Question

An electric dipole of length $2\,cm$, when placed with its axis making an angle of ${60^ \circ }$ with a uniform electric field, experiences a torque of $6\sqrt 3 \,Nm$. Calculate the potential energy of the dipole, if it has a charge of $\pm 2\,nc$.

Answer 1

Hint
The potential energy of the dipole is determined by using the torque formula in the electric dipole and the potential energy formula. By dividing these two formulae, then the potential energy of the dipole can be determined.
The torque of the electric field is given by,
$\Rightarrow \tau = PE\sin \theta$
Where, $\tau$ is the torque of the electric field, $P$ is the electric dipole moment, $E$ is the electric field and $\theta$ is the angle of the electric dipole.
The potential energy of the electric dipole is given by,
$\Rightarrow U = - PE\cos \theta$
Where, $U$ is the potential energy of the electric dipole, $P$ is the electric dipole moment, $E$ is the electric field and $\theta$ is the angle of the electric dipole.

Complete step by step answer
Given that, The length of the electric dipole is, $l = 2\,cm$,
The angle of the electric dipole is, $\theta = {60^ \circ }$,
The torque by the electric field is, $\tau = 6\sqrt 3 \,Nm$,
The charge of the electric field is, $Q = \pm 2\,nc$.
Now, The torque of the electric field is given by,
$\Rightarrow \tau = PE\sin \theta \,..................\left( 1 \right)$
The product of the electric dipole moment is equal to the $2Ql$, substitute this term in the equation (1), then the equation (1) is written as,
$\Rightarrow \tau = 2Ql\sin \theta \,.................\left( 2 \right)$
Now, The potential energy of the electric dipole is given by,
$\Rightarrow U = - PE\cos \theta \,..................\left( 3 \right)$
The product of the electric dipole moment is equal to the $2Ql$, substitute this term in the equation (3), then the equation (3) is written as,
$\Rightarrow U = - 2Ql\cos \theta \,..................\left( 4 \right)$
Now by dividing the equation (2) with the equation (4), then
$\Rightarrow \dfrac{\tau }{U} = \dfrac{{2Ql\sin \theta }}{{ - 2Ql\cos \theta }}$
By cancelling the terms in the above equation, then the above equation is written as,
$\Rightarrow \dfrac{\tau }{U} = - \dfrac{{\sin \theta }}{{\cos \theta }}$
From the trigonometry, the $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$, substitute in the above equation, then
$\Rightarrow \dfrac{\tau }{U} = - \tan \theta$
By substituting the angle in the above equation, then
$\Rightarrow \dfrac{\tau }{U} = - \tan {60^ \circ }$
From the trigonometry, the value of $\tan {60^ \circ } = \sqrt 3$, substitute in the above equation, then
$\Rightarrow \dfrac{\tau }{U} = - \sqrt 3$
By rearranging the terms in the above equation, then
$\Rightarrow U = - \dfrac{\tau }{{\sqrt 3 }}$
By substituting the torque value in the above equation, then
$\Rightarrow U = - \dfrac{{6\sqrt 3 }}{{\sqrt 3 }}$
By cancelling the same terms in the above equation, then
$\Rightarrow U = - 6\,joules$
Thus, the above equation shows the potential energy.

Note
The potential energy of the electric dipole is directly proportional to the electric dipole moment and the electric field in the dipole and the angle of the electric dipole. The torque of the electric field is directly proportional to the electric dipole moment and the electric field in the dipole and the angle of the electric dipole.