Question

A metal cube of side 5 cm is charged with 6 μC. The surface charge density on the cube is

• A. \(0.125 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
• B. \(0.25 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
• C. \(4 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
• D. \(0.4 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)

Answer 1

The Correct Option is D

Surface charge density ($\sigma$) is defined as the electric charge per unit area.

The side length of the object is 5 cm, which converts to 0.05 m.

The surface area (A) is calculated as 6 times the square of the side length: A = 6 × side². Substituting the side length, A = 6 × (0.05 m)² = 0.015 m². The total charge (Q) is 6 μC, equivalent to 6 × 10^-6 C.

Therefore, the surface charge density is calculated as: $\sigma = \frac{6 \times 10^{-6} \text{ C}}{0.015 \text{ m}^2} = 0.4 \times 10^{-3} \text{ C m}^{-2}$.