For physics enthusiasts, engineering aspirants, and students preparing for highly competitive exams like the IIT-JEE, I.E. Irodov’s "Problems in General Physics" is nothing short of legendary. The book is famous for its intricate problems that rigorously test a student’s fundamental grasp of physics concepts. In the realm of Electrodynamics, Problem 3.47 stands out as a beautiful synthesis of macroscopic electrostatic theory applied to microscopic molecular interactions.

In this article, we will break down Irodov Problem 3.47 step-by-step.

The Problem Statement

I.E. Irodov Problem 3.47: Find the interaction force between two water molecules separated by a distance l = 10 nm if their electric moments are oriented along the same straight line. The moment of each molecule equals p = 0.62 * 10^-29 C * m.

At first glance, this problem seems to be about chemistry or molecular physics. However, by modeling the complex water molecules purely as electric dipoles, it becomes an elegant electrostatics problem.

Core Concepts: Understanding the Physics

Before diving into the mathematical derivation, it is crucial to understand the conceptual tools required to unlock this problem.

1. The Electric Dipole

The strength and orientation of a dipole are measured by its dipole moment. By convention, the dipole moment vector points from the negative charge to the positive charge. Water molecules (H2O) are naturally polar. The oxygen atom is highly electronegative, pulling the shared electrons closer to itself and creating a partial negative charge, while the hydrogen atoms carry a partial positive charge. This innate charge separation gives the water molecule a permanent dipole moment.

2. Electric Field on the Axis of a Dipole

A dipole creates a persistent electric field in the space around it. For a point located along the central axis of the dipole, at a distance much larger than the separation between the charges within the dipole itself, the resulting electric field is directed along that same axis.

3. Potential Energy of a Dipole in an External Field

When a second dipole is placed in the electric field generated by the first dipole, it experiences a potential energy. If the dipoles are aligned along the same straight line, this interaction energy depends strictly on the strength of the field and the orientation of the second dipole's moment.

4. Force as the Negative Gradient of Potential Energy

In conservative force fields like electrostatics, the physical force exerted on an object is simply the negative derivative (or spatial gradient) of its potential energy with respect to its position. This gradient relationship is the golden key to solving Problem 3.47.

Step-by-Step Mathematical Solution

Let us assign the two water molecules as Dipole 1 and Dipole 2. They are separated by a distance x (which will eventually be evaluated at l) and have identical dipole moments p.

Step 1: Determine the electric field created by Dipole 1.

Dipole 1 generates an electric field at the location of Dipole 2. Since Dipole 2 is situated on the axial line of Dipole 1, we use the standard formula for the axial electric field of a short dipole:

$$E_1 = \frac{1}{4\pi\varepsilon_0} \frac{2p}{x^3}$$

Here, x is the variable distance from the center of Dipole 1.

Step 2: Formulate the potential energy of Dipole 2.

Dipole 2 is situated in the electric field E1. The potential energy U of a dipole placed in an external electric field is calculated using the dot product of its dipole moment and the electric field vectors:

$$U = -\vec{p}_2 \cdot \vec{E}_1$$

Since the problem explicitly states that the electric moments are oriented along the same straight line, the angle between them is 0 degrees (meaning they are parallel and collinear). The dot product simplifies to standard scalar multiplication:

$$U = -p E_1$$

Substituting the expression for the electric field from Step 1 into this equation gives us the interaction potential energy as a function of distance x:

$$U(x) = -p \left( \frac{1}{4\pi\varepsilon_0} \frac{2p}{x^3} \right) = -\frac{1}{4\pi\varepsilon_0} \frac{2p^2}{x^3}$$

Step 3: Calculate the interaction force.

The force F between the two dipoles is the negative derivative of their interaction potential energy with respect to the separation distance x.

$$F = -\frac{dU}{dx}$$

Taking the derivative of our potential energy function:

$$F = -\frac{d}{dx} \left( -\frac{1}{4\pi\varepsilon_0} \frac{2p^2}{x^3} \right)$$

$$F = \frac{1}{4\pi\varepsilon_0} (2p^2) \frac{d}{dx} (x^{-3})$$

$$F = \frac{1}{4\pi\varepsilon_0} (2p^2) (-3x^{-4})$$

$$F = -\frac{3p^2}{2\pi\varepsilon_0 x^4}$$

The negative sign mathematically indicates that the force is attractive; the two dipoles pull towards one another when oriented in the same direction. For our final answer, we are interested in the magnitude of this force at the specific given distance l.

$$|F| = \frac{3p^2}{2\pi\varepsilon_0 l^4}$$

To make the final calculation easier, we can rewrite this using the standard Coulomb constant by multiplying the numerator and denominator by 2:

$$|F| = 6 \left( \frac{1}{4\pi\varepsilon_0} \right) \frac{p^2}{l^4}$$

Step 4: Plug in the numerical values.

Now we substitute the provided problem values into our derived formula:

• Dipole moment p = 0.62 * 10^-29 C * m
• Distance l = 10 nm = 10 * 10^-9 m = 10^-8 m
• Coulomb's constant = 9 * 10^9 N * m^2 / C^2

$$F = 6 \times (9 \times 10^9) \times \frac{(0.62 \times 10^{-29})^2}{(10^{-8})^4}$$

$$F = 54 \times 10^9 \times \frac{0.3844 \times 10^{-58}}{10^{-32}}$$

$$F = 20.7576 \times 10^{-17} \text{ N}$$

Rounding to the appropriate significant figures, the magnitude of the interaction force is approximately 2.1 * 10^-16 N.

Physical Interpretation: Beyond the Numbers

While calculating the final number is satisfying, the true beauty of physics lies in interpreting the result.

A force of 2.1 * 10^-16 N might seem infinitesimally small to our macroscopic senses, but on the molecular scale, this is a highly significant interaction. This calculation physically demonstrates a fundamental type of intermolecular force known as a dipole-dipole interaction (specifically, a Keesom force in chemistry).

Notice the inverse-fourth-power relationship in our derived force formula ($F \propto 1/l^4$). This is the most critical takeaway. Unlike Coulomb's law for point charges, which falls off as the inverse square of the distance ($1/r^2$), the force between two dipoles drops off much more rapidly. This mathematically proves why dipole-dipole interactions are strictly short-range forces. They play a dominant role only when molecules are very close together, such as in liquid water, contributing heavily to water's unusually high boiling point, surface tension, and overall cohesive properties.

Conclusion

Irodov Problem 3.47 is an excellent exercise in applying foundational electrostatic principles to atomic-scale phenomena. By deliberately linking the concepts of electric fields, potential energy, and spatial derivatives, we can accurately quantify the invisible forces that govern the behavior of water molecules. Whether you are prepping for a competitive exam or just brushing up on your electrodynamics, mastering derivations like this will profoundly deepen your intuition for how the physical universe operates at a microscopic level.

Keep practicing, and remember that every complex physics problem is simply a series of basic concepts waiting to be properly connected.