JEE Advanced 2023 PYQ | Electrostatics | Problem 1 Solution | By Shivendra Sir

Introduction

Hello aspirants! If your goal is cracking JEE Advanced, then solving and deeply understanding past year questions is non-negotiable. These problems reveal the mindset of IIT examiners and train you to think beyond formulas.

In this article, we will break down JEE Advanced 2023 Electrostatics Problem 1 (Paper 2)—a question that beautifully blends vector concepts with electrostatics. At first glance, it may seem tricky due to coordinate changes, but once approached logically, it becomes surprisingly simple.

As always, remember the core philosophy: understand the physics first, then apply the math.

Problem Overview

We are given an electric dipole formed by charges +q and –q placed at specific coordinates. The electric potential at a distant point P is given as $V_{0}$ V0. Then, the charges are shifted to new positions, and we are asked to find the new potential at the same point.

This is not just a formula-based question—it tests your understanding of:

Dipole moment direction
Vector representation
Approximation techniques

Concepts You Must Know

To solve this problem efficiently, you need clarity on the following:

1. Short Dipole Approximation

When the observation point is far away compared to the separation of charges, the dipole behaves like a “point dipole.” This simplifies calculations significantly.

2. Potential Due to a Dipole (Vector Form)

Instead of using angle-based formulas, we use the vector form:

$V = \frac{k (\vec{p} \cdot \vec{r})}{r^{3}}$ V=r3k(p⋅r)

This is faster and reduces calculation errors.

3. Dipole Moment Definition

The dipole moment vector always points:

From negative charge to positive charge
Magnitude = charge × separation

Step-by-Step Solution

Let’s divide the problem into two parts: initial configuration and final configuration.

Initial Configuration

Step 1: Identify Charge Positions

Negative charge at (0, –2) mm
Positive charge at (0, 2) mm

Clearly, the dipole lies along the y-axis.

Step 2: Dipole Moment Vector

The displacement from negative to positive charge is along +y direction: ${\vec{p}}_{1} = q \cdot (4 \hat{j})$ p1=q⋅(4j^)

Step 3: Position Vector of Point P

Point P is at (100, 100) mm: $\vec{r} = 100 \hat{i} + 100 \hat{j}$ r=100i^+100j^

Step 4: Calculate Initial Potential

Using the dot product: ${\vec{p}}_{1} \cdot \vec{r} = q (4 \hat{j}) \cdot (100 \hat{i} + 100 \hat{j})$ p1⋅r=q(4j^)⋅(100i^+100j^)

Only the $\hat{j} \cdot \hat{j}$ j^⋅j^ term contributes: $= 4 q \times 100 = 400 q$ =4q×100=400q

So, $V_{0} = \frac{k \cdot 400 q}{r^{3}}$ V0=r3k⋅400q

Final Configuration

Now the charges are shifted:

Negative charge → (1, –2) mm
Positive charge → (–1, 2) mm

Step 5: New Dipole Moment

Calculate displacement from –q to +q:

x-direction: –1 – 1 = –2
y-direction: 2 – (–2) = 4

So, ${\vec{p}}_{2} = q (- 2 \hat{i} + 4 \hat{j})$ p2=q(−2i^+4j^)

Step 6: Compute New Potential

${\vec{p}}_{2} \cdot \vec{r} = q (- 2 \hat{i} + 4 \hat{j}) \cdot (100 \hat{i} + 100 \hat{j})$ p2⋅r=q(−2i^+4j^)⋅(100i^+100j^) $= (- 2 \times 100) + (4 \times 100)$ =(−2×100)+(4×100) $= - 200 + 400 = 200$ =−200+400=200

Thus, $V_{new} = \frac{k \cdot 200 q}{r^{3}}$ Vnew=r3k⋅200q

Final Result

Comparing both:

Initial potential = $\frac{400 k q}{r^{3}}$ r3400kq
New potential = $\frac{200 k q}{r^{3}}$ r3200kq

Hence, $V_{new} = \frac{V_{0}}{2}$ Vnew=2V0

Important Insights

This problem teaches a powerful lesson:

The magnitude of dipole moment alone doesn’t determine potential
Direction relative to position vector matters equally

The reduction to half occurs due to the change in orientation of the dipole.

Common Mistakes Students Make

1. Ignoring Vector Method

Using trigonometry here wastes time and increases error chances.

2. Wrong Dipole Direction

Always remember: negative → positive.

3. Overcomplicating the Problem

Many students try full charge potential calculation instead of using dipole approximation.

Exam Strategy

Look for symmetry and approximation opportunities
Always convert geometry into vectors
Use dot product for faster calculation
Avoid unnecessary expansions

Why This Question Matters

This is a classic JEE Advanced level conceptual problem because it checks:

Vector clarity
Physical understanding
Application of approximation

If you can solve such problems confidently, you are on the right track for a top rank.

Conclusion

The JEE Advanced 2023 Electrostatics Problem 1 highlights the importance of thinking in vectors rather than blindly applying formulas. With the right approach, even complex-looking problems become manageable.

Practice more such PYQs and focus on understanding the why behind each step. That’s the real key to mastering Physics.

Keep learning, stay consistent, and trust the process.

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Monday, 13 April 2026