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(A) The surface charge density $\sigma$ is defined as the total charge $Q$ divided by the total surface area $A$ of the object.For a cube with side le

Vedclass · Accepted Answer

$4 	imes 10^2 \, \mu C/m^2$

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(A) The surface charge density $\sigma$ is defined as the total charge $Q$ divided by the total surface area $A$ of the object.For a cube with side length $a = 5 \, cm = 0.05 \, m$,the total surface area is $A = 6a^2$.$A = 6 	imes (0.05 \, m)^2 = 6 	imes 25 	imes 10^{-4} \, m^2 = 150 	imes 10^{-4} \, m^2 = 0.015 \, m^2$.The total charge is $Q = 6 \, \mu C = 6 	imes 10^{-6} \, C$.The surface charge density is $\sigma = \frac{Q}{A} = \frac{6 	imes 10^{-6}}{6 	imes 25 	imes 10^{-4}} \, C/m^2$.$\sigma = \frac{10^{-6}}{25 	imes 10^{-4}} \, C/m^2 = \frac{1}{25} 	imes 10^{-2} \, C/m^2 = 0.04 	imes 10^{-2} \, C/m^2 = 4 	imes 10^{-4} \, C/m^2$.Converting to $\mu C/m^2$: $\sigma = 4 	imes 10^{-4} 	imes 10^6 \, \mu C/m^2 = 4 	imes 10^2 \, \mu C/m^2$.

If a metal cube of side $5 \, cm$ has a charge of $6 \, \mu C$,then the surface charge density is

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