Surface Charge Density on Irregular Conductor

❓ Question

Electric charges are transferred to an irregular metallic disc as shown in figure.

If:

$$\sigma_1,\sigma_2,\sigma_3,\sigma_4$$

are charge densities at given points, choose the correct answer from the options below.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Surface charge density
👉 Curvature effect in conductors
👉 Charge distribution on irregular conductors.

Main idea:

$$\sigma \propto \frac{1}{R}$$

Smaller radius of curvature means larger surface charge density.

Sharper region stores more charge.

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🔷 Step 1 — Curvature Rule 💯

For conductors in electrostatic equilibrium:

$$E=\frac{\sigma}{\varepsilon_0}$$

and electric field becomes stronger near sharp points.

Therefore:

$$\text{Sharper point} \rightarrow \text{Higher charge density}$$

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🔷 Step 2 — Identify Sharpest Region

From figure:

Top point near:

$$\sigma_1$$

is the sharpest region.

Hence:

$$\sigma_1$$

is maximum.

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🔷 Step 3 — Identify Smoothest Region

The broader curved region near:

$$\sigma_3$$

has largest radius of curvature.

So charge density there is minimum.

Thus:

$$\sigma_3$$

is minimum.

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🔷 Step 4 — Compare Remaining Points

Between:

$$\sigma_2 \text{ and } \sigma_4$$

the region near:

$$\sigma_4$$

is slightly sharper.

Hence:

$$\sigma_4>\sigma_2$$

Overall order:

$$\sigma_1>\sigma_4>\sigma_2>\sigma_3$$

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🔷 Step 5 — Correct Option

Correct statements are:

$$A,\ C \text{ and } D$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Charge uniformly distributed assume kar lena

❌ Radius of curvature concept ignore kar dena

❌ Bigger area means bigger density assume kar lena

Remember:

$$\boxed{ \text{Sharper point} \rightarrow \text{More charge density} }$$

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✅ Final Answer

$$\boxed{\sigma_1>\sigma_4>\sigma_2>\sigma_3}$$

Hence correct option:

$$\boxed{\text{A, C and D only}}$$

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