Question

A point charge $q$ of mass $m$is suspended vertically by a string of length$l$. A point dipole of dipole moment vector $p$ 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 $𝑁\times (𝑚𝑔ℎ),$ 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, F/sin θ is the same for all forces, where F is any one of the forces and θ is the angle between the other two forces)

Answer 1

Step 1. Given data:

The charge is $q$ and mass is $m$

Length of string $l$

Dipole moment $p$

Work done $w=N\times mgh$

Step 2. Dipole moment in the field of charge $q$

From the diagram,

$\alpha +2\beta =\pi \Rightarrow \beta =\frac{\pi -\alpha }{2}$

where $\alpha$ is the angle when charge is moved due to dipole moment

$\beta$ is the other two angles of the triangle thus formed

Work done in moving charge is, $w=V_f-V_i$

$\Rightarrow w=mgh+\frac{kpq}{d^2}-0$

where $h$ is the vertical displacement in charge, $q$ is the charge, $d$ is the displacement in charge position

and $k=\frac{1}{4\pi {\epsilon }_0}$,${\epsilon }_0$ is the permittivity

Step 3. Equating the coplanar forces using sign law, we get

$\frac{T}{\sin (\alpha +\beta )}=\frac{mg}{\sin (\alpha +\beta )}=\frac{2kpq}{d^3\sin 2\beta }$

$\Rightarrow \frac{mg}{\sin \left({\displaystyle \frac{\pi +\alpha }{2}}\right)}=\frac{2kpq}{d^3\sin \alpha }$

where $T$ is the tension in the string, $m$ is mass of charge, $g$ is acceleration due to gravity, $q$ is the charge, and $d$ is the displacement in charge position

$\Rightarrow \frac{kpq}{d^2}=\frac{mgd\sin \alpha }{2\cos {\displaystyle \frac{\alpha }{2}}}=\frac{mg\sin \alpha 2l\sin {\displaystyle \frac{\alpha }{2}}}{2\cos \frac{\alpha }{2}}$

$\Rightarrow \frac{kpq}{d^2}=\frac{mg2\sin {\displaystyle \frac{\alpha }{2}}\cos {\displaystyle \frac{\alpha }{2}}\times 2l\sin {\displaystyle \frac{\alpha }{2}}}{2\cos \frac{\alpha }{2}}=2mg\sin \frac{\alpha }{2}l\sin \frac{\alpha }{2}$ $\left(1\right)$

From diagram in triangle BDC, $\cos \beta =\frac{h}{d}=\frac{h}{2l\sin \frac{\alpha }{2}}$

$\Rightarrow \cos \left(\frac{\pi -\alpha }{2}\right)=\frac{h}{2l\sin \frac{\alpha }{2}}$

$\Rightarrow {\sin }^2\frac{\alpha }{2}=\frac{h}{2l}$

Putting in equation $\left(1\right)$

$\Rightarrow \frac{kpq}{d^2}=2mgl\times \frac{h}{2l}=mgh$

So, the work done is,

$\Rightarrow w=mgh+mgh=2mgh=Nmgh$

$\Rightarrow N=2$

Hence, the value of $N$ is $\mathbf{2}$.