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

3 × 10–3 J

6 × 10–3 J

9 × 10–3 J

1.2 × 10–2 J

Answer 1

Answer:

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

Answer 2

Given:

Dipole moment of an electric dipole, $p=3*10^{-6} Cm$

Angle made by $p$ with uniform electric field, $\theta =90$°

External electric field, $E=2*10^{3}N/C$

To Find:

We have to find the work done to turn the dipole by $90$°

Solution:

Work done in rotating the dipole is given by,

$W=pE(cos\theta_{1}-cos\theta_{2})$

For stable equilibrium,

$W=pE(1-cos\theta)$

By putting the value from the given data,

$W=3*10^{-6}*2*10^{3}(1-cos90)$

$W=6*10^{-3}(1-0)=6*10^{-3} J$

Hence the work done to turn on the dipole is $6*10^{-3}J$

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