JEE MainPhysicsElectrostatics
An electric dipole placed in a uniform electric field experiences a torque of 3.0 10⁻² N m when its dipole moment is directed at an angle of 30^ to the field. The work done by an external agent to rotate the dipole slowly from its stable equilibrium position to an angle of 180^ with the field is:
Options
- A6.0 10⁻² J
- B3.0 10⁻² J
- C12.0 10⁻² J
- D4 3 10⁻² J
Correct answer
C. 12.0 10⁻² J
Step-by-step solution
The magnitude of the torque on a dipole in a uniform electric field is given by: = pE Given that = 3.0 10⁻² N m at = 30^ , we can find the maximum torque pE : pE = 30^ = 3.0 10⁻² 0.5 = 6.0 10⁻² J The stable equilibrium position of the dipole corresponds to ₁ = 0^ . The work done in rotating the dipole from an initial angle ₁ to a final angle ₂ is given by: W = pE ( ₁ - ₂) Substitute ₁ = 0^ and ₂ = 180^ : W = pE ( 0^ - 180^ ) W = pE (1 - (-1)) = 2pE Now, substitute the value of pE calculated earlier: W = 2 (6.0 10⁻²