Question

Calculate the amount of work done in rotating a dipole, of dipole moment 3 × 10^-8 Cm, from its position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of intensity 10^4 N/C. [CBSE (F) 2011]

January 14, 2024

Answer

Dipole moment (p) = 3 × 10^-8 Cm

The electric field of intensity (E) = 10⁴ N/C

At stable equilibrium (θ1) = 0º

At unstable equilibrium (θ2) = 180º

Work done in a rotating dipole is given by

W = PE (cos θ1 – cos θ2)

= (3 × 10^-8)(10⁴) (cos 0º – 180º)

= 3 × 10^-4 [1 – (-1)] = 6 × 10^-4 J

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