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If a cube of side 5 cm has a change of 6 microculombs, then the surface charge density is

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$4 \times 10^{2} C / m^{2}$

$4 \times 10^{2} μ C / m^{2}$

$4 \times 10^{3} C / m^{2}$

$4 \times 10^{3} μ C / m^{2}$

answer is A.

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Detailed Solution

Here, $q = 6 μ C, L = 5 c m = \frac{1}{20} m .$

Area of 6 facesof the cube $= 6 L^{2}$

Surface charge density $= \frac{q}{6 L^{2}} = \frac{6}{6 {(1 / 20)}^{2}}$

$= 4 \times 10^{2} μ C / m^{2}$

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