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Physics Question on Electrostatics

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

A

0.125×10−3 C m−20.125 \times 10^{-3} \, \text{C} \, \text{m}^{-2}

B

0.25×10−3 C m−20.25 \times 10^{-3} \, \text{C} \, \text{m}^{-2}

C

4×10−3 C m−24 \times 10^{-3} \, \text{C} \, \text{m}^{-2}

D

0.4×10−3 C m−20.4 \times 10^{-3} \, \text{C} \, \text{m}^{-2}

Answer

0.4×10−3 C m−20.4 \times 10^{-3} \, \text{C} \, \text{m}^{-2}

Explanation

Solution

Surface charge density (σ\sigma) = Charge (Q)Surface Area (A)\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

σ=6×10−6 C0.015 m2=0.4×10−3 C m−2\sigma = \frac{6 \times 10^{-6} \text{ C}}{0.015 \text{ m}^2} = 0.4 \times 10^{-3} \text{ C m}^{-2}