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

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Correct answer: C. 1.5 × 10^ -3 J

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 = \Delta U = U_f - U_i

The potential energy of a dipole in an electric field is given by

U = -pE \cos \theta

.

W = -pE \cos \theta_2 - (-pE \cos \theta_1) = pE (\cos \theta_1 - \cos \theta_2)

Given

p = 6.0 \times 10^{-8} \text{ C m}

,

E = 5.0 \times 10^4 \text{ N/C}

,

\theta_1 = 120^{\circ}

, and

\theta_2 = 180^{\circ}

(anti-parallel).

W = (6.0 \times 10^{-8}) \times (5.0 \times 10^4) (\cos 120^{\circ} - \cos 180^{\circ})

W = 3.0 \times 10^{-3} \left(-\dfrac{1}{2} - (-1)\right)

W = 3.0 \times 10^{-3} \times \dfrac{1}{2} = 1.5 \times 10^{-3} \text{ J}

Answer:

1.5 \times 10^{-3} \text{ J}

An electric dipole of moment 6.0 × 10^ -8 C m is placed in a...

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