Question

A cone made of conducting material is given a charge $ Q $. $ \sigma_1, \sigma_2, \sigma_3 $ and $ \sigma_4 $ are charge densities at four points $ P, Q, R $ and $ S $. $ P $ is at the vertex of the cone and $ Q, R, S $ are at the periphery of the base. Choose the correct option.

• A. \( \sigma_1 > \sigma_2 > \sigma_3 > \sigma_4 \)
• B. \( \sigma_1 > \sigma_2 = \sigma_3 = \sigma_4 \)
• C. \( \sigma_1 < \sigma_2 = \sigma_3 < \sigma_4 \)
• D. \( \sigma_1 = \sigma_2 > \sigma_3 > \sigma_4 \)

Hint:

In conducting materials, charge density is inversely related to the surface area. The sharpest regions (like the vertex of a cone) have the highest charge density.

Answer 1

The Correct Option is B

In conducting materials, charge density is highest at points with the sharpest curvature (vertex of the cone) and decreases as the surface area increases (towards the base).
At the vertex \( P \), the charge density \( \sigma_1 \) is the highest.
The charge densities at points \( Q, R, S \), which are symmetrically located at the periphery of the base, are equal because the surface area at these points is larger compared to the vertex. Thus, the relationship between the charge densities is: \[ \sigma_1 > \sigma_2 = \sigma_3 = \sigma_4 \]
Final Answer: The correct option is (2), \( \sigma_1 > \sigma_2 = \sigma_3 = \sigma_4 \).