An electric dipole of moment 6.0 10⁻⁸ C m is placed in a uniform electric field of magnitude 5.0 10^4 N/C . Initially, the dipole is held at an angle of 120^ with respect to the field direction. The work done by an external agent to slowly rotate the dipole until it becomes anti-parallel to the electric field is:

Step-by-step solution

The work done by an external agent to slowly rotate an electric dipole in a uniform electric field is equal to the change in its potential energy. W = U = U_f - U_i The potential energy of a dipole in an electric field is given by U = -pE . W = -pE ₂ - (-pE ₁) = pE ( ₁ - ₂) Given p = 6.0 10⁻⁸ C m , E = 5.0 10^4 N/C , ₁ = 120^ , and ₂ = 180^ (anti-parallel). W = (6.0 10⁻⁸) (5.0 10^4) ( 120^ - 180^ ) W = 3.0 10⁻³ (- 1 2 - (-1) ) W = 3.0 10⁻³ 1 2 = 1.5 10⁻³ J Answer: 1.5 10⁻³ J