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Question

A dipole \( p = 8 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \(
180^\circ \) in a field \( E = 1 \times 10^5 \, \text{N/C} \). What is the work done?

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Explanation

**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,

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