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Class 12 › Physics › Electric Charge and Field
CBSE2010Class Class 12 · Physics2 Marks · Short✅ Verified
Calculate the amount of work done in turning an electric dipole of dipole moment 3 × 10⁻⁸ C·m from its position of unstable equilibrium to the position of stable equilibrium in a uniform electric field of intensity 10³ N/C.
✅ Answer & Solution
<b>Given:</b><br>• Dipole moment p = 3 × 10⁻⁸ C·m<br>• Electric field E = 10³ N/C<br><br><b>Concept:</b><br>• <b>Stable equilibrium:</b> p parallel to E (θ = 0°), minimum potential energy<br>• <b>Unstable equilibrium:</b> p antiparallel to E (θ = 180°), maximum potential energy<br><br><b>Formula:</b><br>W = pE(cosθ<sub>1</sub> - cosθ<sub>2</sub>)<br><br><b>Given direction of rotation:</b><br>From unstable (θ<sub>1</sub> = 180°) → to stable (θ<sub>2</sub> = 0°)<br><br><b>Calculation:</b><br>W = pE(cos 180° - cos 0°)<br>W = pE(-1 - 1)<br>W = -2pE<br><br><b>Substitute values:</b><br>W = -2 × (3 × 10⁻⁸) × (10³)<br>W = -6 × 10⁻⁵ J<br><br><b>Answer:</b><br>Work done <b>by the field</b> = -6 × 10⁻⁵ J<br>Magnitude of work = <b>6 × 10⁻⁵ J = 60 µJ</b><br><br><b>Note:</b> Negative sign indicates that work is done by the field (energy is released as the dipole moves to a more stable position).
