Practice question
Question
A dipole \( p = 9 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \(
180^\circ \) in a field \( E = 5 \times 10^4 \, \text{N/C} \). What is the work done?
Explanation
**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,
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.