Work done by a rotating electric dipole in uniform electric field

Derivation of Work done by a rotating electric dipole in uniform electric field :

Let us consider an electric dipole AB, made up of two charges $+q$ and $-q$, placed at a very small distance $l$ in a uniform electric field $\overrightarrow{E}$. If dipole $AB$ rotates at an angle $θ$ from its equilibrium position. If $A'$ and $B'$ are the new positions of a dipole in the electric field. Then the force on a $+q$ charge particle due to the 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)$

So work done by a force on a $+q$ charge particle to bring it 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 |=\frac{l}{2}$

$ \left | \overrightarrow{CO} \right |=\frac{l}{2}\:cos \theta$

Now substitute the values in equation (6). So equation (6) can be written as in magnitude form →

$ W=2qE(\frac{l}{2}-\frac{l}{2}\:cos\theta )$

$ W=2qE\left(\frac{l}{2}\right)(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

$W_{max}=2pE$

Alternative Method

Let us consider an electric dipole $AB$ rotates from angle $\theta_{1}$ to $\theta_{2}$ by applying torque on an elecric dipole $AB$ in uniform electric field $\overrightarrow{E}$, if change in angle is $d\theta$, then Work done

$dW= \tau d\theta$

$dW= pE sin \theta d\theta \quad \left\{ \because \tau= p E sin\theta \right\}$

Now integrate the above equation from angle $\theta_{1}$ to $\theta_{2}$

$\int dW= \int_{\theta_{1}}^{\theta_{2}} pE sin \theta d\theta$

$\int dW= pE \int_{\theta_{1}}^{\theta_{2}} sin \theta d\theta$

$ W= pE \left[ - cos \theta \right]_{\theta_{1}}^{\theta_{2}}$

$ W= pE \left[ -cos {\theta_{2}} + cos{\theta_{1}}\right]$

$ W= pE \left[ cos {\theta_{1}} - cos{\theta_{2}}\right]$

This is the equation of work done by rotating an electric dipole from angle $\theta_{1}$ to $\theta_{2}$ in a uniform electric field