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 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=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:p1=q⋅(4j^)
Step 3: Position Vector of Point P
Point P is at (100, 100) mm:r=100i^+100j^
Step 4: Calculate Initial Potential
Using the dot product:p1⋅r=q(4j^)⋅(100i^+100j^)
Only the j^⋅j^ term contributes:=4q×100=400q
So,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,p2=q(−2i^+4j^)
Step 6: Compute New Potential
p2⋅r=q(−2i^+4j^)⋅(100i^+100j^)=(−2×100)+(4×100)=−200+400=200
Thus,Vnew=r3k⋅200q
Final Result
Comparing both:
- Initial potential = r3400kq
- New potential = r3200kq
Hence,Vnew=2V0
Important Insights
This problem teaches a powerful lesson:
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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