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#dipole rotation

5 public questions tagged with this topic.

Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 9 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. Work done: W = p E (cos θ₀ - cos θ₁) = 9 × 10⁻⁹ × 5 × 10⁴ × (cos 90° - cos 180°) . W = 9 × 10⁻⁹ × 5 × 10⁴ × (0 - (-1)) = 9 × 10⁻⁹ × 5 × 10⁴ × 1 = 4.5 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r,

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

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Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 5 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 5 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 180°) . W = 5 × 10⁻⁹ × 2 × 10⁵ × (0 - (-1)) = 5 × 10⁻⁹ × 2 × 10⁵ × 1 = 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

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

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Potential Energy of System of Charges

A dipole \( p = 8 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 8 × 10⁻⁹ × 1 × 10⁵ × (cos 0° - cos 180°) . W = 8 × 10⁻⁹ × 1 × 10⁵ × (1 - (-1)) = 8 × 10⁻⁹ × 1 × 10⁵ × 2 = 1.6 × 10⁻³ J . Using V = kQ/r,

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

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PHYSICS

A dipole p = 8 × 10⁻¹⁰C m is rotated from θ = 0° to 90° in a field E = 10⁶N/C . What is the work done?

Given: A dipole p = 8 × 10⁻¹⁰C m is rotated from θ = 0° to 90° in a field E = 10⁶N/C . What is the work done? These values define the system as per NCERT data. Formula: Work done: W = p E (cos θ_0 - cos θ_1) = 8 × 10⁻¹⁰ × 10⁶ × (cos 0° - cos 90°) = 8 × 10⁻⁴ × (1 - 0) = 8 × 10⁻⁴J .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

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