## Potential due to an electric dipole:

Let us consider charge $-q$ is placed at point $P$ and charge $+q$ is placed at point $Q

*Figure:8.a*

Electric potential at point $R$ due to electric dipole would be sum of potential due to both the charges $+q$ and $-q$. So,

………………….. (1)

Draw two perpendiculars $PC$ and $QD$from point $P$ and $Q$ as shown.

From triangle $POC$ $\cos \theta =OC/OP=OC/a$

Or, $OC=a cos\theta $

Similarly, $OD=a cos\theta $ Now,

$r_{1}=QR\approx RD=OR-OD=r-a cos\theta $

And

$r_{2}=PR\approx RC=OR+OC=r+a cos\theta $

From equation (1), we get,

(Consider $r>>a$)

The magnitude of dipole is,$\vert p\vert =2qa$

So, the above equation becomes,

Where ${r}$ is the unit vector along the vector ${OR}$. From the above equation, we can see that the electric potential due to electric dipole does not only depend on $r$ but also depends on the angle between position vector $r$ and dipole moment $p$.

## Work done in rotating an electric dipole in an electric field:

Consider a dipole placed in a uniform electric field. If the dipole is rotated from its equilibrium position, work has to be done.

*Figure:8.b*

Let the dipole of the moment $p$ is rotated through an angle $\theta $ from its equilibrium position.

The torque acting on the dipole is,

$\tau =pE sin\theta $

Work done when the dipole makes a small angle $d\theta $ is given by,

$dW=pE sin\theta d\theta $ Total work done is,$W=_{0}^{\theta }{}pE sin\theta d\theta $

$W=pE\lbrack 1-cos\theta \rbrack $

This is the expression for work done in rotating an electric dipole placed in a uniform electric field $E$ through an angle $\theta $ from its equilibrium.