ = 0 , equatorial plane). Using V = kQ/r, U

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

An electric dipole with moment \( p = 2 \times 10^{-9} \, \text{C m} \) lies along the z-axis. What is the potential at

**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. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (2 × 10⁻⁹/4²) = 9 × 10⁹ × (2 × 10⁻⁹/16) = 1.125 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

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

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 2 \t

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = -p E cos θ = -6 × 10⁻⁹ × 2 × 10⁵ × cos 45° . cos 45° = (1/√(2)) ≈ 0.707 , so U = -6 × 10⁻⁹ × 2 × 10⁵ × 0.707 = -8.48 × 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 > Potential Energy of System of Charges

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Potential Energy of System of Charges

An electric dipole with moment \( p = 5 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at

**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. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (5 × 10⁻⁹/3²) = 9 × 10⁹ × (5 × 10⁻⁹/9) = 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 =

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

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Potential Energy of System of Charges

A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 5 \t

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. U = -p E cos θ = -6 × 10⁻⁹ × 5 × 10⁴ × cos 45° . cos 45° = (1/√(2)) ≈ 0.707 , so U = -6 × 10⁻⁹ × 5 × 10⁴ × 0.707 = -2.121 × 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 > Potential Energy of System of Charges

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Potential Energy of System of Charges

An electric dipole with moment \( p = 6 \times 10^{-10} \, \text{C m} \) lies along the x-axis. What is the potential at

**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. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (6 × 10⁻¹⁰/5²) = 9 × 10⁹ × (6 × 10⁻¹⁰/25) = 2.16 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U =

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

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Electric Potential and Potential Difference

An electric dipole with moment \( p = 4 \times 10^{-9} \, \text{C m} \) lies along the x-axis. What is the potential at

**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. Along the dipole axis ( θ = 180° ): V = -(1/4 π ε₀) (p/r²) . V = -9 × 10⁹ × (4 × 10⁻⁹/2²) = -9 × 10⁹ × (4 × 10⁻⁹/4) = -9 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 > Electric Potential and Potential Difference

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Torque on Current Loop, Magnetic Moment and Galvanometer

A square loop of side \( 0.15 \, \text{m} \) with 50 turns carries \( 1 \, \text{A} \) in a magnetic field of \( 0.6 \,

**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Torque tau = N I A B sin θ , where A = 0.15 × 0.15 = 0.0225 m² . tau = 50 × 1 × 0.0225 × 0.6 × sin 30° = 0.675 × 0.5 =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Torque on Current Loop, Magnetic Moment and Galvanometer

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