Question

The dipole moment of an electric dipole is 3.0 × 10–6 C-m. The dipole is initially aligned with a uniform external electric field of 2 × 103 N/C in stable equilibrium position. The work done to turn the dipole by 90°, will be


a)3 × 10–3 J



b)6 × 10–3 J



c)9 × 10–3 J



d)1.2 × 10–2 J

Answer 1

Answer:

Work done in rotating the dipole = 6 × 10^-3 J

Answer 2

Given - Dipole moment : 3*10-⁶ Cm.

Electric field : 2*10³ N/C

Angle to which dipole is to be turned : 90°.

Find - Work done to turn the dipole in external electric field by 90°

Solution - The formula that can be used to calculate the work done to turn the dipole in external electric field by 90° - W = pE(1- cos θ). The formula has work done represented by W, dipole moment represented by p, electric field represented by E.

Keeping the values in equation-

W = 3*10-⁶*2*10³ (1-cos 90°)

As we know, cos 90° is 0, thus, keeping the value.

W = 6*10-³ (1 - 0)

W = 6*10-³ J.

Hence, the work done is b)6 × 10-³ J.