Derivation:
Let us consider, An electric dipole AB, made up of two charges $+q$ and $-q$, is placed at a very small distance $2l$ in a uniform electric field $\overrightarrow{E}$. If dipole $AB$ rotates at angle $θ$ from its equilibrium position. If $A'$ and $B'$ are the new position of a dipole in the electric field. Then force on $+q$ charge particle due to electric field→
$ \overrightarrow{F_{+q}}=q\overrightarrow{E}\qquad (1)$
Then force on $-q$ charge particle due to electric field→
$ \overrightarrow{F_{-q}}=q\overrightarrow{E}\qquad(2)$
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| Work done by rotating an electric dipole |
So work done by a force on $+q$ charge particle to bring from position $A$ to position $A'$→
$ \overrightarrow{W_{+q}}=\overrightarrow{F_{+q}}·\overrightarrow{AC}$
$ \overrightarrow{W_{+q}}=q\overrightarrow{E}· \overrightarrow{AC}\qquad(3)$
Similarly, work done by the force on $-q$ charge particle to bring from position $B$ to position $B'$→
$ \overrightarrow{W_{-q}}=\overrightarrow{F_{-q}}·\overrightarrow{BD}$
$ \overrightarrow{W_{-q}}=q\overrightarrow{E}· \overrightarrow{BD}\qquad (4)$
So the total work is done by the force on the dipole→
$ \overrightarrow{W}=\overrightarrow{W_{+q}}\:+\:\overrightarrow{W_{-q}}$
$\overrightarrow{W}=q\overrightarrow{E}·(\overrightarrow{AC}+\overrightarrow{BD})\qquad (5)$
From the figure, There is symmetry so
$ \overrightarrow{AC}=\overrightarrow{BD}$
So from equation $(5)$
$ \overrightarrow{W}=q\overrightarrow{E}·(2\overrightarrow{AC})$
$ \overrightarrow{W}=2q\overrightarrow{E}(\overrightarrow{AO}-\overrightarrow{CO})\qquad (6)$
From figure→
$ \left | \overrightarrow{AO} \right |=l$
$ \left | \overrightarrow{CO} \right |=l\:cos \theta$
Now substitute the values in equation (6). So equation (6) can be written as in magnitude form →
$ W=2qE(l-l\:cos\theta )$
$ W=2qEl(1-cos\theta )$
$ W=pE(1-cos\theta )$
The above expression shows that work is done on a rotating electric dipole in a uniform electric field.
Case (i):
If $\theta=0^{\circ}$, Then work done will be minimum
$W_{min}=0$
Case (ii):
If $\theta=90^{\circ}$, Then work done
$W=pE$
Case (iii):
If $\theta=180^{\circ}$, Then work done will be maximum
$W_{max}=2pE$
