Work Done Bringing Point Dipole Near Suspended Charge

To find the value of $N$, we analyze the equilibrium of the suspended charge $q$ and calculate the total work done by the external agent to bring the dipole $\vec{p}$ from infinity to its final position.

1. Geometry of the System

Let the suspension point be $O$, the initial position of the charge be $A$ (at a vertical distance $l$ below $O$), and the final position of the charge be $B$.

• The distance from the suspension point $O$ to the dipole at $A$ is $l$.
• The length of the string is $OB = l$.
• The angle made by the string with the vertical is $\angle AOB = \alpha$.

In the isosceles triangle $\Delta OAB$ formed by sides $OA = l$ and $OB = l$:

• The distance $r$ between the dipole at $A$ and the charge at $B$ is: $r = 2l \sin\left(\frac{\alpha}{2}\right)$
• The vertical height $h$ by which the mass $m$ is raised is: $h = l - l\cos\alpha = 2l \sin^2\left(\frac{\alpha}{2}\right)$

Thus, the increase in gravitational potential energy of the mass $m$ is: $\Delta U_{\text{gravity}} = mgh = 2mgl \sin^2\left(\frac{\alpha}{2}\right)$

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2. Equilibrium of Charge $q$ (Lami's Theorem)

Three forces act on the charge $q$ at position $B$:

1. Gravitational force $\vec{F}_g = m\vec{g}$, acting vertically downwards.
2. Tension $\vec{T}$, acting along the string towards $O$ (making an angle $\alpha$ with the vertical).
3. Electrostatic force $\vec{F}_e$, acting along the line $AB$ (away from the dipole at $A$).

From the geometry of the isosceles triangle $\Delta OAB$, the line $AB$ makes an angle $\frac{\alpha}{2}$ with the horizontal, so the angle between $\vec{F}_e$ and the downward vertical is $\left(90^\circ + \frac{\alpha}{2}\right)$.

The angles between the three forces are:

• Angle between $\vec{T}$ and $m\vec{g}$: $180^\circ - \alpha$
• Angle between $\vec{T}$ and $\vec{F}_e$: $90^\circ + \frac{\alpha}{2}$
• Angle between $\vec{F}_e$ and $m\vec{g}$: $90^\circ + \frac{\alpha}{2}$

Applying Lami's Theorem: $\frac{F_e}{\sin(180^\circ - \alpha)} = \frac{mg}{\sin\left(90^\circ + \frac{\alpha}{2}\right)}$

$\frac{F_e}{\sin\alpha} = \frac{mg}{\cos\left(\frac{\alpha}{2}\right)}$

Using the identity $\sin\alpha = 2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)$: $F_e = mg \cdot \frac{2\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\alpha}{2}\right)}{\cos\left(\frac{\alpha}{2}\right)} = 2mg \sin\left(\frac{\alpha}{2}\right)$

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3. Electrostatic Potential Energy of the System

The electrostatic potential energy $U_e$ of a point charge $q$ in the field of a point dipole $\vec{p}$ at a distance $r$ along its axis varies as: $U_e(r) = \frac{C}{r^2}$

The magnitude of the electrostatic force is related to the potential energy by: $F_e = \left|-\frac{dU_e}{dr}\right| = \frac{2C}{r^3} = \frac{2U_e}{r}$

Therefore, the electrostatic potential energy $U_e$ in the final configuration is: $U_e = \frac{1}{2} F_e r$

Substituting $F_e = 2mg \sin\left(\frac{\alpha}{2}\right)$ and $r = 2l \sin\left(\frac{\alpha}{2}\right)$: $U_e = \frac{1}{2} \left(2mg \sin\left(\frac{\alpha}{2}\right)\right) \left(2l \sin\left(\frac{\alpha}{2}\right)\right) = 2mgl \sin^2\left(\frac{\alpha}{2}\right) = mgh$

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4. Total Work Done

Since the process is performed quasi-statically ($\Delta K = 0$), the work done by the external agent $W_{\text{ext}}$ is equal to the total change in potential energy of the system: $W_{\text{ext}} = \Delta U_{\text{gravity}} + U_e$

$W_{\text{ext}} = mgh + mgh = 2mgh$

Comparing this with $W_{\text{ext}} = N \times (mgh)$, we get: $N = 2$