Question

An electric dipole of length $1cm$ which is kept 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 charge $\pm 2nC$.

Answer 1

Hint: The potential energy of the dipole is the dot product of momentum of the dipole and the electric field acting. The torque experienced is given as the cross product of the momentum of the dipole and the electric field experiencing. This all will help you in solving this question.

Complete step by step answer:
 first of all let us mention what all are given in the question. The length of the electric dipole is given as,
$l=1cm$
The angle at which the dipole is placed from the axis is given as,
$\theta =60{}^\circ $
Torque experienced at this point can be written as,
$\tau =6\sqrt{3}Nm$
Charge existing in this dipole can be written as,
$q=\pm 2nC$
The potential energy of the electric dipole can be found using the equation,
$U=-\vec{P}\cdot \vec{E}$
Where $P$be the momentum of the dipole and $E$ be the electric field experienced there.
The dot product can be written as,
$U=-PE\cos \theta $
The torque acting can be written in the form of the cross product which can be given as,
$\vec{\tau }=\vec{P}\times \vec{E}$
That is,
$\vec{\tau }=PE\sin \theta $
We can divide the equation of potential energy to the torque of the body. This can be written as,
$\dfrac{U}{\tau }=\dfrac{-PE\cos \theta }{PE\sin \theta }$
This can be simplified as,
$U=-\tau \cot \theta $
Substituting the values in it will give,
$U=-6\sqrt{3}\times \cot 60{}^\circ =-6J$

Note: The potential energy of a dipole is always negative. This is because of the alignment of the dipole. The positive end is kept at the point where there is high potential and the negative end is kept at the point where the potential is lesser.