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(a) Deduce the expression for the torque acting on a dipole of dipole moment $\vec{p}$ placed in a uniform electricfield $\vec{E}$ . Depict the direction of the torque. Express it in the vector form.

(b) Show that the potential energy of a dipole making angle $\theta$ with the direction of the field is given by $u(\theta)=-\vec{p}.\vec{E}.$ Hence find out the amount of work done in rotating it from the position of unstable equilibrium to the stable equilibrium.

Answers (1)

(a)

Magnitude of Torque $\tau =F\times 2a\; \sin \theta$

$=qE\; 2a\; \sin \theta$

$=PE\; \sin \theta$

Vector forme $=P\times E$

The direction of the torque is perpendicular to P and E

(b) The amount of work done by external torque in rotating the dipole is stored as potential energy.

$w=\int_{\theta _{o}}^{\theta }\tau d\; \theta =\int_{\theta _{o}}^{\theta _{1}}PE\; \sin \theta d\; \theta$

$=PE\left ( \cos \theta _{o}-\cos \theta _{1} \right )$

When $\theta _{o}\rightarrow \frac{\pi }{2}$ and $\theta _{1}\rightarrow \theta$

$w=-PE\cos \theta$

For unstable equilibrium $\theta _{o}=180^{o}$

For stable equilibrium $\theta _{1}=0^{o}$

$w=PE\left ( \cos \theta _{o}-\cos \theta _{1} \right )$

$=PE\left ( \cos 180-\cos 0 \right )$

$=-2PE$

Posted by

Safeer PP

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