Electric charge is transferred to an irregular metallic disk as shown in the figure. If the surface charge densities at four points are σ₁, σ₂, σ₃, and σ₄, then which of the following options is correct?

 

⚡ Surface Charge Distribution on Irregular Conductor – Electrostatics | JEE Physics | Doubtify JEE

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📌 Question:

Electric charge is transferred to an irregular metallic disk as shown in the figure.
If the surface charge densities at four points are σ₁, σ₂, σ₃, and σ₄, then which of the following options is correct?

🖼️ Question Image:

🧠 Solution Image:

🔍 Concept:

When charge is given to a conductor, it distributes itself in such a way that:

• The entire surface becomes an equipotential, and

• Surface charge density (σ) depends on the local curvature of the surface.

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⚙️ Important Law Used:

Higher curvature ⇒ Higher charge density

Sharper points or edges accumulate more charge, since field lines concentrate there.

Mathematically,

$$E \propto \frac{σ}{\varepsilon_0} \quad \text{and} \quad E \text{ is greater where curvature is high}$$

✅ Solution Explanation:

Let’s assume the disk has varying curvature at points where σ₁ to σ₄ are marked.

• The point with maximum curvature (sharpest tip or edge) will have the highest charge density.

• The point with least curvature (flat or bulged region) will have the lowest charge density.

Hence, rank the charge densities based on curvature:

$$σ_1 > σ_2 > σ_3 > σ_4 \quad \text{(Assuming curvature order)}$$

📌 Correct Answer: Depends on actual figure, but option will be based on the curvature vs. σ relationship.

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🧠 Pro Tip:

This is a very common concept in Electrostatics – especially useful in JEE Advanced level questions.

Always remember:

• Curved/sharp edges concentrate more charge

• Flat areas spread the charge

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🎥 Video Solution:

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