Question

Find the amount of work done in rotating a dipole of dipole moment 3×10−3 cm from its position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of intensity 104NC−1.

• A. (A) 50J
• B. (B) 60J
• C. (C) 80J
• D. (D) 70J

Answer 1

Answer: (B)

(B) 60J

Answer 2

Explanation:
Given that:Dipole Moment, P=3×10−3 cmElectric Field Intensity, E=104NC−1We know that,In rotating the dipole from the position of stable equilibrium by an angle θ, the amount of work done is given by,W=PE(1−cos⁡θ)For unstable equilibrium, θ=180∘∴W=PE(1−cos⁡180∘)[∵cos⁡180∘=−1]=2PE=2×3×10−3×104J=60 JHence, the correct option is (B).