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(B) According to Gauss's Law,the total electric flux $\phi_{	ext{net}}$ through a closed surface is given by $\phi_{	ext{net}} = \frac{q}{\epsilon_0

Vedclass · Accepted Answer

$36 \pi 	imes 10^3 \ Nm^2 / C$

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(B) According to Gauss's Law,the total electric flux $\phi_{	ext{net}}$ through a closed surface is given by $\phi_{	ext{net}} = \frac{q}{\epsilon_0}$.Since the charge $q = 6 \mu C = 6 	imes 10^{-6} \ C$ is placed at the centre of a cube,the flux is distributed equally through all $6$ faces.Therefore,the flux through each face is $\phi_{	ext{face}} = \frac{\phi_{	ext{net}}}{6} = \frac{q}{6 \epsilon_0}$.Given $\frac{1}{4 \pi \epsilon_0} = 9 	imes 10^9 \ Nm^2 C^{-2}$,we have $\frac{1}{\epsilon_0} = 36 \pi 	imes 10^9 \ Nm^2 C^{-2}$.Substituting the values: $\phi_{	ext{face}} = \frac{6 	imes 10^{-6}}{6} 	imes (36 \pi 	imes 10^9) = 10^{-6} 	imes 36 \pi 	imes 10^9 = 36 \pi 	imes 10^3 \ Nm^2 / C$.

$A$ $6 \mu C$ charge is placed at the centre of a cube. What will be the electric flux through each face of the cube? (Take $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ Nm^2 C^{-2}$)

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