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Question

A dipole \( p = 7 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ
\) in a field \( E = 2 \times 10^5 \, \text{N/C} \). What is the work done?

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Explanation

**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 =

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