If you are preparing for advanced physics exams like JEE Advanced or олимпиad-level problems, then solving questions from Problems in General Physics by I.E. Irodov is almost a must. These problems are known for testing deep conceptual clarity, and Problem No. 3.35 from Electrostatics is a perfect example.

In this blog post, we will go through the I.E. IRODOV Problem No. 3.35 Solution Electrostatics step by step, understand the key concepts, and learn how symmetry plays a powerful role in simplifying calculations.

📘 Problem Statement (Irodov 3.35)

We are given a uniformly charged ring of radius $R$ R carrying a total charge $Q$ Q. The task is to find the electric potential at a point on the axis of the ring, located at a distance $x$ x from its center.

🧠 Understanding the Concept

This problem is based on electric potential due to continuous charge distribution and uses two major ideas:

Superposition Principle
Symmetry of Charge Distribution

Unlike electric field, electric potential is a scalar quantity, which makes calculations much easier. We don’t need to resolve vectors—just add contributions directly.

⚙️ Step-by-Step Solution

Step 1: Consider a Small Charge Element

Let us take a small charge element $dq$ dq on the ring.

The electric potential due to this small element at point $P$ P is: $dV = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq}{r}$ dV=4πε01⋅rdq

where $r$ r is the distance between $dq$ dq and point $P$ P.

Step 2: Determine the Distance $r$ r

From geometry, every point on the ring is at the same distance from point $P$ P because of symmetry.

Using the Pythagorean relation: $r = \sqrt{R^2 + x^2}$ r=R2+x2

This is the key simplification — all charge elements are equidistant from point PPP.

Step 3: Substitute into Potential Expression

Now, $dV = \frac{1}{4\pi\varepsilon_0} \cdot \frac{dq}{\sqrt{R^2 + x^2}}$ dV=4πε01⋅R2+x2dq

Step 4: Integrate Over the Ring

To find total potential: $V = \int dV = \frac{1}{4\pi\varepsilon_0 \sqrt{R^2 + x^2}} \int dq$ V=∫dV=4πε0R2+x21∫dq

Since total charge on the ring is $Q$ Q: $\int dq = Q$ ∫dq=Q

✅ Final Answer:

$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{\sqrt{R^2 + x^2}}$ V=4πε01⋅R2+x2Q

✨ Key Insights from This Problem

1. Symmetry Simplifies Everything

Because the ring is symmetric, every charge element contributes equally in terms of distance. This avoids complicated integration.

2. Scalar Nature of Potential

Unlike electric field, potential does not require vector addition. This makes such problems easier to handle.

3. Standard Result

This is a very important standard formula used in many advanced problems, including:

Electric field derivation on axis of ring
Disk and sphere potential problems
Capacitor analysis

🎯 Why This Problem is Important for JEE Advanced

The I.E. IRODOV Problem No. 3.35 Solution Electrostatics is not just about solving one question. It builds a strong foundation for:

Continuous charge distributions
Use of symmetry in physics
Integration-based derivations
Advanced electrostatics modeling

In exams like JEE Advanced, similar concepts are often twisted into new problems. If you understand this deeply, you can solve many variations easily.

🚀 Pro Tips for Students

Always check for symmetry before starting integration
Prefer potential approach over electric field when possible
Practice similar problems like:
- Potential due to a disk
- Field on axis of ring
- Charged rod problems

🧾 Conclusion

The I.E. IRODOV Problem 3.35 Solution beautifully demonstrates how powerful symmetry and basic electrostatics concepts can simplify what initially looks like a complex problem.

By understanding this solution, you not only master one problem but also gain a strategy to tackle many similar problems in electrostatics.

Keep practicing and try deriving the electric field on the axis of the ring as your next challenge—it’s a natural extension of this problem!