A point charge q of mass m is suspended vertically by a string of length l. A point dipole of dipole

Q: Why is the initial potential energy zero?

The problem states the dipole is brought from infinity. At an infinite distance, the electrostatic interaction between the dipole and the charge is zero, hence the initial potential energy of the system is zero.

Q: How did we determine the angle $\theta = \alpha/2$?

The suspension point, the dipole, and the charge form an isosceles triangle with two sides of length $l$. The angle at the top is $\alpha$. The base angles are $(180^\circ - \alpha)/2 = 90^\circ - \alpha/2$. The angle $\theta$ is the angle the base makes with the horizontal, which geometrically evaluates to $\alpha/2$.

Q: Why did we use Lami's theorem instead of resolving forces into components?

While resolving forces into horizontal and vertical components works perfectly fine, Lami's theorem is a powerful shortcut whenever exactly three coplanar forces keep an object in equilibrium. It directly relates the magnitudes of the forces to the angles between them, saving algebraic steps.

Q: Why is the electric field $E = 2kp/r^3$?

The dipole $\vec{p}$ is pointing directly towards the charge $q$. This means the charge $q$ lies on the axial line of the dipole. For a short dipole, the electric field at an axial point at distance $r$ is given by $2kp/r^3$.

Q: Is it a coincidence that the electrostatic energy equals the gravitational energy?

It's not a coincidence, but rather a beautiful consequence of the inverse-cube nature of the dipole's electric field and the specific geometry of the isosceles triangle setup. The $r$ factor from the field perfectly cancels with the $r$ in the potential energy, leaving a term that geometrically maps to the height $h$.

Question

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

Sigma Percentile · Accepted Answer

### The Symphony of Electrostatics and Mechanics

This problem is a beautiful testament to how seemingly disparate branches of physics—electrostatics and classical mechanics—can intertwine to create a perfectly balanced system. We are tasked with finding the work done to bring a dipole from infinity to a specific equilibrium position near a suspended charge. Let's break down this elegant setup.

### Decoding the Geometry

Before diving into the physics, we must understand the spatial layout. The suspension point $O$, the dipole at $P$, and the charge $q$ at $Q$ form a triangle. Because the string length is $l$ and the dipole is placed at a distance $l$ directly below the suspension point, we have an **isosceles triangle** with two sides of length $l$.

The distance between the dipole and the charge is the base of this triangle, which we can call $r$. Using basic trigonometry, we find:

$r = 2l \sin\left(\frac{\alpha}{2}ight)$

Furthermore, if we look at the height $h$ of the charge above the horizontal line passing through the dipole, geometry tells us that $h = l - l \cos\alpha$. Using the half-angle formula, this becomes $h = 2l \sin^2\left(\frac{\alpha}{2}ight)$. Notice the beautiful relationship here: $h = r \sin\left(\frac{\alpha}{2}ight)$. This geometric link will be our secret weapon later.

### The Energy Landscape

The core of the question asks for the work done by an external agent. According to the work-energy theorem, this work equals the total change in the potential energy of the system:

$W = \Delta U = U_f - U_i$

Since the dipole is brought from infinity, the initial interaction energy is zero ($U_i = 0$). The final energy $U_f$ consists of two parts: the gravitational potential energy gained by the charge ($mgh$) and the electrostatic potential energy between the dipole and the charge. Because the dipole points directly at the charge, the charge lies on its axis, making the electrostatic potential $V = \frac{kp}{r^2}$. Thus, the energy is:

$W = mgh + \frac{kpq}{r^2}$

Our mission is now clear: we need to express the electrostatic term $\frac{kpq}{r^2}$ in terms of $mgh$.

### The Force Equilibrium

To find the missing link, we turn to mechanics. The charge $q$ is in perfect equilibrium under the influence of three coplanar forces:
1.  **Gravity:** $mg$ acting downwards.
2.  **Tension:** $T$ acting along the string.
3.  **Electrostatic Force:** $qE$ acting repulsively along the line connecting the dipole and the charge.

Whenever three coplanar forces keep an object in equilibrium, **Lami's Theorem** is the most elegant tool to use. It states that the ratio of each force to the sine of the angle between the other two forces is constant.

Applying Lami's theorem to $mg$ and $qE$:

$\frac{mg}{\sin(90^\circ + \frac{\alpha}{2})} = \frac{qE}{\sin(180^\circ - \alpha)}$

Using trigonometric identities, $\sin(90^\circ + \frac{\alpha}{2}) = \cos(\frac{\alpha}{2})$ and $\sin(180^\circ - \alpha) = \sin\alpha = 2\sin(\frac{\alpha}{2})\cos(\frac{\alpha}{2})$. Substituting these in, the $\cos(\frac{\alpha}{2})$ terms cancel out beautifully, leaving us with:

$qE = 2mg \sin\left(\frac{\alpha}{2}ight)$

### The Grand Unification

Now, we bring electrostatics back into the picture. The electric field $E$ of a dipole at an axial distance $r$ is $E = \frac{2kp}{r^3}$. Substituting this into our force equation:

$q\left(\frac{2kp}{r^3}ight) = 2mg \sin\left(\frac{\alpha}{2}ight)$

Let's rearrange this to isolate the electrostatic potential energy term $\frac{kpq}{r^2}$:

$\frac{kpq}{r^2} = mg \left(r \sin\left(\frac{\alpha}{2}ight)ight)$

Remember our secret geometric weapon from the beginning? We established that $r \sin\left(\frac{\alpha}{2}ight)$ is exactly equal to the height $h$! Substituting $h$ into the equation yields a stunning result:

$\frac{kpq}{r^2} = mgh$

The electrostatic potential energy is perfectly equal to the gravitational potential energy. It is not a coincidence; it is a direct consequence of the inverse-cube nature of the dipole force and the geometry of the isosceles setup.

### Final Calculation

Substituting this revelation back into our total work equation:

$W = mgh + mgh = 2mgh$

The problem states that the work done is $N 	imes (mgh)$. Comparing the two expressions, we arrive at our final answer:

**$N = 2$**

This problem is a masterclass in physics problem-solving, showing how geometric constraints and physical laws dance together to produce a clean, integer result.

Enter Numerical Value:

Visualized Solution

$G eo m e t r i c A na l y s i s$

$W or k - E n er g y T h eor e m$

$F r ee B o d y D ia g r am$

$A ppl y in g L am i^{'} s T h eor e m$

$E l ec t r os t a t i c F or ce & E n er g y$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Electric Dipole

The Symphony of Electrostatics and Mechanics

Decoding the Geometry

The Energy Landscape

The Force Equilibrium

The Grand Unification

Final Calculation

Similar Questions

Two point charges $+ q$ and $- q$ are held fixed at $(- d, 0)$ and $(d, 0)$ respectively of a $x$ - $y$ co-ordinate system. Then

Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle as shown in the figure.

A point dipole $p = - p_{0} \hat{x}$ is kept at the origin. The potential and electric field due to this dipole on the Y-axis at a distance $d$ are, respectively [Take, $V = 0$ at infinity]

An electric dipole is formed by two equal and opposite charges $q$ with separation $d$ . The charges have same mass $m$ . It is kept in a uniform electric field $E$ . If it is slightly rotated from its equilibrium orientation, then its angular frequency $ω$ is