1. Understanding the Problem and Resolving Contradictions

The problem asks for the work done in bringing a dipole from infinity to a specific position near a suspended charge q, causing the charge to move to a new equilibrium position. The final configuration is shown in a diagram.

A critical analysis of the forces is required. The electric field E of a dipole p at a position r (where the angle between p and r is θ) has radial ($E_r$) and tangential ($E_θ$) components: $E_r = \frac{1}{4\pi\epsilon_0} \frac{2p \cos\theta}{r^3}, \quad E_\theta = \frac{1}{4\pi\epsilon_0} \frac{p \sin\theta}{r^3}$ The force on the charge q is $F_e = qE$.

The diagram shows the dipole moment p to be perpendicular to the line joining the dipole and the charge q. This corresponds to θ = 90°. In this case, $E_r = 0$ and $E_\theta = \frac{p}{4\pi\epsilon_0 y^3}$. The force $F_e$ would be purely tangential (perpendicular to the line joining them). However, applying equilibrium conditions with such a force leads to a physical contradiction (Tension = -mg).

Alternatively, the problem states the "charge moves away", implying a repulsive force that acts along the line joining the dipole and the charge. This requires the force to be radial ($F_e = F_r$), which means $F_\theta = 0$, implying θ = 0 or 180°. This contradicts the diagram's depiction of p's orientation.

Given that a physical equilibrium must be possible, we resolve this contradiction by assuming the force $F_e$ is indeed radial (repulsive), and the diagram is misleading about the orientation of p. This is a common situation in complex physics problems where diagrams can be illustrative.

2. Equilibrium Analysis

Let's analyze the forces on the charge q in its final equilibrium position. The forces are:

1. Tension T along the string.
2. Gravitational force mg acting vertically downwards.
3. Electrostatic repulsive force $F_e$ acting along the line joining the dipole and the charge.

The system is in equilibrium, so the net force is zero. We can use Lami's theorem as suggested by the note, or resolve forces into components.

From the geometry:

• The string makes an angle 2α with the vertical.
• The line DQ (from dipole to charge) makes an angle α with the horizontal. Thus, it makes an angle 90°-α with the vertical.

The angles between the forces are:

• Angle between T and mg: 180° - 2α
• Angle between mg and $F_e$: 180° - (90° - α) = 90° + α
• Angle between $F_e$ and T: (90° - α) + 2α = 90° + α (Check: Sum of angles = (180°-2α) + (90°+α) + (90°+α) = 360°. Correct.)

Applying Lami's theorem: $\frac{F_e}{\sin(180° - 2α)} = \frac{mg}{\sin(90° + α)}$ $\frac{F_e}{\sin(2α)} = \frac{mg}{\cos(α)}$ $F_e = mg \frac{\sin(2α)}{\cos(α)} = mg \frac{2\sin(α)\cos(α)}{\cos(α)} = 2mg\sin(α)$ This gives us the magnitude of the electrostatic force in terms of mg and α.

3. Work-Energy Calculation

The work done (W) by an external agent in bringing the dipole from infinity to its final position is equal to the change in the total energy of the system. Since the process starts and ends at rest, the change in kinetic energy is zero. $W = \Delta E = \Delta PE_g + \Delta PE_e$ where $ΔPE_g$ is the change in gravitational potential energy of the charge q, and $ΔPE_e$ is the change in electrostatic potential energy of the charge-dipole system.

Change in Gravitational Potential Energy ($ΔPE_g$): The charge q is raised by a vertical height h. $h = l - l\cos(2α) = l(1 - \cos(2α)) = l(1 - (1 - 2\sin^2α)) = 2l\sin^2α$ $\Delta PE_g = mgh = 2mgl\sin^2α$

Change in Electrostatic Potential Energy ($ΔPE_e$): The initial electrostatic potential energy is zero as the dipole is at infinity. The final potential energy $PE_e$ for a charge-dipole system with a radial force is: $PE_e = \frac{kqp}{y^2}$ where $k = 1/(4πε_0)$. The corresponding radial force is $F_e = \frac{2kqp}{y^3}$. From these two expressions, we can relate $PE_e$ and $F_e$: $PE_e = \frac{y}{2} F_e$ Substituting the expression for $F_e$ from our equilibrium analysis: $PE_e = \frac{y}{2} (2mg\sinα) = ymg\sinα$

4. Geometric Analysis

We need to find the distance y. The problem configuration implies that the dipole is placed at the initial position of the charge q. Let the suspension point S be at (0, l) and the initial position of q (and final position of the dipole D) be at the origin (0,0). The final position of the charge Q is: $Q = (l\sin(2α), l - l\cos(2α))$ The distance y = DQ is: $y = \sqrt{(l\sin(2α) - 0)^2 + (l(1-\cos(2α)) - 0)^2}$ $y = \sqrt{l^2\sin^2(2α) + l^2(1-\cos(2α))^2} = l\sqrt{(2\sinα\cosα)^2 + (2\sin^2α)^2}$ $y = l\sqrt{4\sin^2α\cos^2α + 4\sin^4α} = l\sqrt{4\sin^2α(\cos^2α + \sin^2α)} = \sqrt{4l^2\sin^2α} = 2l\sinα$

5. Final Calculation

Now we can find the total work done W. $W = \Delta PE_g + PE_e = mgh + ymg\sinα$ Substitute the expressions for h and y: $W = mg(2l\sin^2α) + (2l\sinα)mg\sinα$ $W = 2mgl\sin^2α + 2mgl\sin^2α = 4mgl\sin^2α$ We are given that $W = N \times (mgh)$. We know $mgh = 2mgl\sin^2α$. $4mgl\sin^2α = N \times (2mgl\sin^2α)$ $N = \frac{4mgl\sin^2α}{2mgl\sin^2α} = 2$ The value of N is 2.