91Ó°ÊÓ

The given potential energies of the four orientations are: ()

1)$−5U_0$

2)$−7U_0$

3)$3U_0$

4)$5U_0$

The electric dipole moment of the body is the vector quantity used for measuring the separation between the positive and negative charges that consist of the dipole. Due to opposite charges, the body undergoes orientation in a uniform electric field. Thus, it experiences torque at the given position that results in an orientation of the body to get it in a stable position. Now, this torque gives rise to the potential energy associated with the orientation.

The potential energy of the dipole associated with its orientations,

$U=−pE\cos \mathrm{θ}\text{ â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\text{i}\right)$

The torque associated with the dipole orientation,

$\mathrm{Ï„}=pE\sin \mathrm{θ}\text{ â¶Ä‰}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(\text{ii}\right)$

From equation (i), we get the relation of the angle and the potential energy as:

$\begin{array}{c}U=−pE\cos \mathrm{θ}\\ =pE\cos (\mathrm{π}−\mathrm{θ})\mathrm{..................}\left(a\right)\end{array}$

Thus, from the above equation, we get that the higher is the potential energy of the dipole, the higher is the angle between electric dipole moment and electric field$\mathrm{θ}$.

Rank of the potential energies as per the data:$4>3>1>2$

Hence, the rank of the orientations will be$4>3>1>2$ .

Now, using equation (ii) ($\mathrm{τ}=pE\sin \mathrm{θ}$), we can say that the value of torque will be minimum for angles between$0^0\text{to}{180}^0$and will be maximum when$\mathrm{θ}$will be at${90}^0$ .

Now, using these values in equation (a) of part (a), we can get that the potential energy at$\mathrm{θ}=\mathrm{π}/2$as:

data-custom-editor="chemistry" $\begin{array}{c}U=pE\cos (\mathrm{π}−\mathrm{π}/2)\\ =pE\cos \mathrm{π}/2\\ =0\end{array}$

Similarly, if$\mathrm{θ}=\mathrm{π}$, the potential energy will be maximum and if$\mathrm{θ}=0$, the potential energy will be minimum.

Thus, the term near to zero value will give maximum torque.

Thus, the rank of potential energies closer to zero value (considering only the magnitudes$3>1=4>2$):

Hence, the rank value according to magnitude of torque is $3>1=4>2$.