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 $N \times( mgh )$, 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, $\frac{F}{\sin \theta}$ is the same for all forces, where $F$ is any one of the forces and $\theta$ is the angle between the other two forces)
IIT 2020, Advanced
Download our app for free and get started
$U _{ i }=0$
$U _{ f }=\frac{ kq P }{\left(2 \ell \sin \frac{\alpha}{2}\right)^2}+ mgh$
Now, from $\triangle OAB$
$\alpha+90-\theta+90-\theta=180$
$\Rightarrow \alpha=2 \theta$
From $\triangle ABC : h =2 \ell \sin \left(\frac{\alpha}{2}\right) \sin \theta$
$h =2 \ell \sin \left(\frac{\alpha}{2}\right) \sin \left(\frac{\alpha}{2}\right)$
$\Rightarrow h =2 \ell \sin ^2\left(\frac{\alpha}{2}\right)$
Now charge is in equilibrium at point $B$.
So, using sine rule
$\Rightarrow \frac{ mg }{\sin \left[90+\frac{\alpha}{2}\right]}=\frac{ qE }{\sin [180-2 \theta]}$
$\Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{\sin 2 \theta}$
$\Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{\sin \alpha} \Rightarrow \frac{ mg }{\cos \frac{\alpha}{2}}=\frac{ qE }{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}$
$\Rightarrow qE = mg 2 \sin \left(\frac{\alpha}{2}\right)$
$\Rightarrow \frac{ q 2 lq }{\left[2 \ell \sin \frac{\alpha}{2}\right]^3}= mg 2 \sin $$\left(\frac{\alpha}{2}\right) \Rightarrow \frac{ kpq }{\left[2 \ell \sin \frac{\alpha}{2}\right]^2}= mg \sin \left(\frac{\alpha}{2}\right) \times\left(2 \ell \sin \frac{\alpha}{2}\right)$
$\Rightarrow \frac{ kpq }{\left[2 \ell \sin \frac{\alpha}{2}\right]^2}= mgh \Rightarrow \operatorname{substituting~this~in~equation~(i)~}$
$\Rightarrow U$
$W =\Delta U = Nmgh = N =2$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A parallel plate capacitor is to be designed, using a dielectric of dielectric constant $5$, so as to have a dielectric strength of $10^9\;Vm^{-1}$ . If the voltage rating of the capacitor is $12\;kV$, the minimum area of each plate required to have a capacitance of $80\;pF$ isView Solution
- 2The equivalent capacitance of three capacitors of capacitance ${C_1},{C_2}$ and ${C_3}$ are connected in parallel is $12$ units and product ${C_1}.{C_2}.{C_3} = 48$. When the capacitors ${C_1}$ and ${C_2}$ are connected in parallel, the equivalent capacitance is $6$ units. Then the capacitance areView Solution
- 3On rotating a point charge having a charge $q$ around a charge $Q$ in a circle of radius $r$. The work done will beView Solution
- 4The radius of nucleus of silver (atomic number $=$ $47$) is $3.4 \times {10^{ - 14}}\,m$. The electric potential on the surface of nucleus is $(e = 1.6 \times {10^{ - 19}}\,C)$View Solution
- 5As shown in the figure, a point charge $Q$ is placed at the centre of conducting spherical shell of inner radius a and outer radius $b$. The electric field due to charge $Q$ in three different regions I, II and III is given by: $( I : r < a , II : a < r < b , III : r > b )$View Solution
- 6The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ isView Solution
- 7When air in a capacitor is replaced by a medium of dielectric constant $K$, the capacityView Solution
- 8A charge of $5\,C$ is given a displacement of $0.5\,m$. The work done in the process is $10\,J$. The potential difference between the two points will be.......$V$View Solution
- 9A $6\,\mu F$ capacitor is charged from $10\;volts$ to $20\;volts$. Increase in energy will beView Solution
- 10View SolutionWhen a dielectric material is introduced between the plates of a charged condenser then electric field between the plates