Question

3. Calculate the amount of work done in turning an electric dipole of dipole moment 3 x 10^-8C-m from its position of unstable equilibrium to the stable equilibrium, in a uniform electric field intensity 10^3 N/C. ​

Answer 1

Explanation:- given dipole moment 'P ' = 3× 10⁻⁸ c-m

• Electric field 'E' = 10³N/C

For unstable equilibrium θ₁ = 180°

For stable equiblirium θ₂ = 0°

work done to rotating dipole is 'W' = PE(cosθ₁ -cosθ₂)

• W = 3× 10⁻⁸ × 10³ (cos180° - cos0°)

• W = 3× 10⁻⁵ (-1-1)

• W = 3× 10⁻⁵ (-2) = -6× 10⁵JAnswer .

Here negative (-ve) sign represents that work is done by the system itself .