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Obtain the expression for potential energy of an electric dipole placed with its axis at an angle $(\theta)$ to an electric field $(\overrightarrow{E})$ . What is the minimum value of potential energy?

Answers (1)

$\vec{\varsigma }= \vec{P}\times \vec{E}$
$\vec{\varsigma }= pE\sin \theta$
where, $\vec{\varsigma }= torque$
$p= momentum$
Now
work done $dw= \varsigma \cdot d\theta$
$dw= PE\sin \theta d\theta$
on integrating both side,we have
$w=\int_{\theta 1}^{\theta 2}dw= \int_{\theta 1}^{\theta 2}PE\sin \theta d\theta$
$w= pE\left [ \cos \theta _{1}-\cos \theta _{2} \right ]$
The minimum value depends on the choice of angle theta

if $\theta _{1}= 0\: and \, \theta _{2}= \theta$
then, $w= pE\left ( 1-\cos \theta \right )$

The minimum value of potential energy occurs when $\theta = \frac{\pi}{2}$ means potential in the equatorial plane is zero.

$if \ \theta_1=\frac{\pi}{2}\ and \ \ \theta_2=0$

Then the minimum value is

$-PE$

Posted by

Safeer PP

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