Question

Potential energy of a dipole in an external field
Electric potential energy of an Electric dipole in dipole from zero energy position to the desired position (The zero enrgy position to be one when the dipole When an electric dipole of dipole moment p is on it such that $P=\underset{P}{\overset{}{\to }}\times \underset{E}{\overset{}{\to }}$
This torque tends to rotate the dipole let dW the small amount of work done. turning the dipole through an angle $d\theta$
$dW=Text.d\theta$
$=pEsin\theta d\theta$
W is the work done in turning the dipole from $0^{o}to\theta$ then,
$W={\int }_0^{W}dW$
= ${\int }_{{90}^o}^{\theta }pEsin\theta d\theta$

Answer 1

The force experienced by the charges is given as $$qE$$ and$$+qE$$ , as can be seen in the figure.

$\tau =\overrightarrow{p}\times \overrightarrow{E}$

The amount of work done by the external torque

$W={\int }_{{\theta }_0}^{{\theta }_1}{\tau }_{sxt}\left(\theta \right)d\theta ={\int }_{{\theta }_0}^{{\theta }_1}pE\sin \theta d\theta$ $=pE(\cos {\theta }_0-\cos {\theta }_1)$

As we know that the work done in bringing a system of charges from infinity to the given configuration is defined as the potential energy of the system.

The potential energy $U\left(\theta \right)$

$U\left(\theta \right)=pE(\cos {\theta }_0-\cos {\theta }_1)$