Question

A point charge $q$ of mass $m$ is suspended vertically by a string of length $l$. A point dipole of dipole moment $\overrightarrow{p}$ is now brought towards $q$ from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is $N\times \left(mgh\right)$, where $g$ is the acceleration due to gravity, then the value of $N$ is ____ . (Note that for three coplanar forces keeping a point mass in equilibrium, $\frac{F}{\sin \theta }$ is the same for all forces, where $F$ is any one of the forces and $\theta$ is the angle between the other two forces)

Answer 1

Initial potential energy of system is zero, considering reference gravitational P.E to be zero there.
$U_i=0$
$U_f=\frac{kqp}{{\left(2l\sin \frac{\alpha }{2}\right)}^2}+mgh$ .... (i)
Now, from $△OAB$
$\alpha +{90}^{\circ }–\theta +{90}^{\circ }–\theta ={180}^{\circ }$
$\Rightarrow \alpha =2\theta$
From $△ABC$ : $h=2l\sin \left(\frac{\alpha }{2}\right)\sin \theta$
$h=2l\sin \left(\frac{\alpha }{2}\right)\sin \left(\frac{\alpha }{2}\right)$
$\Rightarrow h=2l{\sin }^2\left(\frac{\alpha }{2}\right)$
Now charge is in equilibrium at point $B$.
So, using sine rule
$\Rightarrow \frac{mg}{\sin [{90}^{\circ }+\frac{\alpha }{2}]}=\frac{qE}{\sin [{180}^{\circ }-2\theta ]}$
$\Rightarrow \frac{mg}{\cos \frac{\alpha }{2}}=\frac{qE}{\sin 2\theta }$
$\Rightarrow \frac{mg}{\cos \frac{\alpha }{2}}=\frac{qE}{\sin \alpha }\Rightarrow \frac{mg}{\cos \frac{\alpha }{2}}=\frac{qE}{2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}}$
$\Rightarrow qE=mg\times 2\sin \left(\frac{\alpha }{2}\right)$
using the formula of electric field due to a dipole,
$\Rightarrow \frac{q\times 2kp}{{\left[2l\sin \frac{\alpha }{2}\right]}^3}=mg\times 2\sin \left(\frac{\alpha }{2}\right)\Rightarrow \frac{kpq}{{\left[2l\sin \frac{\alpha }{2}\right]}^2}=mg\sin \left(\frac{\alpha }{2}\right)\times \left(2lsin\frac{\alpha }{2}\right)$
$\Rightarrow \frac{kpq}{{\left[2l\sin \frac{\alpha }{2}\right]}^2}=mgh$
$\Rightarrow$substituting this in equation $\left(i\right)$,
$U_f=mgh+\frac{kpq}{{\left[2l\sin \frac{\alpha }{2}\right]}^2}$
$\Rightarrow U_f=2mgh$
Now work done by external in bringing dipole to this position,
$W=\Delta U=2mgh$
Comparing it with given relation in question,
$N=2$