A metal cube of side 5 cm is charged with 6 μC. The surface charge density on the cube is
- \(0.125 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
- \(0.25 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
- \(4 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
- \(0.4 \times 10^{-3} \, \text{C} \, \text{m}^{-2}\)
The Correct Option is D
Solution and Explanation
Surface charge density ($\sigma$) = $\frac{\text{Charge (Q)}}{\text{Surface Area (A)}}$
Side = 5 cm = 0.05 m.
A = 6 × side2 = 6 × (0.05)2 = 0.015 m2 Q = 6 μC = 6 × 10-6 C
$\sigma = \frac{6 \times 10^{-6} \text{ C}}{0.015 \text{ m}^2} = 0.4 \times 10^{-3} \text{ C m}^{-2}$
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